Rohit

CLASS 11th CHAPTER -6 Thermodynamics |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 6 Thermodynamics. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

NCERT Solutions for Class 11 Chemistry Part 1

Chemistry Part 1 Class 11 will introduce you to basic concepts of subjects, about atoms, chemical bonding and other topics. There are 7 chapters through which learners can study in an organized manner and improve their studying habits.

Class 11th Chapter -6 Thermodynamics | NCERT CHEMISRTY SOLUTION |

NCERT Exercises

Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(a) used to determine heat changes
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only.
Answer.
(b): State function is a property of the system whose value depends only upon the state of the system and is independent of the path or the manner by which the state is reached.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is
(a) ΔT = 0
(b) Δp = 0
(c) g = 0
(d) w=0
Answer..
(c): Adiabatic system does not exchange heat with the surroundings.

Question 3.
The enthalpies of all elements in their standard states are
(a) unity
(b) zero
(c) <0
(d) different for each element.
Answer.
(b) : By convention, the standard enthalpy of formation of every element in its standard state is zero.

Question 4.
ΔU° of combustion of methane is – X kJ mol^-1. The value of ΔH° is
(a) = ΔU°
(b) >ΔU°
(c) <ΔU°
(d) =0
Answer.
(c) : CH_4(g)+2O₂ → CO_{2(g)+2H2Ol
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT}

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1,-393.5 kJ mol^-1 and-285.8 kJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
Solution.

Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(a) possible at high temperature
(b) possible only at low temperature
(c) not possible at any temperature
(d) possible at any temperature
Solution.
(d): A + B → C + D + q, ΔS = +ve
Here, ΔH = -ve
ΔG = ΔH – TΔS
For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 S of work is done by the system. What is the change in internal energy for the process?
Solution.
Heat absorbed by the system (q) = 701 J
Work done by the system (w) = -394 J
According to first law of thermodynamics,
ΔU = q + w = 701 + (-394) = 701 – 394 = 307 J

Question 8.
The reaction of cyanamide, NH₂CN_(g), with dioxygen was carried out in a bomb calorimeter, and AU was found to be -742.7 kJ mol 1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN_(g) + $\frac { 3 }{ 2 } { O }_{ 2(g) }$ → N₂+CO_2(g)+H₂O_l
Solution.

Question 9.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Solution.
Mass of Al = 60 g
Rise in temperature, ΔT = 55 – 35 = 20°C
Molar heat capacity of Al = 24 J mo^-1 K^-1
Specific heat capacity of Al = $\frac { 24 }{ 27 } J{ g }^{ -1 }{ K }^{ -1 }$
∴ Energy required = m x c x ΔT
= $60\times \frac { 24 }{ 27 } \times 20=\frac { 28800 }{ 27 } =1066.67\quad J$
= 1.068kJ or 1.07kJ

Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C.
C_p[H₂O_(l)] = 75.3 J mol^-1 K^-1,
C_p[H₂O_(s)] = 36.8 J mol^-1 K^-1
Solution.

Question 11.
Enthalpy of combustion of carbon to CO₂ is -393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Solution.
C + O₂ → CO₂; ΔT = -393.5 kJ
∵ When 44 g of COz is formed from carbon and dioxygen gas, heat released = 393.5 kj
∴ When 35.2 g of CO₂ is formed from carbon and dioxygen gas, heat released
= $\frac { 393.5\times 35.2 }{ 44 } =\frac { 138551.2 }{ 44 } =314.8kJ$
Thus, ΔH = -314.8 kJ

Question 12.
Enthalpies of formation of CO_(g), CO_2(g), N₂O_(g) and N₂O_4(g) are -110, – 393, 81 and 9.7 kJ mol ^-1 respectively. Find the value of Δ_rf for the reaction
N₂O_4(g) + 3CO_(g) → N₂O_(g) + 3CO_2(g)
Solution.

Question 13.
Given: N_2(g) + 3H_2(g) → 2NH_3(g); Δ_rH° = -92.4 kJ mol^-1 What is the standard enthalpy of formation of NH₃ gas?
Solution.
N_2(g) + 3H_2(g) → 2NH_3(g) ; Δ_rH° = -92.4 kJ mol^-1
∴ Standard enthalpy of formation of NH_3(g)
= $\frac { -92.4 }{ 2 } =\quad -46.2\quad kJ/mol$

Question 14.
Calculate the standard enthalpy of formation of CH₃OH_l from the following data :

Solution.

Question 15.
Calculate the enthalpy change for the process
CCl_4(g) → C_(g) + 4Cl_(g)
and calculate bond enthalpy of C – Cl in CCl_4(g).
Δ_vapH°(CCl₄) = 30.5 kJ mol^-1,
Δ_fH°(CCl₄) = -135.5 kJ mol^-1,
Δ_aH°(C) = 715.0 kJ mol^-1, where Δ_aH° is enthalpy of atomisation Δ_aH°(Cl₂) = 242 kJ mol^-1
Solution.

Question 16.
For an isolated system, ΔU = 0, what will be ΔS?
Solution.
When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.

Question 17.
For the reaction at 298 K, 2A + B → C,
ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.
Solution.
According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.

Question 18.
For the reaction, 2Cl_(g) → Cl_2(g), what are the signs of ΔH and ΔS?
Solution.
ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.

Question 19.
For the reaction
2A_(g) + B_(g) → 2D_(g), ΔU° = -10.5 kJ and ΔS° = – 44.1 JK^-1.
Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.
Solution.
ΔH° = ΔU° + Δn_(g)RT

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of ΔG° ?
R = 8.314J K^-1 mol^-1, T= 300 K.
Solution.
We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol

Question 21.
Comment on the thermodynamic stability of NO_(g), given

Solution.
Since Δ_(r)H° is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But Δ_(r)H° is negative for the formation of NO₂. So, NO₂ is stable.

Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H₂O_l is formed under standard conditions.
Δ_(r)H° = – 286 kJ mol^-1.
Solution.

Rohit0 comments

CLASS 11th CHAPTER -5 States of Matter |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 5 States of Matter. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -5 States of Matter | NCERT CHEMISRTY SOLUTION |

NCERT Exercises

Question 1.
What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?
Solution.
According to Boyle’s law, at constant temperature,

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas of 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C.What would be its pressure?
Solution.
Since temperature and amount of gas remain constant, therefore, Boyle’s law is applicable.

Question 3.
Using the equation of state PV = nRT; show that at a given temperature density of a gas is proportional to gas pressure P.
Solution.
According to ideal gas equation for ‘n’ moles of a gas,

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Solution.

Question 5.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Solution.

Question 6.
The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?
Solution.
The reaction between aluminium and caustic soda is

Question 7.
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C?
Solution.

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution.
Partial pressure of hydrogen gas

Question 9.
Density of a gas is found to be 5.46 g/dm³ at 27°C and at 2 bar pressure. What will be its density at STP?
Solution.
We know that PV = nRT

Question 10.
34.05 mL of phosphorus vapours weigh 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Solution.
Molar mass of phosphorus, M

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Solution.
Suppose the number of moles of gas present at 27°C in flask of volume V at pressure P is n₁ then assuming ideal gas behaviour,

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar.
(R – 0.083 bar dm³ K^-1 mol^-1)
Solution.

Question 13.
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Solution.
Number of moles of dinitrogen

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second?
Solution.
Time taken to distribute 10¹⁰ grains = 1 sec.
Time taken to distribute 6.023 x 10²³ grains
$=\frac { 1\times 6.022\times { 10 }^{ 23 } }{ { 10 }^{ 10 } } =6.022\times { 10 }^{ 23 }\quad sec$
$=\frac { 6.022\times { 10 }^{ 13 } }{ 60\times 60\times 24\times 365 } =1.90956\times { 10 }^{ 6 }\quad years$

Question 15.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C.
R = 0.083 bar dm³ K^-1 mol^-1
Solution.
Partial pressure of oxygen gas,

Question 16.
Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air =1.2 kg m^-3 and R = 0.083 bar dm³ K^-1 mol^-1)
Solution.

Question 17.
Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure.
(R = 0.083 bar L K^-1 mol^-1)
Solution.
According to ideal gas equation, PV= nRT

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Solution.
Let molar mass of gas = M

Question 19.
A mixture of dihydrogen and dioxygen at one bar pressurecontains 20% by weight of dihydgrogen. Calculate the partial pressure of dihydrogen.
Solution.
Let the total mass of the mixture be 100 g

Question 20.
What would be the SI unit for the quantity pV²T²/n?
Solution.

Question 21.
In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Solution.
According to Charles’ law,
Volume of gas at 1°C,
${ V }_{ t }={ V }_{ 0 }\left[ 1+\frac { t }{ 273 } \right]$ (∵ V₀ = Volume at 0°C)
Volume of gas at -273°C, $V={ V }_{ 0 }\left[ 1+\frac { 273 }{ 273 } \right] =0$
Thus, -273°C is the lowest possible temperature because below this temperature, the volume will become negative, and that is meaningless. This lowest temperature is called absolute zero temperature.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Solution.
Higher the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction, CO₂ has stronger intermolecular forces than methane.

Question 23.
Explain the physical significance of van der Waals parameters.
Solution.
van der Waals constant ‘a’ is a measure of the magnitude of the attractive forces among the molecules of a gas, while constant ‘b’ is a measure of effective size of the gas molecules.

Rohit0 comments

CLASS 11th CHAPTER -4 Chemical Bonding and Molecular Structure |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 4 Chemical Bonding and Molecular Structure. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -4 Chemical Bonding and Molecular Structure| NCERT CHEMISRTY SOLUTION |

NCERT Exercises

Question 4.1
Explain the formation of a chemical bond.
Answer.
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond. The formation of chemical compounds takes place as a result of combination of atoms in different ways. The atoms interact with each other on account of following reasons :
(i) Lowering of energy of combining atoms : The process of bonding between the atoms decreases the energy of the combining atoms and gives rise to the formation of a system which has lower energy and hence has greater stability.
(ii) Lewis octet rule : Octet theory of valency or electronic theory of valency states that in the formation of a chemical bond, atoms interact with each other by losing, gaining or sharing of electrons so as to acquire a stable outer shell of eight electrons. The tendency of atoms to achieve eight electrons in their outermost shell is known as Lewis octet rule.

Question 4.2
Write Lewis dot symbols for atoms of the following elements:
Mg, Na, B, O, N, Br.
Answer.
Lewis dot symbols :

Question 4.3
Write Lewis symbols for the following atoms and ions:
S and S^2-; Al and Al³⁺; H and H^–
Answer.
Lewis symbols:

Question 4.4
Draw the Lewis structures for the following molecules and ions:
H²S, SiCl⁴, BeF², ${ CO }_{ 3 }^{ 2- }$ , HCOOH.
Answer.
Lewis structures :

Question 4.5
Define octet rule. Write its significance and limitations.
Answer.
Octet rule : The tendency of atoms in molecules to have eight electrons in their valence shells (two for hydrogen atom) is known as the octet rule.
Significance :

It explains the formation of most of the compounds.
It is quite useful for understanding the structures of most of the organic compounds.

Limitations:

In some compounds, the number of electrons surrounding the central atom is less than eight e.g., LiCl, BeH₂ and BCl₃.
In molecules with an odd number of electron like nitric oxide (NO) and nitrogen dioxide (NO₂), the octet rule is not satisfied for all the atoms.
In a number of compounds there are more than eight valence electrons around the central atom. This is termed as the expanded octet. Some of the examples of such compounds are PF₅, SF₆, H₂SO₄ and a number of coordination compounds.

Question 4.6
Write the favourable factors for the formation of ionic bond.
Answer.
The following factors facilitate the formation of an ionic bond between a metal and a non-metal:

Ionization energy : Lesser the ionization energy, greater is the ease of formation of a cation.
Electron affinity : High electron affinity favours formation of an anion.
Lattice Energy : It is defined as the amount of energy released when cations and anions are brought close to each other from infinity to their respective equilibrium sites in the crystal lattice to form one mole of the ionic compound. The higher the magnitude of the lattice energy, the greater is the tendency of the formation of an ionic bond.

Question 4.7
Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.
Answer.
According to VSEPR theory, the shape of molecule depends upon the number of valence shell electron pairs (bonded or non bonded) around the central atom.
(i) BeCl₂: Presence of two bond pairs and lack of lone pairs gives it linear shape. Cl: Be : Cl
(ii) BCl₃ : Presence of three bond pairs and lack of lone pairs gives it trigonal planar geometry.

(iii) SiCl₄ : Presence of four bond pairs and lack of lone pairs gives it tetrahedral geometry.

(iv) AsF₅ : Presence of five bond pairs and lack of lone pairs gives it trigonal bipyramidal geometry.

(v) H₂S : Presence of two bond pairs and two lone pairs gives it angular or bent shape.

(vi) PH₃ : Presence of three bond pairs and one lone pair gives it trigonal pyramidal shape.

Question 4.8
Although geometries of NH3and H20 molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer.

Bond angle in NH3 is 107.5° and in H₂O, angle is 104.5°. In H₂O, oxygen has two lone pairs and in NH₃, nitrogen has only one lone pair. In H₂O, two lone pairs on oxygen repel each other and the bonded pairs more strongly and cause them to come closer thereby reducing the angle.

Question 4.9
How do you express the bond strength in terms of bond order?
Answer.
Greater the bond order, shorter is the bond length and so, higher is the bond strength.

Question 4.10
Define the bond length.
Answer.
The equilibrium distance between the centres of the nuclei of two atoms bonded together is termed as bond length or bond distance.
Bond length in ionic compounds = ${ r }_{ c }^{ + }+{ r }_{ a }^{ - }$
Bond length in covalent compounds (AB) = r_A+ r_B

Question 4.11
Explain the important aspects of resonance with reference to the ${ CO }_{ 3 }^{ 2- }$ ion.
Answer.
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons, known as the canonical forms of the hybrid are taken, which describe the molecule accurately. This phenomenon is known as resonance.
Resonance hybrid structures of carbonate ion are shown as follows :

The relative position of all the atoms in each of the canonical forms is same.
The number of unpaired and paired electrons in each of the canonical forms is same.
Negative charge is residing on more electronegative atom.
As a result of resonance, bond length and bond order is same.

Question 4.12
H₃PO₃ can be represented by structures (i) and(ii) shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.

Answer.
These two cannot be called as canonical forms as relative position of hydrogen is changed. In resonance, canonical forms should differ only in the position of electrons and not in the position of atoms.

Question 4.13
Write the resonance structures for SO₃, NO₂ and ${ NO }_{ 3 }^{ - }$ .
Answer.
SO₃

NO₂

Question 4.14
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
(a) K and S
(b) Ca and O
(c) Al and N.
Answer.

Question 4.15
Although both CO₂ and H₂O are triatomic molecules, the shape of H20 molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.
Answer.
Even though both CO₂ and H₂O are triatomic, but CO₂, has zero dipole moment. It implies that CO, has a symmetrical structure which is possible only when it is linear. However, H₂0 has a net dipole moment suggesting that the molecule is non-linear.

Question 4.16
Write the significance/applications of dipole moment.
Answer.
Applications of dipole moment are :

To decide polarity of the molecule : Molecules having zero dipole moment are said to be non-polar molecules and those having µ_R ≠ 0 are polar in nature.
To determine percentage of ionic character :
Percentage of ionic character
$=\frac { Experimental\quad value\quad of\quad dipole\quad moment }{ Theoretical\quad value\quad of\quad dipole\quad moment } \times 100$
To determine geometry of molecules: The values of dipole moments provide valuable information about the structures of molecules.
- CO₂, CS₂ are linear as values of their dipole moments are zero.
- H₂O is non-linear as it has a net dipole moment.

Question 4.17
Define electronegativity. How does it differ from electron gain enthalpy?
Answer.
The relative tendency of a bonded atom to attract the shared electron pair towards itself is called electronegativity while electron gain enthalpy is the energy change that occurs for the process of adding an electron to a gaseous isolated atom to convert it into a negative ion i.e., to form, a monovalent anion. Electron gain enthalpy and electronegativity both measure the power of attracting electrons, but electron gain enthalpy is concerned with an isolated gaseous atom while electronegativity is concerned with the atom in combination.

Question 4.18
Explain with the help of suitable example polar covalent bond.
Answer.
In a hetero-atomic molecule the shared pair of electrons between the two atoms gets displaced more towards the electronegative atom. The resultant covalent bond is polar covalent bond e.g., in H – F the electron pair is displaced towards fluorine due to highly electronegative F atom. Hence H develops a partial +ve charge while F develops a partial
-ve charge $\overset { \delta + }{ H } -\overset { \delta - }{ F }$ .

Question 4.19
Arrange the bonds in order of increasing ionic character in the molecules : LiF, K₂O, N₂, SO₂ and CIF3.
Answer.
As per Fajan’s rule, N₂ < SO₂ < CIF₂ < K₂O < LiF

Question 4.20
The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid

Answer.
In the given figure, hydrogen has valency of two, which is not possible and also octet of oxygen is not complete.
So, the correct structure is

Question 4.21.
Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH₄ is not square planar?
Answer.
In CH₄ only bond pairs are there and no lone pairs are present so, to attain most stabilized structure it arranges itself in tetrahedral geometry with greater bond angle of 109° rather than in square planar geometry with lesser bond angle of 90°.

Question 4.22
Explain why BeH₂molecule has a zero dipole moment although the Be—H bonds are polar?
Answer.
The dipole moment in case of BeH₂ is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.

Question4. 23
Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer.
Both the molecules NH₃ and NF₃ have pyramidal shape with a lone pair of electron on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH₃ (4.90 x 10^-30 C m) is greater than that of NF₃ (0.8 x 103 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N—H bonds whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of the three N—F bonds. The orbital dipole because of lone pair decreases the effect of the resultant N—F bond moment which results in low dipole moment of NF₃ as represented below :

Question 4.24
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
Answer.
Hybridisation can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.
Salient features of hybridisation : The main features of hybridisation are as under :

The number of hybrid orbitals is equal to the number of the atomic orbitals that undergo hybridisation.
The hybrid orbitals are always equivalent in energy and shape.
The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
These hybrid orbitals are directed in space in some preferred directions to have minimum repulsion between electron pairs and thus have a stable arrangement. Therefore, the type of hybridisation indicates the geometry of the molecules.

Important conditions for hybridisation :

Only those orbitals that are present in the valence shell of the atom undergo hybridisation.
The orbitals undergoing hybridisation should have almost equal energy.
Promotion of electron is not essential condition prior to hybridisation.
It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Shapes of hybrid orbitals :

(i) sp-hybridisation : One ‘s’ and one ‘p’ orbital of an atom intermix giving two sp-hybrid orbitals making an angle of 180° with each other. Each hybrid orbital has 50% s and 50% p-orbital character.
The molecules having such type of hybridisation in their central atom have linear geometry with bond angles 180°. Examples : BeCl₂, C₂H₂, CO₂, HgCl₂, CS₂, N₂O, BeH₂, etc.

(ii) sp²-hybridisation : One ‘s’ and two ‘p’ orbitals of an atom intermix giving three sp2-hybrid orbitals, which are directed towards the corners of an equilateral triangle making an angle of 120° with each other. Each hybrid orbital has 33.3% s and 66.7% p-orbital character. The molecules having such type of hybridisation in their central atom are expected to have triangular planar geometry with bond angles 120°.
Examples : BCl₃, BF₃, SO₃, C₂H₄, Graphite, SO₂, etc.

(iii) sp²-hybridisation : One ‘s’ and three ‘p’ orbitals of an atom intermix giving four sp²-hybrid orbitals, which are directed towards the corners of a regular tetrahedron making an angle of 109°28′. Each hybrid orbital has 25% s- and 75% p-orbital character. The molecules having such type of hybridisation in their central atom are expected to have tetrahedral geometry with bond angles 109°28′.
Examples : CH₄, C₂H₆, CCl₄, H₂O, NH₃, diamond, etc.

Question 4.25
Describe the change in hybridisation (if any) of the Al atom in the following reaction.
Alcl₃ + cl^– → ${ AlCl }_{ 4 }^{ - }$
Answer.
There is change in hybridization from sp² → sp³. In AlCl₃, Al is sp² hybridised. AlCl₃ thus has three sp² hybrid orbitals and a vacant unhybrid p-orbital. When ${ AlCl }_{ 4 }^{ - }$ is formed, Al undergoes sp³ hybridisation and thus Cl^– overlaps with the fourth orbital.

Question 4.26.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF₃ + NH₃ → F₃B.NH₃
Answer.
In BF₃, B is sp² hybridised and has one vacant p-orbital which gets filled by accepting a lone pair of electron present on the N-atom of NH₃. Nitrogen in this adduct acts as donor atom and BF₃ acts as an acceptor. So, hybridization of B in BF₃ changes from sp² to sp³ whereas there is no change in hybridization of N in NH₃ and in the adduct.

Question 4.27
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
Answer.
C₂H₄ molecule

C₂H₂ molecule

Question 4.28.
What is the total number of sigma and pi bonds in the following molecules?
(a) C₂H₂
(b) C₂H₄
Answer.

Question 4.29.
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s
(b) 1s and 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s
Answer.
A bond formed between two atoms by the overlap of singly occupied orbitals along their axes (end to end overlap) is called c bond. Hence, 1s and 1s, 1s and 2p_x and 1s and 2s will form σ bonds. π bonds are formed by the sidewise or lateral overlapping of p-orbitals. Hence, p_yand p_y orbitals will form π bond.

Question 4.30.
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃CH₃
(b) CH₃CH = CH₂
(c) CH₃CH₂OH
(d) CH₃CHO
(e) CH₃COOH
Answer.
(a) sp³, sp³
(b) sp³, sp², sp²
(c) sp³, sp³
(d) sp³, sp²
(e) sp³, sp²

Question 4.31
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer.
Bond pairs are the pairs of electrons taking part in bonding e.g.,

A pair of electrons in a molecule which is not shared by any of the two constituent atoms i.e., does not take part in the direct bonding is called a lone pair e.g.,

Question 4.32.
Distinguish between a sigma and a pi bond.
Answer.
Comparison between σ and π bonds

	Sigma bond	Pi bond
1.	The bond is formed by the overlap of orbitals along their axes (end to end overlap). It includes s-s and p-p overlapping.	The bond is formed by sidewise overlapping of orbitals (lateral overlapping). It includes p-p overlapping.
2.	It is a strong bond.	It is a weak bond.
3.	There can be free rotation of atoms around this bond.	Free rotation is not possible around this bond.
4.	The shape of the molecule is determined by these bonds.	These bonds do not affect the shape of the molecule.
5.	o-electrons are referred to as localized electrons.	71-electrons are referred to as mobile electrons.

Question 4.33.
Explain the formation of H2 molecule on the basis of Valence Bond Theory.
Answer.

We can consider two hydrogen atoms A and B approaching each other having nuclei N_A and N_A and electrons present in them are represented by e_A and e_A. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms begin to approach each other, new attractive and repulsive forces begin to operate.
Attractive forces arise between :

nucleus of one atom and its own electron that is N_A – e_A and N_B – e_B.
nucleus of one atom and electron of other atom i.e., N_A – e_B, N_B – e_A.

Similarly repulsive forces arise between :

electrons of two atoms like e_A – e_B.
nuclei of two atoms N_A – N_B.
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.
Experimentally it has been found that the magnitude of new attractive forces is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Ultimatelv a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm.
Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than isolated hydrogen atoms. The energy so released is caljed bond enthalpy, which corresponds to the minimum in the curve depicted in fig. Conversely, 435.8 kj of energy is required to dissociate one mole of H2 molecule.
H_2(g) + 435.8 kj mol^-1 → H_(g) + H_(g)

Question 4.34
Write the important conditions required for the linear combination of atomic orbitals to
form molecular orbitals.
Answer.
Necessary conditions for linear combination of atomic orbitals :

The symmetry and relative energies of combining atomic orbitals _A and _B must be close to each other.
The extent of overlap between the charge clouds of _A and _B should be large,
Atomic orbitals of the same sign should overlap.

Question 4.35
Use molecular orbital theory to explain why Be₂ molecule does not exist.
Solution.
Beryllium atom has electronic configuration 1s² 2s². So, in Be₂ there are a total of eight electrons which must be filled in four molecular orbitals.
M.O. configuration of Be₂ → σ1s²σ*1s² σ2s² σ*2s²
B.O. = 1/2(4 – 4) = 0. Since the bond order is zero it shows that molecular beryllium will not exist.

Question 4.36
Compare the relative stability of the following species and indicate their magnetic properties; O₂, ${ O }_{ 2 }^{ + }$ , ${ O }_{ 2 }^{ - }$ (superoxide), ${ O }_{ 2 }^{ 2- }$ (peroxide).
Answer.

Question 4.37
Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer.
The plus ancj minus sign indicate the sign of the wave function in the various lobes and not that of the nuclear or electronic charges. The ‘+’ sign (or upward crest) represents one phase and thesign represents second phase of the wave function. Bonding molecular orbital is formed by combination of ‘+’ with ‘+’ and with part of the electron waves whereas antibonding molecular orbital is formed by combination of ‘+’ with part of the electron waves.
As orbitals are represented by wave functions, a plus sign in an orbital represents a +ve wave function and a minus sign represents a -ve wave function.

Question 4.38
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
Answer.
Formation of PCl₅ (sp³d hybridisation):
The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented here.

Now the five orbitals (i.e., one s, three p and one d orbital) are available for hybridisation to yield a set of five sp³d hybrid orbitals which are directed towards the five corners of a trigonal bipyramid as depicted in the figure.

All the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl₅ the five sp³d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P-Cl sigma bonds. Three P-Cl bonds lie in one plane and make angle of 120° with each other; these bonds are termed as equatorial bonds. The remaining two P-Cl bonds, one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds.

Question 4.39
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer.
Hydrogen bond can be defined as an attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Following conditions should be fulfilled by a molecule for the formation of hydrogen bond.
(a) Hydrogen atom should be linked to a highly electronegative atom, such as F, O or N through a covalent bond.
(b) The size of the electronegative atom bonded covalently to the hydrogen atom should be small.
Types of H-bonding:
Intermolecular hydrogen bond : When hydrogen bond is formed between the hydrogen atom of one molecule and the electronegative atom of another molecule of the same compound (such as HF) or different compounds (such as water and alcohol), it is known as intermolecular hydrogen bond.
>
Intermolecular Hydrogen bond : When hydrogen bond is formed between the hydrogen atom and an electronegative atom, present in the same molecule, it is known as intramolecular hydrogen
bond.

H-bonding is stronger than van der Waals forces.

Question 4.40
What is meant by the term bond order?
Calculate the bond order of: N₂,O₂, ${ O }_{ 2 }^{ + }$ and ${ O }_{ 2 }^{ - }$
Answer.
Bond order of a molecule or an ion is a measure of the strength or stability of the bond. Numerically, it is half of the difference between number of electrons in bonding (N_b) and in anti-bonding (N_a) molecular orbitals.
Bond order = $\frac { 1 }{ 2 } ({ N }_{ b }-{ N }_{ a })$
If B.O. is zero or negative, the bond is highly unstable, i.e. the molecule does not exist. Higher bond order means high bond dissociation energy and greater stability. Greater the bond order, lesser is the bond length.
M.O. configuration of N₂ :

Rohit0 comments

CLASS 11th CHAPTER -3 Classification of Elements and Periodicity in Properties |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 3 Classification of Elements and Periodicity in Properties. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -3 Classification of Elements and Periodicity in Properties | NCERT CHEMISRTY SOLUTION |

NCERT Exercises

Question 3.1
What is the basic theme of organisation in the periodic table?

Answer.
By 1865, number of identified elements was 63 and with such a large number of elements it was very difficult to study individually the chemistry of all these elements and their innumerable compounds individually. To ease out this problem, scientists searched for a systematic way to organize their knowledge by classifying the elements. Here comes the basis of periodic table. Various elements have now been divided into different groups on the basis of similarities in chemical properties. This has made the study simpler as now the properties of elements are studied in groups rather than individually.

Question 3.2
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to it?

Answer.
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. So, he did not stick to his criteria, he ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

Question 3.3
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Answer.
According to Mendeleev’s periodic law, the physical and chemical properties of the elements are periodic functions of their atomic masses, but according to modern periodic law, the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 3.4
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer.
In the sixth period, the orbitals to be filled are 6s, 4f, 5d and 6p. The complete filling of these orbitals require 2 + 14 + 10 + 6 = 32 electrons, hence, the sixth period of the periodic table should have 32 elements.

Question 3.5
In terms of period and group where would you locate the element with Z= 114?
Answer.
The outermost electronic configuration of element (₁₁₄Z) is
[Rn] 5f¹⁴ 6d¹⁰7s²7p². It has n = 7, so period → 7 It belongs to p-block so,
group number = 10 + 4 = 14.

Question 3.6
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer.
₁₇Cl → It belongs to the third period. So, outermost shell is n = 3. Its configuration is [Ne] 3s²3p⁵.
Therefore, its atomic number = 17.

Question 3.7
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
Answer.
(i) Lawrencium → ₁₀₃Lr
(ii) Seaborgium → ₁₀₆Sg

Question 3.8
Why do elements in the same group have similar physical and chemical properties?
Answer.
In a group, the chemical properties of the elements remain nearly the same due to same valence shell configuration.

Question 3.9
What does atomic radius and ionic radius really mean to you?
Answer.
Atomic radius: It is the distance between the centre of the nucleus and outermost shell where electrons are present.
Ionic radius : It is the distance between the nucleus and outermost shell of an ion.

Question 3.10
How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer.
Variation of atomic radii in a period :
As we move from left to right across a period, there is regular decrease in atomic radii of representative elements. This can be explained on the basis of effective nuclear charge which increases gradually in a period, i.c, electron cloud is attracted more strongly towards nucleus as the effective nuclear charge becomes more and more along the period. The increased force of attraction brings contraction in size.

Variation of atomic radii in a group : Atomic radii in a group increase as the atomic number increases. The increase in size is due to extra energy shell which outweighs the effect of increased nuclear charge.

Question 3.11
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F ^–
(ii) Ar
(iii) Mg²⁺
(iv) Rb⁺
Answer.
Isoelectronic species are those which have same number of electrons.
(i) F ^– has 10 electrons. Therefore, the species N³⁺, O^2-, Ne, Na⁺, Mg²⁺, etc., are isoelectronic with F ^–.
(ii) Ar has 18 electrons. Therefore, the species P^3-, S^2-, Cl^–, K⁺, Ca²⁺, etc., are isoelectronic with Ar
(iii) Mg²⁺ has 10 electrons. Therefore, the species N^3-, O^2-, Ne, Na⁺, etc., are isoelectronic with Mg²⁺.
(iv) Rb⁺ has 36 electrons. Therefore, the species Br^–, Kr, Sr²+, etc., are isoelectronic with Rb⁺

Question 3.12
Consider the following species :
N^3-, O^2-, F^–, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Answer.
(a) All these are isoelectronic species as they are having same number of electrons i.c., 10.
(b) As Z/e decreases, size increases so, order should be,
N^3- > O^2- > F ^– > Na⁺ > Mg²⁺ > Al³⁺.

Question 3.13
Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer.
A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 3.14
What is the significance of the terms – ‘isolated gaseous atom’and’ground state’while defining the ionization enthalpy and electron gain enthalpy? [Hint: Requirements for comparison purposes.]
Answer.
In the definition of ionization enthalpy and electron gain enthalpy, isolated gaseous atom is required for comparison purposes. Ionization energy is the minimum amount of energy required to remove most loosely bound electron from an isolated atom in the gaseous state of an element so as to convert it into gaseous monovalent positive ion. Electron gain enthalpy is the energy change accompanying the process of adding an electron to a gaseous isolated atom to convert it into a negative ion, i.e., a monovalent anion.

Both the above mentioned processes are carried out on an isolated gaseous atom, which in turn is obtained from either the excitation of a ground state atom (in case the element is monoatomic) or atomisation of polyatomic elements.

The force with which an electron is attracted by the nucleus is appreciably affected by presence of other atoms in the neighbourhood. Since in the gaseous state the atoms are widely separated, therefore these interatomic forces are minimum.
The term ground state means that the atom must be present in the most stable state.

Question 3.15
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1. [Hint: Apply the idea of mole concept to derive the answer]
Answer.
Energy of the electron in the ground state of H-atom, E₁ = -2.18 x 10^-18 J
Ionisation energy = E_∞ – E_n
Ionisation enthalpy per mole of atomic hydrogen = (E_∞ – E₁)NA
= [0 – (- 2.18 x 10^-18)] x 6.023 x 10²³
= 2.18 x 6.023 x 10⁵ J/mol = 13.13 x 10⁵ j/mol
= 1.313 x 10⁶ J/mol

Question 3.16
Among the second period elements the actual ionization enthalpies are in the order Li <B <Be< C< 0 < N< F < Ne. Explain why?
(i) Be has higher ∆_iH than B
(ii) O has lower ∆_iH than N and F?
Answer.
(i) An s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2 p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.
(ii) O has lower ionisation energy than N because N (1s² 2s² ${ 2p }_{ x }^{ 1 }\quad { 2p }_{ y }^{ 1 }\quad { 2p }_{ z }^{ 1 }$ ) has extra stable electronic configuration whereas O (1s² 2s² ${ 2p }_{ x }^{ 1 }\quad { 2p }_{ y }^{ 1 }\quad { 2p }_{ z }^{ 1 }$ ) does not. O has lower ionisation energy than F because O has larger size than F.

Question 3.17
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer.
The electronic configurations of Na and Mg are :
Na : 1s² 2s² 2p⁶ 3s¹
Na⁺ : 1s² 2s² 2p⁶
Mg : 1s² 2s² 2p⁶ 3s²
Mg⁺ : 1s² 2s² 2p⁶ 3s¹
Mg²⁺ : 1s² 2s² 2p⁶

The 1st ionization enthalpy of Na is lesser than that of Mg because Mg has an extra stable configuration and smaller size, so, a larger amount of energy would be required to remove an electron from the 3s orbital, which has a pair of electrons.

The 2^nd ionization enthalpy of Na is more than that of Mg because Na⁺ has an extra stable configuration (complete octet), whereas Mg⁺ does not have an extra stable configuration.

Question 3.18
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer.

We have to consider two factors :
(i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of shielding or screening of the valence electron from the nucleus by the intervening core electrons. As we descend the group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. The increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.

Question 3.19
The first ionization enthalpy values (in kJ mol^-1) of group 13 elements are :
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend?
Answer.
(i) Al has lower ionization enthalpy than B due to larger size.
(ii) Ga has slightly higher ionization enthalpy than Al due to ineffective shielding by 3d electrons.
(iii) In has lower ionization enthalpy than Ga due to larger size.
(iv) Tl has higher ionization enthalpy than In due to ineffective shielding by 4f electrons.

Question 3.20
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl
Answer.
(i) O or F : F has more negative electron gain enthalpy than O due to smaller size, higher nuclear charge and greater possibility of attaining the nearest stable noble gas configuration by gaining one electron.
(ii) F or Cl : Cl has more negative electron gain enthalpy because in F the incoming electron is added to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level. In Cl, the added electron goes to n = 3 quantum level and occupies a larger region of space and electron-electron repulsion experienced is far less.

Question 2.21
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Answer.
The second electron gain enthalpy of O is +ve. This is because energy has to be supplied to convert O^–_(g) to O^2-_(g) in order to overcome the repulsive forces.

Question 3.22
What is the basic difference between the terms electron gain enthalpy and electronegativity?
Answer.

	Electron gain enthalpy	Electronegativity
1.	It provides a measure of the ease with which an atom adds an electron to form an anion.	It is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself.
2.	It has an absolute value.	It is not a measurable quantity.
3.	Its periodicity is not regular in a period or in a group.	The periodicity is regular in a period but not so regular in a group.
4.	Its units are electron volts per atom or kilojoules per mole or kilocalories per mole.	It has no units but is merely a number.

Question 3.23
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Answer.
The statement is not correct because electronegativity of an element varies with the state of hybridisation and oxidation state of the element.

Question 3.24
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron.
Answer.
The distance between the nucleus and the outermost shell of an ion is known as ionic radius.
(a) The gain of an electron leads to the formation of an anion. The radius of the anion is larger than the atomic radius of its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge which acts on more electrons so, each electron is held less tightly and thereby the electron cloud expands.

(b) The removal of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same and since it is now acting on lesser number of electrons and pulls them closer, the ion is smaller.

Question 3.25
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer.
Two isotopes of the same element have the same first ionization enthalpies because of same effective nuclear charges.
e.g., $_{ 17 }^{ 35 }{ Cl }$ and $_{ 17 }^{ 37 }{ Cl }$ have same ionization enthalpies.

Question 3.26
What are the major differences between metals and non-metals?
Answer.

	Metals	Non-metals
1.	They are usually solids at room temperature. Mercury is an exception.	They may be solids or gases at room temperature.
2.	They have high melting and boiling points.	They have low melting and boiling points.
3.	They are good conductors of heat and electricity.	They are not good conductors of heat and electricity.
4.	They are malleable and ductile.	They are neither malleable nor ductile.
5.	They have 1 to 3 electrons in their valence shells.	They have 4 to 7 electrons in their valence shells.
6.	They have low ionisation energies.	They have high ionisation energies.
7.	They have low electron affinities.	They have high electron affinities.
8.	They are electropositive.	They are electronegative.
9.	They are brittle and non-lustrous.	They are hard and lustrous.
10.	They are sonorous.	They are non sonorous.

Question 3.27

Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Answer.
(a) Fluorine
(b) Magnesium
(c) Oxygen
(d) Group 17 (Halogens) :

F, Cl – Non metals and gases
Br – Non metal and liquid
I – Shows metallic lustre.

Question 3.28
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer.
The trend Li < Na < K < Rb < Cs is observed for chemical reactivity because upon descending the group the ionization energy of alkali metals decreases i.e., it is easy for them to lose an electron from their valence shell and attain the nearest stable noble gas configuration. The trend F > Cl > Br > I is observed for chemical reactivity in halogens because the standard reduction potential decreases as we descend the group. F being the most electronegative readily accepts an e~ to complete its octet and hence the trend.

Question 3.29
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer.
General outer electronic configuration :

Question 3.30
Assign the position of the element having outer electronic configuration
(i) ns² np⁴ for n = 3
(ii) (n – 1)d²ns² for n = 4, and
(iii) (n-2)f ⁷(n-1)d¹ns² for n = 6, in the periodic table.
Answer.

Question 3.31
The first (∆_iH₁) and the second (∆_iH₁) ionization enthalpies (in kJ mol^-1) and the (∆H₁) electron gain enthalpy (in kJ mol-1) of a few elements are given below:

Which of the above elements is likely to be:
(a) the least reactive element
(b) the most reactive metal
(c) the most reactive non-metal
(d) the least reactive non-metal
(e) the metal which can form a stable binary halide of the formula MX₂(X = halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X= halogen)?
Answer.
(a) V : The element V has highest first ionization enthalpy and positive electron gain enthalpy hence, it is least reactive.
(b) II : The element II has the least first ionization enthalpy hence, it is most reactive metal.
(c) III; The element III has very high negative electron gain enthalpy hence, it is most reactive non-metal.
(d) IV : Element IV has high negative electron gain enthalpy but ionization energy is not that high hence, it is least reactive non-metal.
(e) VI : The first and second ionization energies of element VI indicate that it can form a stable binary halide.
(f) I : The element 1 has very low value of first ionization energy but very high second ionization energy. Hence, it will form a stable covalent halide of the formula MX.

Question 3.32
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer.
(a) Li₂O
(b) Mg₃N₂
(c) AlI₃
(d) SiO₂
(e) PF₅
(f) LuF₃

Question 3.33
In the modern periodic table, the period indicates the value of
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Answer.
(c) Principal quantum number

Question 3.34
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (/) for the last subshell that received electrons in building up the electronic configuration.
Answer.
(b): The d-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.

Question 3.35
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Answer.
(c): Nuclear mass

Question 3.36
The size of isoelectronic species – F^–, Ne and Na⁺ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Answer.
(a): Nuclear charge (Z).

Question 3.37
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Question 3.38
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer.
(d) K > Mg > Al > B

Question 3.39
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F > N > C > Si > B
Answer.
(c) F>N>C>B>Si

Question 3.40
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) CI>F>0>N
(d) O > F > N > Cl
Answer.
(b) F > O > Cl > N

Rohit0 comments

CLASS 11th CHAPTER – 2 Structure of Atom |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 2 Structure of Atom. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -2 Structure of Atom NCERT CHEMISRTY SOLUTION |

Page No: 65

Exercises
2.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Answer
(i) Mass of one electron = 9.11×10^-31kg
∴ 1g = 10^-3 kg = (1/9.11×10^-31)×10^-3electrons = 1.09⁸×10²⁷

(ii) Mass of one electron = 9.11×10^-31kg

∴ Mass of 1 mole of electrons = (9.11×10^-31)×(6.022×10²³)= 5.48×10^-7kg
Charge on one electron = 1.602×10^-19coulomb
∴ Charge on 1 mole of electrons = (1.602×10^-19)×(6.022×10²³) = 9.65×10⁴coulombs.

2.2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675×10^-27kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed ?

Answer
(i) Electrons present in 1 molecule of methane (CH₄) = 6+4 = 10

∴ Electrons in mol i.e. 6.022×10²³ molecules = 6.022×10²⁴

(ii) (a) Number of atoms in ¹⁴C in 1 mole= 6.022×10²³ atoms
1 atom of ¹⁴C contains = 14-6 = 8 neutrons.

∴ The number of neutrons in 14 g of ¹⁴C = 6.022×10²³×8 neutrons
Number of neutrons in 7 mg = (6.022×10²³×8×7)/14000 = 2.4088×10²¹neutrons

(b) Mass of one neutron = 1.674×10^-27kg
Mass of total neutrons in 7 g of ¹⁴C = (2.4088×10²¹) (1.675 ×10^-27 kg) = 4.035 ×10^-6 kg
(iii) (a) 1 mol of NH₃= 17gNH₃=6.022×10²³ molecules of NH₃

1 atom of NH₃contains = 7+3 = 10 protons

∴ The number of protons in 1 mol of NH₃=6.022×10²⁴protons.

Number of protons in 34 mg of NH₃= (6.022×10²⁴×34)/17×1000 = 1.2044×10²²protons.

(b) Mass of one proton = 1.6726×10^-27kg

∴ Mass of 1.2044×10²²protons = (1.6726×10^-27)×(1.2044×10²²⁾kg = 2.0145×10^-5kg.

No, there will be no effect of temperature and pressure.

2.3. How many neutrons and protons are there in the following nuclei ?
¹³C₆, ¹⁶O₈, ²⁴Mg_12,⁵⁶Fe_26,⁸⁸CSr₃₈
Answer

Nucleus Z A Protons(Z) Neutrons(A-Z)
¹³C₆ 6 13 6 13-6=7
¹⁶O₈ 8 16 8 16-8=8

²⁴Mg₁₂ 12 24 12 24-12=12
⁵⁶Fe₂₆ 26 56 26 56-26=30
⁸⁸CSr₃₈38 88 38 88-38=50

2.4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35.
(ii) Z = 92 , A = 233.
(iii) Z = 4 , A = 9.

Answer

(i) ³⁵Cl₁₇

(ii) ²³³U₉₂

(iii) ⁹Be₄

2.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (ṽ) of the yellow light.

Answer

λ = 580 nm = 580×10^-9m

frequency (ν) = c/λ = 3.0×10⁸ms^-1/580×10^-9m = 5.17×10¹⁴s^-1

wave number (ṽ) = 1/λ = 1/580×10^-9m = 1.72×10⁶m^-1

2.6. Find energy of each of the photons which
(i) correspond to light of frequency 3×10¹⁵ Hz.
(ii) have wavelength of 0.50 Å.

Answer

(i) ν = 3×10¹⁵Hz

E = hν = (6.626×10^-34Js)×(3×10¹⁵s^-1) = 1.988×10^-18J

(ii) λ = 0.50×10^-10m

E = hν = hc/λ = (6.626×10^-34Js)×(3×10¹⁵s^-1)/0.50×10^-10m = 3.98×10^-15J

2.7. Calculate the wavelength, frequency and wave number of a light wave whose period is

2.0×10^-10s.

Answer

Frequency (ν) = 1/Period = 1/2.0×10^-10s = 5×10⁹s^-1.

Wavelength (λ) = c/ν = 3.0×10⁸ms^-1/5×10⁹s^-1= 6.0×10² m

Wave number (ṽ) = 1/λ = 1/6.0×10² m = 16.66 m^-1

2.8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?

Answer

λ = 4000 pm = 4000×10^-12m = 4×10^-9m

E = Nhν = Nhc/λ
∴ N = E×λ/h×c = (1J×4×10^-9m)/(6.626×10^-34Js×3.0×10⁸ms^-1) = 2.012×10¹⁶photons.

2.9. A photon of wavelength 4×10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and
(iii) the velocity of the photoelectron (1 eV = 1.6020×10^-19 J).

Answer
(i) Energy of the photon (E) = hν = hc/λ = (6.626×10^-34Js×3.0×10⁸ms^-1)/4×10^-7m = 4.97×10^-19J
= 4.97×10^-19/1.602×10^-19eV
(ii) Kinetic energy of emission (1/2 mv²) = hν- hνo = 3.10-2.13 = 0.97 eV
(iii) 1/2 mv²= 0.97 eV = 0.97×1.602×10^-19J
⇒ 1/2×(9.11×10^{-31 kg})×v²=0.97×1.602×10^-19J
⇒ v²=0.341×10¹²= 34.1×10¹⁰
⇒ v = 5.84×10⁵ms^-1

2.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Answer

E = Nhν = Nhc/λ = (6.022×10²³mol^-1)×(6.626×10^-34Js×3.0×10⁸ms^-1)/242×10^-9m

= 4.945×10⁵Jmol^-1= 494.5 kJmol^-1

2.11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Answer

Energy emitted by the bulb = 25 watt = 25 Js^-1

Energy of one photon (E) = hν = hc/λ

λ = 0.57µm = 0.57×10^-6m

E = (6.626×10^-34Js×3.0×10⁸ms^-1)/0.57×10^-6m = 3.48×10^-19J

∴ No. of photons emitted per sec = 25 Js^-1/3.48×10^-19J = 7.18×10¹⁹

2.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (νo) and work function (wo) of the metal.

Answer

c = νλ

∴ νo = c/λo = 3.0×10⁸ms^-1/6800×10^-10 m = 4.14×10¹⁴s^-1

Work function (wo) = hνo = 6.626×10^-34Js×4.14×10¹⁴s^-1 = 2.92×10^-19 J

2.13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer

ṽ = R(1/n1²–1/n2²) = 109677(1/2²–1/4²) cm^-1 = 20564.4cm^-1
λ = 1/ν = 1/20564.4 = 486×10^-7cm =486×10^-9m = 486 nm

Page No: 66

2.14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n=1 orbit).

Answer
E_n = -21.8×10^-19/n² Jatom^-1
For ionization from 5th orbit, n₁=5_,n₂= ∞

∴ ΔE = E₂ –E_{1 =} -21.8×10^-19×(1/n2²–1/n1²) = 21.8×10^-19×(1/n1²–1/n2²)

= 21.8×10^-19×(1/5²–1/∞) = 8.72×10^-20J

For ionization from 1st orbit, n₁=1_,n₂= ∞

∴ ΔE’ = 21.8×10^-19×(1/1²–1/∞) = 21.8×10^-19J

ΔE’/ΔE = 21.8×10^-19/8.72×10^-20= 25

Hence, 25 times less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

2.15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Answer

The number of spectral lines produced when an electron in the nthlevel drops down to the ground state is given by n(n-1)/2.

Given, n=6

∴ Number of spectral lines = 6×5/2 = 15

also given by, ∑(n2 – n1) = ∑(6-1) = ∑5 = 5+4+3+2+1 = 15

2.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18×10^-18 J atom^-1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Answer

(i) E_n = -21.8×10^-19/n² J

∴ E₅= -21.8×10^-18/5² J = 8.72×10^-20J

(ii) For H atom, r_n = 0.529×n²Å

∴ r₅= 0.529×5²= 13.225Å = 1.3225 nm

2.17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer

For the Balmer series, n1 = 2. Hence, ṽ = R(1/2²–1/n2²)

ṽ = 1/λ (inversely proportional)

For λ to be maximum, ṽ should be minimum. This can be happened when n2 is minimum i.e. n2 = 3. Hence, ṽ = (1.097×10⁷m^-1) (1/2²–1/3²) = 1.097×10⁷×5/36 m^-1= 1.523×10⁶m^-1

2.18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18×10^-11ergs.

Answer

1erg = 10^-7J

As ground state electronic energy is –2.18×10^-11ergs, this means that E_{n =}-21.8×10^-11/n² ergs.

ΔE = E₅ –E₁= 2.18×10^-11(1/1²–1/5²) = 2.18×10^-11(24/25) = 2.09×10^-11ergs = 2.09×10^-18 J

When electron returns to ground state (n=1), energy emitted = 2.09×10^-11 ergs.

As, E = hν = hc/λ

⇒ λ = hc/E = (6.626×10^-27erg sec) (3.0×10¹⁰cm s^-1)/2.09×10^-11 ergs

= 9.51×10^-6cm = 951×10^-8cm = 951 Å

2.19. The electron energy in hydrogen atom is given by E n = (–2.18×10^-18)/n² J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Answer

ΔE = E_∞–E₂ = 0 – (–2.18×10^-18J atom^-1/2²) = 5.45×10^-19J atom^-1
ΔE = hν = hc/λ
⇒ λ = hc/ΔE = (6.626×10^-34 Js)×(3.0×10⁸ms^-1)/5.45×10^-19J = 3.674×10^-7m = 3.674×10^-5cm

2.20. Calculate the wavelength of an electron moving with a velocity of 2.05×10⁷ ms^-1.

Answer

By de Broglie equation,

λ = h/mv = 6.626×10^-34 Js/(9.11×10^-31 kg) (2.05×10⁷ ms^-1) = 3.55×10^-11m

2.21. The mass of an electron is 9.11×10^-31 kg. If its K.E. is 3.0×10^-25 J, calculate its wavelength.

Answer

K.E. = 1/2 mv²

NCERT Solutions for Class 11th: Ch 2 Structure of Atom

∴ v = √2 K.E./m

= 812 ms^-1

By de Broglie equation, λ = h/mv = 6.626×10^-34 Js/(9.11×10^-31 kg) (812 ms^-1) = 8.967×10^-7 m

2.22. Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na⁺ , K⁺ , Mg²⁺ , Ca²⁺ , S^2- , Ar.

Answer

Notes:

Isoelectronic are the species having same number of electrons.
A positive charge means the shortage of an electron.

A negative charge means gain of electron.

Number of electrons in Na⁺ = 11-1 = 10
Number of electrons in K⁺ = 19-1 = 18
Number of electrons in Mg²⁺ = 12-2 = 10
Number of electrons in Ca²⁺ = 20-2 = 18
Number of electrons in S^2- = 16+2 = 18
Number of electrons in Ar = 18
Hence, the following are isoelectronic species:
1) Na⁺ andMg²⁺ (10 electrons each)
2) K⁺, Ca²⁺, S^2- and Ar (18 electrons each)

2.23. (i) Write the electronic configurations of the following ions: (a) H^– (b) Na⁺ (c) O^2-(d) F^–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s¹ (b) 2p³and (c) 3p⁵ ?

(iii) Which atoms are indicated by the following configurations ? (a) [He]2s¹(b) [Ne]3s²3p³ (c) [Ar] 4s² 3d¹ .

Answer

(i) (a) ₁H = 1s^{1 .}A negative charge means gain of electron.
∴ electronic configuration of H^–=1s²
(b) ₁₁Na = 1s²2s²2p⁶3s^{1 .}A positive charge means the shortage of an electron.

∴ electronic configuration of Na⁺= 1s²2s²2p⁶

(c)₈O = 1s²2s²2p⁴

∴ electronic configuration of O^2-=1s²2s²2p⁶

(d) ₉F = 1s²2s²2p⁵

∴ electronic configuration of F^– =1s²2s²2p⁶

(ii) (a) 3s¹
Completing the electron configuration of the element as 1s²2s²2p⁶3s¹
∴ Number of electrons present in the atom of the element = 2+2+6+1 = 11
∴ Atomic number of the element = 11
(b) 2p³
Completing the electron configuration of the element as 1s²2s²2p³
∴ Number of electrons present in the atom of the element = 2+2+3 = 7
∴ Atomic number of the element = 7
(c) 3p⁵
Completing the electron configuration of the element as 1s²2s²2p⁶3s²3p⁵
∴ Number of electrons present in the atom of the element = 2+2+6+2+5 = 17
∴ Atomic number of the element = 9

(iii) (a) [He]2s¹
electronic configuration = 1s²2s¹
∴ Atomic number of the element = 2+1 = 3
Hence, the element with the electronic configuration [He]2s¹ is lithium (Li).
(b) [Ne]3s²3p³
electronic configuration = 1s²2s²2p⁶3s²3p³
∴ Atomic number of the element = 2+2+6+2+3 = 15
Hence, the element with the electronic configuration [Ne]3s²3p³ is phosphorus (P).
(c) [Ar] 4s²3d¹
electronic configuration = 1s²2s²2p⁶3s²3p⁶4s²3d¹

∴ Atomic number of the element = 2+2+6+2+6+2+1 = 21
Hence, the element with the electronic configuration [Ar] 4s²3d¹ is scandium (Sc).

2.24. What is the lowest value of n that allows g orbitals to exist?

Answer
For g-orbitals, l = 4.

For any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴ For l = 4, minimum value of n = 5

2.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and m_lfor this electron.

Answer

For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (m_l) = -2,-1,0,1,2

2.26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Answer

(i) For neutral atom, number of protons = number of electrons.
∴ Number of protons in the atom of the given element = 29 = Atomic number
(ii) The electronic configuration of the atom with Z=29 is 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

2.27. Give the number of electrons in the species H₂⁺,H₂ and O₂⁺

Answer
H₂⁺= 2-1 = 1 electron
H_{2 =}₁H + ₁H = 2 electrons
O₂⁺= 16-1 = 15 electrons

2.28. (i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers (m_l and l ) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f

Answer
(i) For a given value of n, l can have values from 0 to (n-1).
∴ For n = 3 , l = 0, 1, 2
For a given value of l, m_l can have (2l+1) values.
When l = 0, m = 0
l = 1, m = – 1, 0, 1
l = 2, m = – 2, – 1, 0, 1, 2
l = 3, m = -3, -2, -1, 0, 1, 2, 3
(ii) For 3d orbital, n = 3, l = 2.
∴ For l = 2
m₂ = -2, -1, 0, 1, 2
(iii) 1p is not possible because when n = 1, l = 0. (for p, l = 1)
2s is possible because when n=2, l = 0,1 (for s, l=0)
2p is possible because when n=2, l = 0,1 (for p, l=1)
3f is not possible because when n=3, l = 0, 1 , 2 (for f, l=3)

2.29. Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1, l=0; (b) n=3; l=1 (c) n=4; l=2; (d) n=4; l=3.

Answer
(a) 1s
(b) 3p
(c) 4d
(d) 4f

2.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, m_l = 0, m_s = +½
(b) n = 1, l = 0, m_l = 0, m_s = –½
(c) n = 1, l = 1, m_l = 0, m_s = +½
(d) n = 2, l = 1, m_l = 0, m_s = –½
(e) n = 3, l = 3, m_l = –3, m_s = +½
(f) n = 3, l = 1, m_l = 0, m_s = +½

Answer
(a) Not possible because n≠0
(b) Possible
(c) not possible because when n=1, l≠1
(d) Possible
(e) Not possible because when n=3, l≠3
(f) Possible

Page No: 67

2.31. How many electrons in an atom may have the following quantum numbers?
(a) n = 4, m_s = –½ (b) n = 3, l = 0

Answer
(i) The total number of electrons in n is given by 2n²
n=4, Number of electrons = 2×4²= 32

Half of 32 electrons will have spin quantum number m_s = –½ i.e. 16 electrons

(ii) n=3 and l=0 means it is 3s orbital which can have only 2 electrons.

2.32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Answer

According to Bohr postulate of angular momentum,

mvr = nh/2π ⇒ 2πr = nh/mv …(i)
According to de Broglie equation, λ=h/mv …(ii)
Substituting the value of eqn (ii) in eqn (i) we get,
2πr = nλ
Thus, circumference of the Bohr orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

2.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer

For H-like particles, ṽ = (2π²mZ²e⁴/ch³)×(1/n1²–1/n2²) = RZ²(1/n1²–1/n2²)
∴ For He⁺ spectrum, Balmer transition, n=4 to n=2
ṽ = 1/λ = RZ²(1/2²–1/4²) = R×4×3/16 = 3R/4
For hydrogen spectrum ,
ṽ = 1/λ = R(1/n1²–1/n2²) = 3R/4 ⇒ (1/n1²–1/n2²) = 3/4
which can be true only for n1=1 and n2=2 i.e. transition from n=2 to n=1.

2.34. Calculate the energy required for the process
He⁺(g) → He²⁺(g) + e^–
The ionization energy for the H atom in the ground state is 2.18×10^-18 J atom^-1.

Answer

For H-like particles, E_n= – (2π²mZ²e⁴/n²h²)

For H-atom, I.E. = E-E₁= 0 – (-2π²m×2²×e⁴/1²×h²)

= (4×2π²me⁴/h²) = 4×2.18×10^-18 J = 8.72×10^-18 J

2.35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Answer

Diameter of a carbon atom = 0.15 nm = 0.15×10^-9 m = 1.5×10^-10m

Length along which atoms are to be placed = 20 cm = 2×10^-1m

∴ No. of C-atoms which can be placed along the line = 2×10^-1m/1.5×10^-10m = 1.33×10⁹

2.36. 2×10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Answer

Total length = 2.4 cm

Total number of atoms along the length = 2×10⁸

∴ Diameter of each atom = 2.4 cm/2×10⁸ = 1.2×10^-8cm

∴ Radius of the atom = Diameter/2 = 1.2×10^-8cm/2 = 0.6×10^-8cm

2.37. The diameter of zinc atom is 2.6 Å.Calculate (a) radius of zinc atom in pm and (b)number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Answer

(a) Radius of zinc atom = 2.6Å/2 = 1.3Å = 1.3×10^-10m = 130×10^-12pm

(b) Given length = 1.6 cm = 1.6×10^-2 m

Diameter of one atom = 2.6 Å = 2.6×10^-10m

∴ No. of atoms present along the length = 1.6×10^-2/2.6×10^-10 = 6.154×10⁷

2.38. A certain particle carries 2.5×10^-16 C of static electric charge. Calculate the number of electrons present in it.

Answer

Charge on one electron = 1.602×10^-19C

∴ Number of electrons carrying 2.5×10^-16 C charge = 2.5×10^-16/1.602×10^-19 =1560

2.39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining

X-rays. If the static electric charge on the oil drop is –1.282×10^-18 C, calculate the number of electrons present on it.

Answer

As in the above question,

Number of electrons present in oil drop = –1.282×10^-18/1.602×10^-19 = 8

2.40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?

Answer

Heavy atoms have nucleus carrying large amount of positive charge. Therefor, some α-particles will easily deflected back. Also a number of α-particles are deflected through small angles because of large positive charge.

If light atoms are used, their nuclei will have small positive charge, hence the number of α-particles getting deflected even through small angles will be negligible.

2.41. Symbols ⁷⁹Br₃₅ and ⁷⁹Br can be written, whereas symbols ³⁵Br₇₉ and ₃₅Br are not acceptable. Answer briefly.

Answer
³⁵Br₇₉ is not acceptable because atomic number should be written as subscript, while mass number should be written as superscript. ₃₅Br is not acceptable because atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotopes taken. Hence, it is essential to indicate mass number.

2.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Answer
Mass number = protons+neutrons = p+n = 81 (given)
Let p be x, then neutrons = x + (31.7/100)x = 1.317 x
∴ x + 1.317 x = 81
⇒ 2.317 x = 81
⇒ x = 81/2.317 = 35
Thus, protons = 35 = atomic number.
The symbol of the element is ⁸¹Br₃₅ or ⁸¹₃₅Br

2.43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Answer
Let the number of electrons in the ion = x
Then, number of neutrons = x+(11.1 x/ 100) = 1.111 x
Number of electrons in the neutral atom = x-1 (ion possesses one unit of negative charge)
∴ Number of protons = x-1
Mass number = No. of protons + No. of neutrons
∴ 1.111 x + x – 1 = 37
⇒ 2.111x = 38
⇒ x = 18
∴ No. of protons = Atomic no. = x-1 = 18-1 = 17
The symbol of the ion is ³⁷₁₇Cl^-1
2.44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Answer

Let the number of electrons in the ion = x
Then, number of neutrons = x+(30.4 x/ 100) = 1.304 x
Number of electrons in the neutral atom = x+3 (ion possesses 3 units of positive charge)
∴ Number of protons = x+3
Mass number = No. of protons + No. of neutrons
∴ 1.304 x + x +3 = 56
⇒ 2.304x = 53
⇒ x = 23
∴ No. of protons = Atomic no. = x+3 = 23+3 = 26
The symbol of the ion is ⁵⁶₂₆Fe⁺³
2.45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Answer

The increasing order of frequency is as follows:

Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays

2.46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6×10²⁴, calculate the power of this laser.

Answer
E = Nhν = Nhc/λ = (5.6×10²⁴)×(6.626×10^-34Js×3.0×10⁸ms^-1)/337.1×10^-9m = 3.3×10⁶J

2.47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Answer
λ = 616 nm = 616×10^-9m
(a) Frequency, ν = c/λ = 3.0×10⁸ms^-1/616×10^-9m = 4.87×10¹⁴s^-1

(b) Velocity of the radiation = 3.0×10⁸ms^-1

∴ Distance travelled in 30 s = 30×3×10⁸m = 9.0×10⁹m

(d) No. of quanta in 2J of energy = 2J/32.27×10^-20J = 6.2×10¹⁸

2.48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15×10^-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

Answer

Energy of one photon = hν = hc/λ = (6.626×10^-34Js×3.0×10⁸ms^-1)/600×10^-9m = 3.313×10^-19 J

Total energy received = 3.15×10^-18 J

∴ No. of photons received = 3.15×10^-18 J/3.313×10^-19 J = 9.51 (approx 10)

Page No: 68

2.49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5×10¹⁵ , calculate the energy of the source.

Answer

Frequency = 1/2×10^-19s = 0.5×10⁹s^-1

Energy = Nhν = (2.5×10¹⁵)×(6.626×10^-34Js)×(0.5×10⁹s^-1) = 8.28×10^-10 J

2.50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.

Answer

λ₁ = 589 nm = 589×10^-9 m
∴ ν₁= c/λ₁ = 3.0×10⁸ms^-1/589×10^-9 m = 5.093×10¹⁴s^-1
λ₂ = 589.6 nm = 589.6×10^-9 m
∴ ν₂= c/λ₂ = 3.0×10⁸ms^-1/589.6×10^-9 m = 5.088×10¹⁴s^-1
ΔE = E₂ –E₁= h(ν₂–ν₁) = (6.626×10^-34Js)×(5.093-5.088)×10¹⁴s^-1= 3.31×10^-22 J

2.51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer
(a) Work function (W₀) = hν₀
∴ ν₀= W₀/h = 1.9×1.602×10^-19J/6.626×10^-34Js = 4.59×10¹⁴s^-1(1eV = 1.602×10^-19J)
(b) λ₀ = c/ν₀= 3.0×10⁸ms^-1/4.59×10¹⁴s^-1= 6.54×10^-7 m = 654×10^-9m = 654 nm
(c) K.E. of ejected electron = h(ν–ν₀) = hc (1/λ – 1/λ₀)
= (6.626×10^-34Js×3.0×10⁸ms^-1)×(1/500×10^-9 m – 1/654×10^-9 m)
= (6.626×3.0×10^-26/10^-9)×(154/500×654) J = 9.36×10^-20J
K.E. = 1/2 mv²= 9.36×10^-20J
∴ 1/2×(9.11×10^-31 kg) v²= 9.36×10^-20kgm²s^-2
⇒ v²= 2.055×10¹¹m²s^-2= 20.55×10¹⁰m²s^-2

⇒ v = 4.53×10⁵ms^-1

2.52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

λ (nm) 500 450 400

v×10^-5(cm s^-1) 2.55 4.35 5.35

Answer

Let the threshold wavelength to be λ₀nm = λ₀×10^-9m .

Following equation holds true for photoelectric emission in given case:
K.E. = 1/2 mv²=h(ν–ν₀)
⇒1/2 mv²= hν–hν₀
⇒ hν₀=hν – 1/2 mv²
⇒ hc/λ₀ =hc/λ – 1/2 mv²

(a) Substituting the value of λ and v from the above given data, we get three values of λ₀as,
λ₀₍₁₎= 541 nm
λ₀₍₂₎= 546 nm
λ₀₍₃₎= 542 nm
Threshold frequency = λ_av = {λ₀₍₁₎+λ₀₍₂₎+λ₀₍₃₎}/3 = (541+546+542)/3 = 543 (approx 540)
(b) Part of this question can’t be solved due to incorrect value of v i.e 5.35.
Students can assume this value as 5.20 if they want to solve this question.

2.53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Answer
Energy of the incident radiation = Work function + Kinetic energy of photoelectron
E = hc/λ = (6.626×10^-34Js×3.0×10⁸ms^-1)/(256.7×10^-9 m) = 7.74×10^-19J = 4.83 eV
The potential applied gives kinetic energy to the electron.
Hence, kinetic energy of the electron = 0.35 eV
∴ Work Function = 4.83 eV – 0.35 eV = 4.48 eV

2.54. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×10⁷ms^-1 , calculate the energy with which it is bound to the nucleus.

Answer
Energy of the incident photon= hc/λ = (6.626×10^-34Js×3.0×10⁸ms^-1)/(150×10^-12m) = 13.25×10^-16J
Energy of the electron ejected = 1/2 mv²= 1/2×(9.11×10^-31kg)×(1.5×10⁷ms^-1)²= 1.025×10^-16J
Energy with which the electron was bound to the nucleus = 13.25×10^-16J – 1.025×10^-16J
= 12.225×10^-16J = 12.225×10^-16/1.602×10^-19eV = 7.63×10³eV

2.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29×10¹⁵(Hz) [1/3²–1/n²] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Answer
ν = c/λ = 3.0×10⁸ms^-1/1285×10^-9m = 3.29×10¹⁵(1/3²–1/n²)
⇒ 1/n²=1/9 – (3.0×10⁸ms^-1/1285×10^-9m)×(1/3.29×10¹⁵) = 0.111-0.071 = 0.04 = 1/25
⇒ n²= 25
⇒ n= 5
The radiation corresponding to 1285 nm lies in the infrared region.

2.56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Answer
Radius of nth orbit of H-like particles = 0.529n²/Z Å = 52.9n²/Z pm
r₁= 1.3225 nm = 1322.5 pm = 52.9n₁²
r₂= 211.6 pm = 52.9n₂²/Z
∴ r₁/r_{2 =}1322.5 pm/211.6 pm = n₁²/n₂²
⇒ n₁²/n₂²= 6.25
⇒ n₁/n₂= 2.5
If n₂= 2, n₁ = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
ṽ = 1.097×10⁷m^-1(1/2²–1/5²) = 1.097×10⁷×21/100 m^-1
λ = 1/ṽ = 100/(1.097×21×10⁷) m = 434×10^-9m = 434 nm
It lies in visible range.

2.57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6×10⁶ms^-1 , calculate de Broglie wavelength associated with this electron.

Answer
λ = h/mv = 6.626×10^-34kgm²s^-1/(9.11×10^-31 kg) (1.6×10⁶ms^-1) = 4.55×10^-10m = 455 pm

2.58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Answer
Mass of neutron = 1.675×10^-27 kg
λ = h/mv
⇒ v = h/mλ = 6.626×10^-34kgm²s^-1/(1.675×10^-27 kg) (800×10^-12m) = 4.94×10⁴ms^-1
2.59. If the velocity of the electron in Bohr’s first orbit is 2.19×10⁶ ms^-1, calculate the de Broglie wavelength associated with it.

Answer

λ = h/mv = 6.626×10^-34kgm²s^-1/(9.11×10^-31 kg) (2.19×10⁶ ms^-1) = 3.32×10^-10 m =332 pm

2.60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×10⁵ms^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.

Answer

λ = h/mv = 6.626×10^-34kgm²s^-1/(0.1 kg) (4.37×10⁵ ms^-1) = 1.516×10^-28 m

2.61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is

h/(4π×0.05 nm), is there any problem in defining this value.

Answer

Δx = 0.002 nm = 2×10^-12 m

Δx × Δp = h/4π

∴ Δp = h/4πΔx = 6.626×10^-34kgm²s^-1/(4×3.14×2×10^-12 m) = 2.638×10^-23kgms^-1

Actual momentum = h/(4π×0.05 nm) = h/(4π×5×10^-11 m)

= 6.626×10^-34kgm²s^-1/(4×3.14×5×10^-11 m) = 1.055×10^-24kgms^-1

It cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.

2.62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

1. n = 4, l = 2, m_l = –2 , m s = –1/2

2. n = 3, l = 2, m_l = 1 , m s = +1/2

3. n = 4, l = 1, m_l = 0 , m s = +1/2

4. n = 3, l = 2, m_l = –2 , m s = –1/2

5. n = 3, l = 1, m_l = –1 , m s = +1/2

6. n = 4, l = 1, m_l = 0 , m s = +1/2

Answer

The orbitals occupied by the electrons are:

(1) 4d

(2) 3d

(3) 4p

(4) 3d

(5) 3p

(6) 4p

Same orbitals will have same energy and higher the value of (n+l) higher is the energy,

Their energies will be in order: (5)<(2)=(4)<(6)=(3)<(1)

Page No: 69

2.63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?

Answer

4p electrons, being farthest from the nucleus experience the lowest effective nuclear charge.

2.64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

Answer

(i) 2s is closer to the nucleus than 3s. Hence, 2s will experience larger effective nuclear charge.

(ii) 4d (Reason being the same as above)

(iii) 3p (Reason being the same as above)

2.65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?

Answer

Silicon has greater nuclear charge (+14) than aluminium (+13). Hence, the unpaired 3p electron in case of silicon will experience more effective nuclear charge.

2.66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Answer

(a) ₁₅P = 1s²2s²2p⁶3s²3 p_x¹p_y¹p_z¹.3 unpaired electrons. (in 3p)
(b) ₁₄Si = 1s²2s²2p⁶3s²3 p_x¹p_y¹. 2 unpaired electrons.(in 3p)
(c) ₂₄Cr = 1s²2s²2p⁶3s²3p⁶3d⁵4s¹. 6 unpaired electrons. (5 in 3d and 1 in 4s)
(d) ₂₆Fe = 1s²2s²2p⁶3s²3p⁶3d⁶4s². 4 unpaired electrons. (in 3d)
(e) ₃₆Kr = It is a Noble gas. All orbitals are filled. No unpaired electrons.

2.67. (a) How many sub-shells are associated with n = 4 ? (b) How many electrons will be present in the sub-shells having m_svalue of –1/2 for n = 4 ?

Answer
(a) n=4, l = 0, 1, 2, 3. 4 sub-shells are associated with n = 4
(b) No. of orbitals in the shells = n²= 4²= 16
Each orbitals has one electron with m_s= -1/2. Hence, there will be 16 electrons with m_s= -1/2.

Rohit0 comments

CLASS 11th CHAPTER -1 Some Basic Concepts of Chemistry |Chemistry | NCERT SOLUTION | EDUGROWN

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concept Of Chemisrty includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 1 Some Basic Concept Of Chemisrty. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -1 Some Basic Concepts of Chemistry| NCERT CHEMISRTY SOLUTION |

Page No: 22

Exercises

1.1. Calculate the molecular mass of the following :

(i) H₂O (ii) CO₂(iii) CH₄

Answer

(i) H₂O = (2×Atomic mass of H) + (1×Atomic mass of O)
= [2(1.0084) + 1(16.00)] amu = 2.016 u + 16.00 amu = 18.016 amu

(ii) CO₂ = (1 × Atomic mass of C) + (2 × Atomic mass of O)
= [1(12.011) + 2 (16.00)] amu = 12.011 amu + 32.00 u = 44.01 amu

(iii) CH₄= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011) + 4 (1.008)] amu = 12.011 amu + 4.032 amu = 16.043 amu

1.2. Calculate the mass percent of different elements present in Sodium Sulphate (Na₂SO₄).
Answer
Molar mass of Na₂SO₄= [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g

Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100

∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%

Mass percent of sulphur(S): (32/142) × 100 = 22.54%
Mass percent of oxygen:(O): (64/142) × 100 = 45.07%

1.3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer

% of iron by mass = 69.9 % [Given]

% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu

Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88

Simplest molar ratio = 1.25/1.25 : 1.88/1.25

⇒ 1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe₂O₃.

1.4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer
The balanced reaction of combustion of carbon in dioxygens is:
C(s) + O₂(g) → CO₂ (g)
1mole 1mole(32g) 1mole(44g)

(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon.

(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.

(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

1.5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245g mol^-1.

Answer
0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.

∴No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875

Molar mass of sodium acetate = 82.0245g mol^-1

∴Mass of sodium acetate acquired = 0.1875×82.0245 g = 15.380g

1.6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^-1 and the mass per cent of nitric acid in it being 69%.

Answer
Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO₃) = 1+14+48 = 63g mol^-1
Number of moles in 69 g of HNO₃= 69/63 moles = 1.095 moles
Volume of 100g nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L
∴ Conc. of HNO₃in moles per litre = 1.095/0.07092 = 15.44 M

1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄ )?

Answer
1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
∴ copper can be obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 39.81g

1.8. Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.Given that the molar mass of the oxide is 159.69 g mol^-1

Answer
% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu

Simplest molar ratio = 1.25/1.25 : 1.88/1.25

⇒ 1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe₂O₃.
Mass of Fe₂O_{3 =}(2×55.85) + (3×16.00) = 159.7 g mol^-1
n = Molar mass/Empirical formula mass = 159.7/159.6 = 1(approx)

Thus, Molecular formula is same as Empirical Formula i.e. Fe₂O₃.

1.9. Calculate the atomic mass (average) of chlorine using the following data :

% Natural Abundance Molar Mass

³⁵Cl 75.77 34.9689

³⁷Cl 24.23 36.9659

Answer
Fractional Abundance of ³⁵Cl = 0.7577 and Molar mass = 34.9689
Fractional Abundance of ³⁷Cl = 0.2423 and Molar mass = 36.9659
∴ Average Atomic mass = (0.7577×34.9689)amu + (0.2423×36.9659)
= 26.4959 + 8.9568 = 35.4527

1.10. In three moles of ethane (C₂H₆), calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.

Answer
(i) 1 mole of C₂H₆ contains 2 moles of Carbon atoms
∴ 3 moles of of C₂H₆will contain 6 moles of Carbon atoms
(ii) 1 mole of C₂H₆ contains 6 moles of Hydrogen atoms
∴ 3 moles of of C₂H₆will contain 18 moles of Hydrogen atoms
(iii) 1 mole of C₂H₆ contains Avogadro’s no. 6.02×10²³molecules
∴ 3 moles of of C₂H₆will contain ethane molecule = 3×6.02×10²³= 18.06×10²³molecules.

1.11. What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if its 20 g are dissolved in
enough water to make a final volume up to 2L?

Answer
Molar mass of sugar (C₁₂H₂₂O₁₁) = (12×12)+(1×22)+(11×16) = 342 g mol^-1
No. of moles in 20g of sugar = 20/342 = 0.0585 mole

Volume of Solution = 2L (given)

Molar concentration = Moles of solute/Volume of solution in L = 0.0585mol/2L = 0.0293 mol L^-1 = 0.0293 M

1.12. If the density of methanol is 0.793 kg L^-1 , what is its volume needed for making

2.5 L of its 0.25 M solution?

Answer
Molar mass of methanol (CH₃OH) = (1×12)+(4×1)+(1×16) = 32 g mol^-1= 0.032 kg mol^-1
Molarity of the solution = 0.793/0.032 = 24.78 mol L^-1

Applying, M₁V₁ (Given Solution) = M₂V₂(Solution to be prepared)
24.78×V₁= 0.25×2.5 L
V₁= 0.02522 L = 25.22 mL

1.13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below :
1Pa = 1N m^-2
If mass of air at sea level is 1034 g cm^-2,calculate the pressure in pascal.

Answer
Pressure is the force (i.e. weigh) acting per unit area.
P= F/A = 1034g×9.8ms^-2/cm²
=1034g×9.8ms^-2/cm²× 1kg/1000g × 100cm/1m × 100cm/1m = 1.01332×10⁵ N
Now,
1Pa = 1N m^-2
∴ 1.01332×10⁵ N×m^-2 =1.01332×10⁵ Pa

1.14. What is the SI unit of mass? How is it defined?

Answer
The SI unit of mass is kilogram (kg).

The kg is defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight jar at International Bureau of Weigh and Measures in France.

1.15. Match the following prefixes with their multiples:

Prefixes Multiples

(i) micro 10⁶

(ii) deca 10⁹

(iii) mega 10^-6

(iv) giga 10^-15

(v) femto 10

Answer

Prefixes Multiples

(i) micro 10^-6

(ii) deca 10

(iii) mega 10⁶

(iv) giga 10⁹

(v) femto 10^-15
1.16. What do you mean by significant figures ?

Answer

Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.

For example,

In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be ±1 in the last digit. Hence last uncertain digit is also included in Significant figures.

1.17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer
(i) 15 ppm means 5 parts in million(10⁶⁾parts.
∴ % by mass = 15/10⁶× 100 = 15×10^-4 = 1.5×10^-3%
(ii) Molar mass of chloroform(CHCl₃) = 12+1+(3×35.5) = 118.5 g mol^-1
100g of the sample contain chloroform = 1.5×10^-3g
∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10^-2g
= 1.5×10^-2/118.65 mole = 1.266×10^-4mole
∴ Molality = 1.266×10^-4m.

1.18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Answer
(i) 0.0048 = 4.8×10^-3
(ii) 234, 000 = 2.34×10⁵
(iii) 8008 = 8.008×10³
(iv) 500.0 = 5.000×10²
(v) 6.0012 = 6.0012×10⁰

1.19. How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034

Answer

(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5

1.20. Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Answer

(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810

1.21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
(a) Which law of chemical combination is obeyed by the above experimental data?Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = …………………. mm = …………………. pm
(ii) 1 mg = …………………. kg = …………………. ng
(iii) 1 mL = …………………. L = …………………. dm³
Answer

(a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These masses of dioxygen bears a simple whole number ratio as 2:4:2:5. Hence, the data given will obey the law of multiple proportions.

The statement is as follows two elements always combine in a fixed mass of other bearing a simple ratio to another to form two or more chemical compounds.

(b) (i) 1 km = 1km×1000m/1km×100cm/1m/10mm/1cm = 10⁶mm

1 km = 1km×1000m/1km×1pm/10^-12m⁼10¹⁵pm

(ii) 1 mg = 1mg×1g/1000mg×1kg/1000g = 10^-6 kg

1 mg = 1mg×1g/1000mg×1ng/10^-9g = 10^-6 ng

(iii) 1 mL = 1mL×1L/1000mL = 10^-3 L

1 mL = 1cm³ =1cm³×(1dm×1dm×1dm/10cm×10cm×10cm) = 10³dm³

1.22. If the speed of light is 3.0 × 10⁸ms^-1, calculate the distance covered by light in 2.00 ns.

Answer

Distance covered = Speed×Time = 3.0 × 10⁸ms^-1× 2.00 ns
= 3.0×10⁸ms^-1×2.00 ns×10^-9s/1ns = 6.00×10^-1m = 0.600m

1.23. In a reaction
A + B₂→ AB₂
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B

Answer
(i) According to the reaction, 1 atom of A reacts with 1 molecule of B.

∴200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B.

∴ 2 mol of A will react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting reagent.

(iii) 1 atom of A combines with 1 molecule of B.

∴ All 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric and ther is no limiting reagent.

(iv) 1 mol of atom A combines with 1 mol of molecule B.

∴ 2.5 mol of B will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence, B is the limiting reagent.

(v) 1 mol of atom A combines with 1 mol of molecule B.

∴ 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left. Hence, A is the limiting reagent.

1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N₂(g) + H₂(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00×10³g dinitrogen reacts with 1.00×10³g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Answer
1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).

∴ 2000 g of N₂will react withH₂= 6/28 ×200g = 428.6g. Thus, here N₂is the limiting reagent whileH₂is in excess.

28g of N₂ produce 34g of NH_3.

∴2000g of N₂ will produce = 34/28×2000g = 2428.57 g of NH_3.

(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1000g – 428.6g = 571.4 g

1.25. How are 0.50 mol Na₂CO₃and 0.50 M Na₂CO₃different?

Answer
Molar mass of Na₂CO₃ = (2×23)+12.00+(3×16) = 106 g mol^-1

∴0.50 mol Na₂CO₃means 0.50×106g = 53g

0.50 M Na₂CO₃means 0.50 mol ofNa₂CO₃i.e. 53g of Na₂CO₃are present in 1litre of the solution.

1.26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer

Dihydrogen gas reacts with dioxygen gas as,

2H₂(g) + O₂(g) → 2H₂O(g)

Thus, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

1.27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Answer

(i) 1 pm = 10^-12 m

28.7 pm = 28.7×10^-12 m = 2.87×10^-11 m
(ii) 1 pm = 10^-12 m
∴15.15 pm = 15.15×10^-12 m = 1.515 ×10^-11 m
(iii) 1 mg = 10^-3 g
25365 mg = 2.5365×10⁴×10^-3g
Now,
1 g = 10^-3 kg
2.5365×10 g = 2.5365×10×10^-3 kg
∴25365 mg = 2.5365×10^-2kg

1.28. Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl₂ (g)

Answer
(i) 1 g Au = 1/197 mol = 1/197×6.022×10²³atoms
(ii) 1 g Na = 1/23 mol = 1/23×6.022×10²³atoms
(iii) 1 g Li = 1/7 mol = 1/7×6.022×10²³atoms
(iv) 1 g Cl₂= 1/71 mol = 1/71×6.022×10²³atoms
Thus, 1 g of Li has the largest number of atoms.

1.29. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Answer

Mole fraction of C₂H₅OH = No. of moles of C₂H₅OH/No. of moles of solution
n_C2H5OH = n(C₂H₅OH)/(C₂H₅OH)+n(H₂O) = 0.040 (Given) … 1
We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefor, water is approx. 1L.
No. of moles in 1L of water = 1000g/18g mol^-1= 55.55 moles
Substituting n(H₂O) = 55.55 in equation 1
n(C₂H₅OH)/(C₂H₅OH) + 55.55 = 0.040
⇒ 0.96n(C₂H₅OH) = 55.55 × 0.040
⇒ n(C₂H₅OH) = 2.31 mol
Hence, molarity of the solution = 2.31M

1.30. What will be the mass of one ¹²C atom in g ?

Answer
1 mol of ¹²C atoms = 6.022×10²³atoms = 12g
∴ Mass of 1 atom ¹²C = 12/6.022×10²³g = 1.9927×10^-23g

1.31. How many significant figures should be present in the answer of the following calculations?
(i) 0.02856×298.15×0.112/0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215

Answer

(i) Least precise term i.e. 0.112 is having 3 significant digits.

∴ There will be 3 significant figures in the calculation.

(ii) 5.364 is having 4 significant figures.

∴ There will be 4 significant figures in the calculation.

(iii) Least number of decimal places in each term is 4.

∴ There will be 4 significant figures in the calculation.

1.32. Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:

Isotope Isotopic molar mass Abundance

³⁶Ar 35.96755 g mol^-1 0.337%

³⁸Ar 37.96272 g mol^-1 0.063%

⁴⁰Ar 39.9624 g mol^-1 99.600%

Answer
Molar mass of Ar = ∑p_iA_i
= (0.00337×35.96755)+(0.00063×37.96272)+(0.99600×39.9624) = 39.948 g mol^-1

1.33. Calculate the number of atoms in each of the following

(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Answer

(i) 1 mol of Ar = 6.022×10²³atoms

∴ 52 mol of Ar = 52×6.022×10²³atoms = 3.131×10²⁵atoms

(ii) 1 atom of He = 4 u of He

4 u of He = 1 Atom of He

∴ 52 u of He = 1/4 × 52 = 13 atoms

(iii) 1 mol of He = 4 g = 6.022×10²³atoms

∴ 52 g of He = (6.022×10²³/4) × 52 atoms = 7.8286×10²⁴atoms

1.34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide , 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Answer

Amount of carbon in 3.38 g of CO₂= 12/44 × 3.38 g = 0.9218 g
Amount of hydrogen in 0.690 g H₂O = 2/18 × 0.690 g = 0.0767 g
The compound contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 = 0.9985 g
% of C in the compound = (0.9218/0.9985)×100 = 92.32
% of H in the compound = (0.0767/0.9985)×100 = 7.68
(i) Calculation of empirical formula,
Moles of carbon in the compound = 92.32/12 = 7.69
Moles of hydrogen in the compound = 7.68/1 = 7.68
Simplest molar ratio = 7.69 : 7.68 = 1(approx)
∴ Empirical formula CH
(ii) 10.0 L of the gas at STP weigh = 11.6 g
∴ 22.4 L of the gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx)
∴ Molar mass of gass = 26 g mol^-1
(iii) Mass of empirical formula CH = 12+1 = 13

∴ n = Molecular Mass/Empirical Formula = 26/13 = 2

∴ Molecular Formula = C₂H₂

1.35. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Answer
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO₃ i.e. 100g
∴ 0.6844 g HCl reacts completely with CaCO₃ = 100/73 × 0.6844 g = 0.938 g

1.36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4HCl (aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?

Answer
1 mol of MnO₂= 55+32 g = 87 g
87 g of MnO₂react with 4 moles of HCl i.e. 4×36.5 g = 146 g of HCl.
∴ 5.0 g of MnO₂will react with HCl = 146/87×5.0 g = 8.40 g.

Rohit0 comments

Class 11th Chapter -15 Waves |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 15 Waves includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 15 Waves. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 15 Waves NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -15 Waves | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Here, M = 2.50 kg, T = 200 N, length of the string, 1 = 20 m
Therefore, mass per unit length of the string,

Question 2.
A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top? Given that the speed of sound in air is 340 m s^_1?
(g = 9.8 m s^-2)
Answer:
Here, h = 300 m, g = 9.8 m s ^-2, v = 340 m s^-1if h = time taken by stone to strike the surface of water in the pond, then from

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals to the speed of sound in dry air at 20 °C = 343 m s^_1.
Answer:
Here, l = 12.0 m, M = 2.10 kg, T = ?, ν = 343 m s^-1Mass per unit length ,μ

Question 4.
Use the formula   $v=\sqrt { \cfrac { \Upsilon \rho }{ \rho } }$ to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Answer:
(a) According to Boyle’s law, we know that
PV = constant at a constant temperature … (i)
Now volume of gas, $v=\cfrac { M }{ \rho }$
∴ From (i) and (ii), we get
$v=\cfrac {PM }{ \rho }$ = constant.
Since mass of a gas remains constant.
$v=\cfrac { p }{ \rho }$ = constant.
g is always constant for a given gas or air
∴ from $v=\sqrt { \cfrac { \Upsilon \rho }{ \rho } }$ equation ,we conclude that velocity of sound in air always remaun constant if its temperature is constant.

where M = ρV = molecular weight of the gas
As ϒ,R and M are Constants, So α√T
i.e. velocity of sound in a gas is directly proportional to the square root of its I      temperature, hence we conclude that the velocity of sound in air increases with increase in temperature.
(c) Effect of humidity : The presence of water vapours in air changes the density of air, thus the velocity of sound changes with  humidity of air.
Let ρ_m = density of moist air.
ρ_d = density of dry air.
_υm = velocity of sound in moist air.
_υd = velocity of sound in dry air.

Also we know that density of water vapours is less than the density of dry air i.e. dry air is heavier than water vapours as the molecular mass of water is less than that of N₂ (28) and O₂ (32), so p_m < ρ_d.i.e. velocity of sound in air increases with humidity, i.e. velocity of sound in moist air is greater than velocity of sound in dry air. That is why sound travels faster on rainy day than on a dry day.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination (x – vt) to (x + vt), i.e. y = f (x ± vt). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
(a) (x-vt)²(b) log [(x + vt)/x₀]
(c) 1/(x + vt)
Answer:
No, the converse is not true.
The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, the wave function must have a finite value.
Out of the given functions for y, none of these satisfies this condition, so these functions do not represent a travelling wave.

Question 6.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If this sound meets a water surface, what is the wavelength of
(a) the reflected sound,
(b) the transmitted sound? Speed of sound in air is 340 m s-¹ and in water 1486 ms-¹.
Answer:
Here, υ = 1000 kHz = 1000 x 10³ Hz = 10⁶ Hz;
speed of sound in air,υ_a = 340 m s^_1;
speed of sound in water, v_w = 1,486 m s^_1(a) The reflected sound: After reflection, the ultrasonic sound continues to travel in air. If λ_a is wavelength in air, then

(b) The transmitted sound : The transmitted ultrasonic sound travels in water. If λ_w is wavelength of ultrasonic sound in water, then

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s¹? The operating frequency of the scanner is 4.2 MHz.
Answer:
Here, υ= 1.7 km s^_1 = 1.7 x 10³ m s^_1;
υ= 4.2 MHz = 4.2 x 10⁶ Hz

Question 8.
A transverse harmonic wave on a string is described by
y(x, t) = 3.0 Sin (36 t + 0.018 x + π/4)
where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Answer:
The equation of the form

represents a harmonic wave of amplitude A, wavelength l and travelling from right to left with a
velocity v.
Now, the given equation for the transverse harmonic wave is

(a) Since the equation (i) and (ii) are of the same form, the given equation also represents a travelling wave propagating from right to left. Further, the coefficient of t gives the speed of the wave. Therefore, v = 2000 cm s^_1 = 20 m s^-1(b) Obviously, amplitude, A = 3.0 cm

(c) Initial phase at the origin, φ= $\cfrac { \Pi }{ 4 }$ rad
(d) Least distance between two successive crests in the wave is equal to wavelength.

Question 9.
For the wave described in question no. 8, plot the displacement (y) versus (t) graphs for x = 0,2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:

For different values of t, we calculate y using equation (i). These values are tabulated below : On plotting y versus t graph, we obtain a sinusoidal curve as shown in above figure.

Similar graphs are obtained for x = 2 cm and x = 4 cm.
The oscillatory motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three cases.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m
(b) 0.5 m
(c) λ/2
(d) 3λ/4
Answer:
Given equation of a travelling harmonic wave is
y(x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35) …….(i)
The standard equation of a travelling harmonic wave

Comparing equation (i) and (ii), we get

Question 11.
The transverse displacement of a string (clamped at its two ends) is given by
y(x,f) = 0.06sin $(\cfrac { 2\Pi }{ 3 } x)$ cos(120 πt)
Where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 x 10 ² kg. Answer the following:
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.
Answer:
Here, the equation for transverse displacement is given by
y(x,f) = 0.06sin $\cfrac { 2\Pi }{ 3 } x$ cos 120 πt …………(i)
(a) The displacement, which involves harmonic functions of x and f separately, represents a stationary wave and the displacement, which is harmonic function of the form (υt ± x), represents a travelling wave. Hence, the equation given above represents a stationary wave.
(b) When a wave pulse

Question 12.
(i) For the wave on a string described in question no. 11, do all the points on the string oscillate with the same
(a) frequency,
(b) phase,
(c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
(i) All the points on the string
(a) have the same frequency of oscillation (except at the nodes, where the frequency is zero).
(b) have the same phase between two adjacent nodes,
(c) but not the same amplitude.
(ii) Now, the amplitude of the given wave is
A = 0.06 sin $\cfrac { 2\Pi }{ 3 } x$
At x = 0.375 m, the amplitude is given by
A(at x = 0.375 m) = 0.06 sin $\cfrac { 2\Pi }{ 3 } x$ x 0.375
= 0 06 sin $\frac { \Pi }{ 4 }$ = 0.06 x 0.707 = 0.042 m 4

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent
(1) a travelling wave,
(2) a stationary wave or
(3) none at all:

(a) y=2cos(3x)sin(10t)
(b) y = 2-√x-vt
(c) y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
(d) y = cos x sin t + cos 2x sin 2t

Answer:
(a) This equation has two harmonic functions of each x and t separately, so it represents stationary wave.
(b) This function does not represent any wave as it contains no harmonic function.
(c) It represents progressive/travelling harmonic wave as the arguments of cosine and sin functions are same.
(d) This equation is the sum of two functions cos x sin t and cos 2x sin 2t each representing a stationary wave. Therefore it represents superposition of two stationary waves

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10^-2 kg and its linear mass density is 4.0 x 10 ² kg m^-1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Answer:
Here, υ = 45 Hz, M = 3.5 x 10^-2 kg;
mass/length =μ = 4.0 x 10^-2 kg m^-1Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer:
As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore, the pipe is in resonance with the fundamental note and the third harmonic (79.3 cm is about 3 times 25.5 cm]
In the fundamental mode, $\frac { \Pi }{ 4 }$ = l₁= 25.5 cm
λ = 4 x 25.5 = 102 cm = 1.02 m.
Speed of sound in air,
υ = υλ= 340 x 1.02 = 346.8 m s^-1.

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is speed of sound in steel?
Answer:

A rod clamped in the middle has antinodes (A) at its ends and node (N) at the point of clamping. In fundamental mode, thus the length of the rod is
l= $\frac { \Pi }{ 2 }$ or 2l
Where l = length of rod and also λ = wavelength of the wave
Here, l = 100 cm
υ = 2.53 kHz = 2.53 x 10 ³ Hz
∴ λ= 2 x 100 = 200 cm
If υ be the speed of sound in steel, then
υ = υ λ = 2.53 x 10³ x 200
= 506 x 10³ cm s^_1
= 5.06 x 10³ ms^-1υ= 5.06 km s^_1.

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^_1
Answer:
Here, L = 20 cm = 0.2 m, υ„ = 430 Hz, υ= 340 m s^_1The frequency of n^th normal mode of vibration of closed pipe is

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slight out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Answer:
We know that υ ∝ √T
where u = frequency, T = tension
The decrease in the tension of a string decreases its frequency.
So let us assume that original frequency υ_A of A is more than the frequency υ_B of B.
Thus υ_A– υ_B= ± 6 Hz (given)
and υ_A = 324 Hz
∴ 324 – υ_B = ± 6
or υ_B= 324 ± 6 = 318 Hz or 330 Hz.
On reducing tension of A, Δυ= 3 Hz
If υ_B = 330 Hz and on decreasing tension in A, will be reduced i.e. no. of beats will increase, but this is not so because no. of beats becomes 3.
∴ must be 318 Hz because on reducing the tension in string A, its frequency may be reduced to 321 Hz, thus giving 3 beats with υ_B = 318 Hz.

Question 19.
Explain why (or how):
(a) in a sound wave, a displacement node is a pressure antinode and vice versa,
(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”,
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:
(a) In a sound wave, a node is a point where the amplitude of oscillation i.e. displacement is zero as here a compression and a rarefaction meet and the pressure is maximum, so it is called pressure antinode. While an antinode is a point where the amplitude of oscillation is maximum i.e. displacement is maximum but pressure is minimum. So this point is called pressure node.
Hence displacement node coincides with pressure antinode and displacement antinode with pressure node.
(b) Bats emit ultrasonic waves of large frequencies (small wavelength) when they fly. These ultrasonic waves are received by them after reflection from the obstacle. Their ears are so sensitive and trained that they not only get the information of the distance of the obstacle but also that of the nature of the reflecting surface.
(c) The quality of the sound produced by an instrument depends upon the number of overtones. Since the number of overtones is different in the cases of sounds produced by violin and sitar therefore we can distinguish through them.
(d) Solids possess both the volume elasticity and the shear elasticity. Therefore they can support both longitudinal and transverse waves.
On the other hand, gases have only the volume elasticity and no shear elasticity, so only longitudinal waves can propagate in gases.
(e) A sound pulse is a combination of waves of different wavelengths. In a dispersive medium, the waves of different wavelengths travel with different speeds in different directions e. with different velocities. So the shape of the pulse gets distorted i.e. a plane wave front in a non- dispersive medium does not remain a plane wave front in a dispersive medium.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
(i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 m s^-1.
(b) recedes from the platform with a speed of 10 m s^-1?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms¹.
Answer:
Here, frequency of source of sound,υ = 400 Hz; v = 340 m s^_1 = speed of sound
Speed of source = υ_s = 10m s^_1(i) (a) When the train approaches the platform, the apparent frequency as heard by the observer on the platform will be

(b) When the train recedes from the platform, the apparent frequency as heard by the observer is given by according to the formula :

(ii) The speed of sound in each case remains same i.e.340 ms^-1.

Question 21.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Here, v = 340 m s^-1; υ = 400 Hz
(a) Speed of the wind, v_w = 10 m s^-1As the direction of blow of wind (yard to station) is the same as the direction of sound, therefore, for the observer standing on the platform; velocity of sound,
υ’ =υ + υ_w = 340 + 10 = 350 m s^-1As there is no relative motion between source of the sound and the observer, the frequency of sound will remain unchanged.
Thus, frequency of sound = 400 Hz
Wavelength of sound,

(b) Speed of the observer, υ₀ = 10 m s^-1(towards yard). When the observer moves towards the source of sound, the apparent frequency,

The wavelength of sound waves is not affected due to the motion of the observer and hence the wavelength of the sound waves will remain unchanged.
Speed of sound relative to the observer
= 340 + 10 = 350 m s^-1Therefore, situations (a) and (b) are not equivalent.

Question 22.
A travelling harmonic wave on a string is described by y (x, t) = 7.5 sin (0.0050 + 12t + π/4)
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacement and velocity as the x = 1 cm point at f = 2 s, 5 s and 11s.
Answer:

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation?
If the pulse rate is 1 after every 20 s, that is the whistle is blown for a split of second after every 20 s, is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
(a) A narrow sound pulse such as a short pip by a whistle does not have a definite wavelength or frequency. However, being a sound wave, it has a definite speed.
(b) If a short pip is produced after every 20 s, then frequency of the note produced by the whistle cannot be called 1/20 or 0.05 Hz. We may call 0.05 Hz as the frequency of repetition of the short pip.

Question 24.
One end of a long string of linear mass density 8 x 10 ³ kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t=0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Here, μ = 8.0 x 10^-3 kg m^-1, u = 256 Hz,
T= 90 kg = 90 x 9.8 = 882 N
Amplitude of wave, A =0 cm = 0.05 m.
As, the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by

As, the wave is propagating along positive x direction the equation of the wave is
y(x-t) =A sin(ωt-kx)
= 0.05 sin (1.61 x 10³ t- 4.84 x)
Here x, y, are in meter and t is in second.

Question 25.
A SONAR system fixed in a submarine operator at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Here, μ = 40.0 kHz ;
Speed of sound wave in water, υ = 1, 450 m s^_1The frequency of the waves from SONAR system will undergo a change in the following two steps:
(1) Frequency of waves from SONAR system as received by the enemy submarine moving towards the system :
(2) Frequency of waves from the enemy’s submarine as received by SONAR system : The enemy submarine will reflect the waves of frequency υ’ = 42.756 kHz and will thus act as a source of waves moving with a speed; υ_s = 100 m s^_1 toward the SONAR system (listener). If υ” is the apparent frequency as received by SONAR system, then

Question 26.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about km s^-1, and that of P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer:
Let υ₁,υ₂ be the velocities of S waves and P waves, and t₁ ,t₂ be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earthquake from the seismograph, then
l = υ₁t₁ = υ₂t₂ -(i)
Now, υ₁= 4 km s^_1 and υ₂ = 8 km s^_1∴ 4t₁ =8 t₂ or t₁ = 2t₂ …(ii)
Also, t₁ – t₂ = 4 min = 240 s.
Using (ii), 2t₂ – t₂ = 240 s ;t₂ = 240 s
t₁= 2 x t₂ = 2 x 240 = 480 s
Now, from (i) l = υ₁t₁ = 4 x 480 = 1920 km.
Hence earthquake occurs 1920 km away from the seismograph.

Question 27.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of the sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, υ = 40 kHz,
speed of sound in air,υ = 340 m s^_1∴ v’ = speed of bat
= 0.03 x υ = 0.03 x 340 = 10.20 m s^-1The bat moving towards wall acts as a moving source and for the sound waves reflected from the wall, it acts as a moving observer. Thus the source and observer are approaching each other with same speed. i.e. υ_s = υ₀ = υ’ = 10.2 m s^-1Thus apparent frequency of the reflected sound waves heard by the bat is given by

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 15 Waves and we hope this detailed NCERT Solutions are helpful.

Rohit0 comments

Class 11th Chapter -14 Oscillations |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 14 Oscillations. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 14 Oscillations NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -14 Oscillations | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(a) It is not a periodic motion as the swimmer completes only one trip but if he makes more than one trip and time for each trip is same, the motion can be categorised as periodic.
(b) It is a periodic motion because a freely suspended bar magnet if once displaced from N-S direction and released, it oscillates about this position. Hence it is simple harmonic motion also.
(c) It is also a periodic motion.
(d) It is not a periodic motion.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) The rotation of earth about its axis.
(b) Motion of an oscillating mercury column in a U-tube.
(c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) General vibrations of a polyatomic molecule about its equilibrium position.
Answer:
(a) It is periodic but not simple harmonic motion because it is not to and fro motion about a fixed point.
(b) It is simple harmonic motion.
(c) It is simple harmonic motion.
(d) It is periodic but not simple harmonic motion. A polyatomic molecule has a number of natural frequencies and its general motion is the resultant of simple harmonic motions of a number of different frequencies. The resultant motion is periodic but not simple harmonic motion.

Question 3.
Figure depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Answer:
(a) Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
(b) Represents periodic motion with period equal to 2 s.
(c) Does not represent periodic motion, because it is not identically repeated.
(d) Represents periodic motion with period equal to 2 s.

Question 4.
Which of the following functions of time represent
(a) simple harmonic,
(b) periodic but not simple harmonic, and
(c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(a) Sin ωt-cos ωt
(b) Sin³ωt
(c) 3 cos( $\cfrac { \Pi }{ 4 }$ -2ωt)
(d) Cos ωt +Cos 3ωt+cos 5ωt
(e) exp(-ω² t²)
(f) 1+ωt+ω²t²

Answer:
The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent simple harmonic motion, if it can be written

(d) cos ωt+ cos 3ωt +cos 5 ωt
It represents periodic but not simple harmonic motion. Its time period is
$\cfrac { 2\Pi }{ \omega }$ .It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its
value, cos ωt has a period T = $\cfrac { 2\Pi }{ \omega }$ , cos3 ωt has a period $\cfrac { 2\Pi }{ 3\omega } =\cfrac { T }{ 3 }$ , cos5 ωt has period $\cfrac { 2\Pi }{ 5\omega } =\cfrac { T }{ 5 }$ , the last two forms repeat after any integral multiple of their period. Thus each term in the sum repeats itself after T, and hence the sum is a periodic function with a period $\cfrac { 2\Pi }{ \omega }$ .
(e) exp (-ω² t²) : It is an exponential function which decreases monotonically with increasing time and tends to zero as t →∞ and thus never repeats itself. Therefore it represents non-periodic motion.
(f) 1 + ωt+ (-ω² t² ): It represents non-periodic motion (physically unacceptable because the function tends to infinity as t →∞).

Question 5.
A particle is in linear simple harmonic motion between two points, A and 6 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Answer:

Refer figure, here A and B represent the two extreme positions of a simple harmonic motion. For velocity, the direction from A to B is taken to be positive. For acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
(a) Harmonic motion is momentarily at rest being its extreme position of motion. Therefore, its velocity is zero, acceleration is +ve because it is directed along AP. Force is also +ve, since the force is directed along AP, i.e., +ve direction.
∴ 0, + ,+
(b) At the end B, velocity is zero. Here, acceleration and force are negative as they are directed along BP, i.e., along negative direction.
∴ 0, -, –
(c) At the mid point of AB going towards A, the particle is at its mean position P, with a tendency to move along PA, i.e., -ve direction. Hence, velocity is -ve. Both, acceleration and force are zero.
∴ -, 0,0
(d) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is negative direction.Here velocity, acceleration and force, all are negative.
∴ -, -, –
(e) At 3 cm away from A going towards B, the particle is at R, with a tendency to move along RP, which is positive direction. Here, velocity, acceleration and force, all are positive.
∴ +, +, +
(f) At 4 cm away from B going towards A, the ,particle is at S with a tendency to move along SP which is negative direction,Here, velocity, acceleration and force, all are negative.
∴ -, -, –

Question 6.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) o = -200x²(c) a = -10 x
(d) o=100x³Answer:
A particle is said to be executing simple harmonic motion, if the acceleration ‘a’ produced in it satisfies the following two conditions:
(1) ‘a’ is directly proportional to the displacement (say y) from the mean position, i.e. a ∝ y.
(2) ‘a’ is directed towards mean position i.e. acts opposite to the direction in which y increases.
Mathematically a = -ω²y …(i)
where ω = angular frequency.

(a) a =7x does not satisfy eqn. (i), so it does not represent simple harmonic motion.
(b) a = -200.x², does not satisfy equation (i) hence it does not represent simple harmonic motion.
(c) a = -10x, here x = It satisfies equation (i), so it represents simple harmonic motion.
(d) a = 100x³ does not satisfy equation (i), hence it does not represent simple harmonic motion.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function.
x(t) = A cos(ωt+ φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm s^-1, what are its amplitude and initial phase angle? The angular frequency of the particle is Π s^-1 If instead of the cosine function, we choose the sine function to describe the simple harmonic motion : x = B sin ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Here, at t = 0, x = 1 cm and υ= co cm s^_1;
ω= Π S^_1Given : x = A cos(ωt + φ) … (i)
Since at t = 0, x = 1, we get
1 = A cos(Π x 0 +φ) = A cosφ
The instantaneous particle velocity is given by

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Here, m = 50 kg,
Maximum extension, y = 20 – 0 = 20 cm = 0.2 m
Maximum force, F=mg = 50 x 9.8 = 490 N

Question 9.
A spring having a spring constant 1200 N m^-1is mounted on a horizontal table as shown in figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine
(i) the frequency of oscillations of the mass,
(ii)maximum acceleration of the mass,and
(iii) the maximum speed of the mass.
Answer:
Here, m = 3.0 kg; k = 1200 N m^_1,
A = 2 cm = 0.02 m
(i) The frequency of oscillations of the attached mass is

Question 10.
In previous question let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t= 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for simple harmonic motion differ from each other, in frequency, in amplitude or the initial phase?
Answer:
From previous solution

(a) As time is noted from the mean position, hence using
x = A sin ωt, we get x = 2 sin 20t
(b) At maximum stretched position, the mass is at the extreme right position, with an initial phase of $\cfrac { \Pi }{ 2 }$ rad. Then,

Question 11.
Figures (a) and (b) correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:
In figure
(a) T = 2 s; A = 3 cm
At t = 0, OP makes angle $\cfrac { \Pi }{ 2 }$ with x-axis, i.e. = $\cfrac { \Pi }{ 2 }$ radian. While moving clockwise, here φ = + $\cfrac { \Pi }{ 2 }$ Thus the x-projection of OP at time t will give us the equation of simple harmonic motion, given by

or x = -3 sin ωt (where x is in cm)
In figure
(b) T = 4s;A = 2m At t = 0, OP makes an angle π with the positive direction of x-axis, i.e. φ= π While moving anticlockwise, here φ=+π
Thus the x-projection of OP at time t will give us the equation of simple harmonic motion given by

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t=0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s)
(a) x=-2sin $(3t+\cfrac { \Pi }{ 3 } )$
(b) x=cos $(\cfrac { \Pi }{ 6 } )$
(c) x=3sin $(2{ \Pi }t+\cfrac { \Pi }{ 4 } )$
(d) x=2 cos πt
Answer:

(d) Here, x = 2 cos πt
A = 2 cm, φ = 0 and ω= π rad s^-1Therefore, at t = 0, the particle is at the point P on the right extreme position as shown in figure (iv).

Question 13.
Figure (i) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (ii) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in figure (ii) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in figure (i) and the two masses in figure (ii) are released, what is the period of oscillation in each case?
Answer:
(a) Maximum extension of the spring :
(i) Suppose the maximum extension produced in the spring is y. Then,
F = ky (in magnitude)
or y = F/k
(ii) In this case, each mass relative to the other behaves as if the other mass is fixed. In other words, force F on each mass acts as the force of reaction developed due to force F on the other mass. Therefore, in this case also, maximum extension is given by
y = F/k
(b) Period of oscillation:
In figure
(i), T = 2n T=2π√m/k
To calculate the period of oscillation, the spring in figure
(ii) can be considered as to be equivalent to the two springs, each of length 1/2 and joined at the point O, the center of the string as shown in the figure.

If k’ is force constant of each half, then k’ = 2k (Because, if a spring is cut to half of its length, its force constant becomes double. Therefore,
T = 2π√m/2k

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad min^-1, what is its maximum speed?
Answer:
Stroke of piston = 2 times the amplitude.
Let A = amplitude,
Stroke = 1 m (given)
∴ 1 =2A or A = $\cfrac { 1 }{ 2 }$ m
Now, v_max = = 200 x $\cfrac { 1 }{ 2 }$ = 100 m/min
= $\cfrac { 100 }{ 60 }$ ms^-1= $\cfrac { 5 }{ 3 }$ ms^-1

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 m s^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s^-2)
Answer:

Question 16.
Answer the following questions:
(a) Time period of a particle in S.H.M. depends on the force constant k and mass m of the particle :
T = 2π√ $\cfrac { m }{ k }$ . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√ $\cfrac { l }{ g }$ Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time under gravity?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer:
(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.
(b) The effective restoring force acting on the bob of simple pendulum in displaced position is
F = -mgsinB. When θ is small, sin θ = θ . Then the expression for time period of simple pendulum is given by
T = 2π√ $\cfrac { l }{ g }$
When θ is large sin θ < θ. If the restoring force mgsinθ is replaced by mgQ, this amounts to effective reduction in the value of ‘g’ for large angles and hence an increase in the value of time period T.
(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
(d) We know that gravity disappears under free fall, so frequency is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
Answer:
When the car moves along a circular track, the bob of the simple pendulum will possess centripetal acceleration,
a_c=υ²/R (radially inward along horizontal)
The acceleration due to gravity (g) acts vertically downwards.
Therefore, effective acceleration of the pendulum,

Question 18.
A cylindrical piece of cork of base area A and height h floats in a liquid of density p,. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

where p is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let X be the equilibrium position of a cylinder floating in a given liquid.
A = area of cross-section of the cylindrical piece of cork.
h = height of the cylindrical piece of cork,
ρ = density of the material of the cylindrical cork

ρ_c= density of the liquid in which it floats.
l = length of the cylindrical piece of cork dipping in the liquid upto point P in position X.
W = weight of the cylindrical cork.
W₁ = weight of the liquid displaced by the cork.
V = its volume m = mass of cork = Ahρ.
∴ V= Ah
W=mg = (Vρ)g = (Ahρ)g
W₁ = Area of cross-section of cork x length of cylinder dipping in liquid x density of liquid x g = Alρ₁g
According to the law of flotation,

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the liquid column of mercury in the (U-tube executes simple harmonic motion.
Answer:
The suction pump creates the pressure difference, thus mercury rises in one limb of the U-Tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be explained as:

Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.
Let ρ = density of the mercury.
L = total length of the mercury column in both the limbs.
A = internal cross-sectional area of U-tube.
m = mass of mercury in U-tube = LAρ.
Let the mercury be depressed in left limb to P’ by small distance y, then it rises by the same amount in the right limb to position Q’
∴‍ Difference in levels in the two limbs
= P’Q’ = 2y
∴ Volume of mercury contained in the column of length
2y = A x 2y
m’ = A x 2y x ρ x g
If W = weight of liquid contained in the column of length 2y
Then W=m’g = A x 2y x ρ x g
This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.
∴ F = -2Ayρg = -(2Aρg)y
If a = acceleration produced in the liquid column, then

where h = height of mercury in each limb. Now from (i), it is clear that a ∝ y and -ve sign shows that it acts opposite to y, so the motion of mercury in U-tube is simple harmonic in nature having time period (T) given by

Question 20.
An air chamber of volume V has a neck area of cross section A into which a ball of mass m just fits and can move up and down without any friction as shown in figure. Show that when the ball is pressed down a little and released it executes SHM. Obtain an expression for the time period of oscillations assuming pressure- volume variations of air to be isothermal

Answer:
Consider an air chamber of volume V with a long neck of uniform area of cross section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, the pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Δy

Clearly, F ∝ y, Negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, it will start executing linear SHM in the neck of chamber with C as mean position.
In SHM, the restoring force, F = -ky Comparing (i) and (ii), we have k = BA²/V, which is the spring factor.Now, inertia factor = mass of ball = m.
As,

Question 21.
You are riding an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of
(a) the spring constant k and
(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Answer:
(a) Here mass supported by each wheel, M = 750 kg and x =15 cm = 0.15 m
If k is spring constant of the spring, then restoring force developed on being compressed through a distance x
F = -kx
If M is mass supported by each wheel, then
∴ k x = Mg
$(K=\cfrac { Mg }{ x } =\cfrac { 750\times 9.8 }{ 0.15 } =4.9\times { 10 }^{ 4\quad }N{ m }^{ -1 }$
If b is damping constant for the spring and shock absorber system, then damped amplitude of oscillation is given by
A=A₀e-^bt/2Mwhere T is period of oscillation of the spring, A₀ the initial amplitude of the oscillations and M, the mass supported by it. From the above relation we have
A=A₀e-^bt/2MAs the amplitude of oscillation decreases by 50% during one complete oscillation,

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Answer:
Consider a particle of mass m executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by y = A sin ωt

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -αθ, where J is the restoring couple and 0 the angle of twist)
Answer:
Here, m = 10 kg; R = 15 cm ; T = 1.5 s Moment of inertia of the disc about wire
I = $\cfrac { 1 }{ 2 }$ MR² = – $\cfrac { 1 }{ 2 }$ x 10 x (0.15)² = 0.1125 kg m²When the wire is twisted by rotating the disc and then released, restoring torque will be set up. If a is angular acceleration produced, then restoring torque
τ = lα …(i)
The torsional spring constant k of the wire is defined by the relation
τ = -kθ …(ii)
where θ is the angle of twist From the equations (i) and (ii) we have
lα= -kθ or α= -k/l θ
Since k/I is constant, it follows that angular acceleration is directly proportional to the angle of twist (angular displacement). Hence, the motion executed by the disc is simple harmonic in nature and the period of torsional oscillation is given by

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of body when the displacement is
(a) 5 cm,
(b) 3 cm and
(c) 0 cm.
Answer:
Here, r = 5 cm = 0.05 m; T = 0.2 s;

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the center with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω ,x₀ and v₀.
Answer:
Let the displacement of the particle at any time t be represented by
x = Acos(ωt + φ_O) …(i)
where A = amplitude, φ_O ⁼ initial phase If ν be the velocity of the particle at time t,

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 14 Oscillations and we hope this detailed NCERT Solutions are helpful.

Rohit0 comments

Class 11th Chapter -13 Kinetic Theory |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 13 Kinectic Theory includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 13 Kinectic Theory. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 13 Kinectic Theory NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -13 Kinectic Theory | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the
diameter of an oxygen molecule to be 3 Å
Answer:
Here, the diameter of an oxygen molecule
d = 3 A = 3 x 10^-10 m
The radius of an oxygen molecule

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 liters.
Answer:
Here, R = 8.31 J mol^-1 K ^-1T = 0°C = 273 K; n= 1 mole
P = 1 atm = 0.76 of Hg column
= 0.76 x 13.6 x 10³ x 9.8 = 1.013 x 10⁵ N m²Question 3.
Figure shows plot of PV/T versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T, > T₂ or T, < T₂?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 x 10^-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H₂ = 2.02 u, of 0₂ = 32.0 u, ff = 8.31 J mol^-1 MK^-1.)
Answer:
(a) Since the dotted plot is parallel to P-axis, it tells that value of $\cfrac { pv }{ T } (=nR)$ remains constant even when P is changed.
(b) Curve at temperature T, is more close to the dotted plot than the curve at the temperature T₂. Since the behaviour of a real gas approaches the perfect gas behaviour, as the temperature is increased therefore T₁ > T₂.
(c) If the amount of gas under consideration is 1 mole, then the value of PV/T, where the curves meet PV/T- axis will be R (=8.31 J mole^-1 K^-1). For oxygen, molecular mass is 32.0 g i.e. 32.0 x 10^-3 kg. Since mass of oxygen gas under considerations is 1.0 ^x 10^-3 kg, the value of PV/T, where the curves meet the PV/T-axis is given by
$\cfrac { pv }{ T } =\cfrac { 8.31 }{ 32.0 } \times { 10 }^{ -3 }=0.26j{ k }^{ -1 }$
(d) If we obtained similar plots for 1.0 x 10^-3 kg of hydrogen, the value of PV/T, where the curves meet the y-axis will not be the same. Since molecular mass of hydrogen is 2.02 x 10^-3 kg, the mass of hydrogen that will yield same value of PV/T i.e., 0.26 J K^-1 will be
$\cfrac { 2.02\times { 10 }^{ -3 } }{ 8.31 } \times 0.26=6.32\times { 10 }^{ -5 }kg$

Question 4.
An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^-1 K^-1,
molecular mass of 0₂ = 32 u).
Answer:
Here, initially in the oxygen cylinder
Volume V₂ = 30 liters = 30 x 10^-3 m³Pressure P₁= 15 atm = 15 x 1.013 x 10⁵ Pa
Temperature = 27 + 273 = 300 K
If the cylinder contains,let n₂ moles of the oxygen gas then P₁V₁ = n₁RT,

Initial mass of oxygen in the cylinder
m_x = n₁ molecular weight of 0₂= 18.253 x 32 = 584.1 g
Finally in the oxygen cylinder, let n₂ moles of oxygen be left
Final volume V₂ = 30 x 10^-3 m³Final pressure P₂ = 11 atm = 11 x 1.013 x 10⁵ Pa
Final temperature T₂ = 17 + 273 = 290 K

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Answer:
When the air bubble is at depth of 40 m
Volume of air bubble V₂ = 1.0 cm³ = 1.0 x 10^-6 m³Pressure of air bubble P₁ = 1 atm + 40 m depth of water = 1 atm + h₁ρg
= 1.013 x 10⁵ + 40 x 10³ x 9.8 = 4.93 x 10⁵ Pa
Temperature of air bubble T₁ = 12 + 273 = 285 K
When the air bubbles reaches at the surface of lake, pressure of air bubbles
P₂ = 1 atm = 1.013 x 10⁵ Pa
Temperature of air bubble T₂ = 35°C = 35 + 273 = 308 K
Volume of air bubbles = ?

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Answer:
Here, volume of air molecules, V = 25.0 m³Temperature of air molecules
T = 27°C = 27 + 273 = 300 K
Pressure of air molecules
P = 1 atm = 1.013 x 10^s Pa
Now, PV = nRT

Question 7.
Estimate the average thermal energy of a helium atom at
(i) room temperature (27°C),(ii) the temperature on the surface of the Sun(6000 K),
(iii) the temperature of 10 million kelvin (the typical core temperature in case of a star).
Answer:
The average kinetic energy of the gas at a temperature T

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v_ms the largest?
Answer:
All the three vessels (at the same temperature and pressure) have same volume. So, in accordance with the Avogadro’s law, the three vessels will contain equal number of respective molecules, being equal to Avogadro’s number N = 6.023 x 10²³.

at a given temperature therefore, rms speed of molecules will not be the same in the three cases. As neon has the smallest mass, therefore, rms speed will be the largest in case of neon.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
Here, atomic mass of argon M₁= 39.9
Atomic mass of helium He = 4.0
Let C₁ and C₂ be the rms velocities of argon and helium at temperature T₁K and T₂ K respectively.
We know that rms speed is given by

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Answer:
Here, Pressure of nitrogen = 2 atm
= 2 x 1.013 x 10⁵ N m-² = 2.026 x 10⁵ Nm²Temperature of nitrogen = 17° C
= 17 + 273 = 290 K
Molecular diameter d = 2 x 1= 2Å= 2x 10^-10 m

Question 11.
A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer:
When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end, figure
(a) The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.

∴ P₁ = 76 cm and V₁ = 15 cm³When the tube is held vertically, 15 cm air gets another 9 cm of air filled in the right hand side (in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure, figure
(b) Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 – (76 – h) = h cm of mercury
∴ V₂ = (24 + h) cm³ and P₂ = h cm.

Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23.8 cm. Thus, in the vertical

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
Answer:
According to Graham’s law of diffusion

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n₂ = n₁ exp [-mg (h₂ – h₁)/k_BT]
where n₂,n₁ refer to number density at heights h₂ and /i, respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n₂ = n₁exp [-mg N_A (ρ – ρ’) [h₂– h₁)/(ρRT)]
where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[N_A is Avogadro’s number, and R is the universal gas constant.]
Answer:
According to the law of atmospheres,

where n₂, n₂refer to number density of particles at heights h₂ and h₁ , respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let V = average volume of a suspended particle, r = density of suspended particle, r’ = density of liquid, m = mass of equal volume of liquid displaced.
According to Archimede’s principle, effective weight of one suspended particle = actual weight – weight of liquid displaced = mg- m’g

Question 14.
Given below are the densities of some solids and liquids. Give rough estimates of the sizes of their atoms

Substance	Atomic Mass (u)	Density (10³km^-3)
Carbon(diamond)	12.01	2.22
Gold	197.0	19.32
Nitrogen (liquid)	14.01	1.00
Lithium	6.94	0.53
Fluorine (liquid)	19.00	1.14

Answer:

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 13 Kinetic Theory and we hope this detailed NCERT Solutions are helpful.

Rohit0 comments

Class 11th Chapter – 12 Thermodynamics |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 12 Thermodynamics NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -12 Thermodynamics | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 10⁴ J g^_1 ?
Answer:
Here, mass of water heated,
m = 3000 g min^-1Volume of water heated = 3.0 lit min¹Rise in temperature ΔT = (77 – 27)°C = 50°C
Specific heat of water c = 4.2 J s^-1 °C^-1
Amount of heat used ΔQ = mc ΔT

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature)to raise its temperature by 45°C at constant pressure? (Molecular mass of N₂ = 28;R = 8.3 J mol^-1 K¹.)
Answer:
Mass of nitrogen gas,
m = 2 x 10² kg = 20 g
Molecular mass of nitrogen gas = 28 g
Rise in temperature, ΔT = 45°C
Heat supplied = ?
$n=\cfrac { m }{ 2 } =\cfrac { 20 }{ 28 } =\cfrac { 5 }{ 7 }$
For nitrogen (N₂) which is a diatomic gas, molar specific heat at constant pressure
C_p = $\cfrac { 7 }{ 2 }$ R where R = 8.3 J mol^-1 K^_1Therefore, heat supplied to the gas
ΔQ = nC_pΔT = $\cfrac { 5 }{ 7 }$ x $\cfrac { 7 }{ 2 }$ x R x 45 = $\cfrac { 5 }{ 7 }$ x $\cfrac { 7 }{ 2 }$ x8.3 x 45
ΔQ = 933.8 J = 934 J

Question 3.
Explain why
(a) Two bodies at different temperatures T₁) and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures become equal. The final temperature can be the mean temperature (T₁+ T₂)/2 only when thermal capacities of the two bodies are equal.
(b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.
(c) During driving, the temperature of air inside the tyre increases due to motion. According to Charle’s law, P T. Therefore, air pressure inside the tyre increases.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Answer:
As the gas is completely insulated, the process is adiabatic.

Question 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Answer:
ln first case, the state of the gas changes adiabatically from A to
Therefore, ΔQ = 0, ΔW= – 22.3 J
If ΔH is change in internal energy of the system then,
ΔQ = ΔH + ΔW
0 = ΔH + ΔW
ΔH = – ΔW = – (- 22.3 J) = 22.3 J
In the second case, when the state A is taken to state B, the heat absorbed by the system
ΔQ = 9.35 cal = 9.35 x 4.2 J = 39.27 J = 39.3 J
ΔW = ?
Applying first law of thermodynamics ΔQ = ΔH + ΔW
ΔW =ΔH – ΔQ = 39.3 – 22.3 = 17.0 J

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. The cylinder A contains a gas at standard temperature and pressure, while the cylinder B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?
Answer:
(a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmosphere pressure will become two times. Therefore, pressure will decrease to one-half,i.e.. 0.5 atmosphere.
(b) There will be no change in the internal energy of the gas as no work is done on/ by the gas.
(c) Also, there will be no change in temperature of the gas as gas does no work in expansion.
(d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer:
Useful work done per min (output)
= 5.4 x 10⁸ J
Heat absorbed per min (input) = 3.6 x 10⁹ J

= 0.15 = 0.15 x 100% = 15%
Heat energy wasted per minute = Heat energy
wasted per useful work done per minute
= 3.6 x 10⁹ – 5.4 x 10⁸ = 3.06 x 10⁹ J min^-1

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Answer:
Given that
Heat supplied ΔQ = 100 W = 100 J s
Useful work done ΔW = 75 J s^-1Change internal energy ΔU = ?
According to first law of thermodynamics,
change in internal energy is given by
ΔU = ΔQ – ΔW
ΔU = 100 – 75 = 25 J s-¹ = 25 W

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure.

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Answer:
Total work done by gas equals to area enclosed.
W = Area DEED
= $\cfrac { 1 }{ 2 }$ x EF x DF
= $\cfrac { 1 }{ 2 }$ x (5 – 2) x (600 – 300) =450J

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.
Answer:
Here, T₁ = 36°C = 36 + 273 = 309 K
T₂ = 9°C = 9 + 273 = 283 K
Coefficient of performance of the refrigerator

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 12 Thermodynamics and we hope this detailed NCERT Solutions are helpful.

Author Archives: Rohit

CLASS 11th CHAPTER -6 Thermodynamics |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -6 Thermodynamics | NCERT CHEMISRTY SOLUTION |

CLASS 11th CHAPTER -5 States of Matter |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -5 States of Matter | NCERT CHEMISRTY SOLUTION |

CLASS 11th CHAPTER -4 Chemical Bonding and Molecular Structure |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -4 Chemical Bonding and Molecular Structure| NCERT CHEMISRTY SOLUTION |

CLASS 11th CHAPTER -3 Classification of Elements and Periodicity in Properties |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -3 Classification of Elements and Periodicity in Properties | NCERT CHEMISRTY SOLUTION |

CLASS 11th CHAPTER – 2 Structure of Atom |Chemistry | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -2 Structure of Atom NCERT CHEMISRTY SOLUTION |

CLASS 11th CHAPTER -1 Some Basic Concepts of Chemistry |Chemistry | NCERT SOLUTION | EDUGROWN

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -1 Some Basic Concepts of Chemistry| NCERT CHEMISRTY SOLUTION |

Class 11th Chapter -15 Waves |Physics| NCERT Solution | Edugrown

Class 11th Chapter -15 Waves | NCERT PHYSICS SOLUTION |

Class 11th Chapter -14 Oscillations |Physics| NCERT Solution | Edugrown

Class 11th Chapter -14 Oscillations | NCERT PHYSICS SOLUTION |

Class 11th Chapter -13 Kinetic Theory |Physics| NCERT Solution | Edugrown

Class 11th Chapter -13 Kinectic Theory | NCERT PHYSICS SOLUTION |

Class 11th Chapter – 12 Thermodynamics |Physics| NCERT Solution | Edugrown

Class 11th Chapter -12 Thermodynamics | NCERT PHYSICS SOLUTION |

Search

Recent Post

Author Archives: Rohit

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -6 Thermodynamics | NCERT CHEMISRTY SOLUTION |

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -5 States of Matter | NCERT CHEMISRTY SOLUTION |

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -4 Chemical Bonding and Molecular Structure| NCERT CHEMISRTY SOLUTION |

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -3 Classification of Elements and Periodicity in Properties | NCERT CHEMISRTY SOLUTION |

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -2 Structure of Atom NCERT CHEMISRTY SOLUTION |

NCERT Solutions for Class 11 Chemistry Part 1

Class 11th Chapter -1 Some Basic Concepts of Chemistry| NCERT CHEMISRTY SOLUTION |

Class 11th Chapter -15 Waves | NCERT PHYSICS SOLUTION |

Class 11th Chapter -14 Oscillations | NCERT PHYSICS SOLUTION |

Class 11th Chapter -13 Kinectic Theory | NCERT PHYSICS SOLUTION |

Class 11th Chapter -12 Thermodynamics | NCERT PHYSICS SOLUTION |

Search

Recent Post

Table of Contents