Kinetic Theory of Gases Exercise Questions Solutions of HC Verma Ch-24

By Sachin Kesharwani On Jan 13, 2021

Kinetic Theory of Gases Exercise Questions Solutions of HC Verma Ch-24 Concept of Physics Vol-2 for ISC Class-12. Step by Step Solution of Exercise Questions of Ch-24 Kinetic Theory of Gases HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Kinetic Theory of Gases Exercise Questions Solutions of HC Verma Ch-24 Vol-2

Board	ISC and other board
Publications	Bharti Bhawan Publishers
Ch-24	Kinetic Theory of Gases
Class	12
Vol	2nd
writer	H C Verma
Book Name	Concept of Physics
Topics	Solution of Exercise Questions
Page-Number	34, 35, 36, 37

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Exercise

Kinetic Theory of Gases Exercise HC Verma Questions Solutions

(HC Verma Ch-24 Concept of Physics Vol-2 for ISC Class-12)

(Page-34)

Question-1 :-

Calculate the volume of 1 mole of an ideal gas at STP.

Answer-1 :-

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325×0⁵Pa
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we get

PV = nRT

Question-2 :-

Find the number of molecules of an ideal gas in a volume of 1.000 cm³ at STP.

Answer-2 :-

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022×10²³
Number of molecules in 22.4×10³ cm³of ideal gas at STP = 6.022×10²³
Now,
Number of molecules in 1 cm³ of ideal gas at STP

Question-3 :-

Find the number of molecules in 1 cm³ of an ideal gas at 0°C and at a pressure of 10⁻⁵mm of mercury.

Use R = 8.31 J K^-1 mol^-1

Answer-3 :-

Given:
Volume of ideal gas, V = 1 cm³= 10^-6 m³
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10⁻⁸ m of Hg
Density of ideal gas, ρ = 13600 kgm^-3
Pressure (P) is given by

P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,
Using the ideal gas equation, we get

Number of molecules = N × n
= 6.023 × 10²³×5.874 × 10⁻¹³
= 35.384 × 10¹⁰
= 3.538 × 10¹¹

Question-4 :- (Kinetic Theory of Gases Exercise HC Verma )

Calculate the mass of 1 cm³ of oxygen kept at STP.

Answer-4 :-

We know that 22.4 L of O₂ contains 1 mol O₂ at STP. Thus,

22.4×10³ cm³of O₂= 1 mol O₂

Question-5 :-

Equal masses of air are sealed in two vessels, one of volume V₀and the other of volume 2V₀. If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels.

Use R = 8.31 JK^-1 mol^-1

Answer-5 :-

Let the pressure and temperature for the vessels of volume V₀and 2V₀ be P₁, T₁ and P₂ , T₂, respectively.
Since the two vessels have the same mass of gas, n₁ = n₂= n.

T₁=300K

T₂=600K

Using the equation of state for perfect gas, we get PV=nRT

For the vessel of volume Vo:

P₁Vo=nRT₁…(1)

For the vessel of volume 2 Vo:

P₂(2Vo)=nRT₂…(2)

Dividing eq . (2) by eq . (1), we get

2P₂/P1 = T₂/T₁= 600/300 = 2

⇒ P₂/P₁= 1

⇒ P₂: P₁= 1:1

Question-6 :-

An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10⁻³ mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 10²³ mol⁻¹, density of mercury = 13600 kg m⁻³ and g = 10 m s⁻².

Use R=8.314J K^-1 mol^-1

Answer-6 :-

Given:
Volume of electric bulb, V = 250 cc
Temperature at which manufacturing takes place, T = 27 + 273 = 300 K
Height of mercury, h = 10⁻³ mm
Density of mercury, ρ 13600 kgm⁻³
Avogadro constant, N = 6 × 10²³ mol⁻¹
Pressure (P) is given by

P = ρgh

Using the ideal gas equation, we get

PV=nRT

Question-7 :-

A gas cylinder has walls that can bear a maximum pressure of 1.0 × 10⁶ Pa. It contains a gas at 8.0 × 10⁵ Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

Answer-7 :-

Given:-
Maximum pressure that the cylinder can bear, P_max = 1.0 × 10⁶ Pa
Pressure in the gas cylinder, P₁ = 8.0 × 10⁵ Pa
Temperature in the cylinder, T₁ = 300 K
Let T₂ be the temperature at which the cylinder will break.
Volume is constant. Thus, ………… (Given)
V₁= V₂ = V
Applying the five variable gas equation, we get

Question-8 :- (Kinetic Theory of Gases Exercise HC Verma )

2 g of hydrogen is sealed in a vessel of volume 0.02 m³ and is maintained at 300 K. Calculate the pressure in the vessel.

Use R=8.3J K^-1 mol^-1

Answer-8 :-

Given:
Mass of hydrogen, m = 2 g
Volume of the vessel, V = 0.02 m³
Temperature in the vessel, T = 300 K
Molecular mass of the hydrogen, M = 2 u

No of moles, n =m/M=2/2= 1 mole

Rydberg’s constant, R = 8.3 J/Kmol
From the ideal gas equation, we get
PV = nRT

Question-9 :-

The density of an ideal gas is 1.25 × 10⁻³ g cm⁻³ at STP. Calculate the molecular weight of the gas.

Use R=8.31J K^-1 mol^-1

Answer-9 :-

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas,ρ= 1.25 × 10⁻³ gcm⁻³ =1.25 kgm⁻³
Pressure, P = 1.01325×10⁵ Pa (At STP)
Temperature, T = 273 K (At STP)
Using the ideal gas equation, we get

PV=nRT . . . (1)

Question-10 :-

The temperature and pressure at Simla are 15.0°C and 72.0 cm of mercury and at Kalka these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla.

Use R=8.314J K^-1 mol^-1

Answer-10 :-

Here,
Temperature in Simla, T₁= 15 + 273 = 288 K
Pressure in Simla, P₁ = 0.72 m of Hg
Temperature in Kalka, T₂= 35+273 = 308 K
Pressure in Kalka, P₂ = 0.76 m of Hg
Let density of air at Simla and Kalka be ρ_{1 and}ρ_{2 respectively. Then,}

Taking ratios , we get

Question-11 :-

Figure (Figure 24- E1) shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1:3. Find the ratio of the pressures in the two parts of the vessel.

Use R=8.314J K^-1 mol^-1

(Figure 24- E1)

Answer-11 :-

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,
n₁= n₂= n

Volume of the first part = V
Volume of the second part =3V

It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,
T₁ = T₂ = T
Let pressure of first and second parts be P₁ and P₂, respectively.

For first part:- Applying equation of state, we get

P1V=nRT……….(1)

For second part:- Applying equation of state, we get

P₂(3V)=nRT………….(2)

Dividing eq. (1) by eq. (2), we get

Question-12 :- (Kinetic Theory of Gases Exercise HC Verma )

Find the rms speed of hydrogen molecules in a sample of hydrogen gas at 300 K. Find the temperature at which the rms speed is double the speed calculated in the previous part.

Use R=8.314 JK^-1 mol^-1

Answer-12 :-

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, M₀ = 2 g/mol=0.002 kg /mol
We know,

In the second case, let the required temperature be T.
Applying the same formula, we get

Question-13 :-

A sample of 0.177 g of an ideal gas occupies 1000 cm³ at STP. Calculate the rms speed of the gas molecules.

Answer-13 :-

Here,
V = 10^-3 m³
Density = 0.177 kgm^-3
P = 10⁵pa

Question-14 :-

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10⁻¹⁹J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹.

Answer-14 :-

We know from kinetic theory of gases that the average translational energy per molecule is 3/2 kT.

Now,

Question-15 :-

Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.

Use R=8.314 JK^-1 mol^-1

Answer-15 :-

Here,

Question-16 :-

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0°C. Mass of a helium molecule = 6.64 × 10⁻²⁷ kg and Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹.

Answer-16 :-

Here,
m = 6.64 × 10⁻²⁷ kg
T = 273 K

Average speed of the He atom is given by

We know,
Momentum = m × V_avg
= 6.64 × 10⁻²⁷ × 1201.35
= 7.97 × 10⁻²⁴
= 8 × 10⁻²⁴ kg-m/s

Question-17 :-

The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

Use R = 8.314 JK^-1 mol^-1

Answer-17 :-

Mean velocity is given by

Let temperature for H and He respectively be T₁ and T₂, respectively.
For hydrogen:
M_H = 2g = 2 × 10^-3kg
For helium:
M_He= 4 g = 4 × 10^-3kg
Now,

Question-18 :- (Kinetic Theory of Gases Exercise HC Verma )

At what temperature the mean speed of the molecules of hydrogen gas equals the escape speed from the earth?

Use R = 8.314 JK^-1 mol^-1

Answer-18 :-

Mean speed of the molecule is given by

For escape velocity of Earth :-

Let r be the radius of Earth

Multiplying numerator and denominator by R, we get

Kinetic Theory of Gases Exercise HC Verma Questions Solutions

(HC Verma Ch-24 Concept of Physics Vol-2 for ISC Class-12)

(Page-35)

Question-19 :-

Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.

Use R = 8.314 JK^-1 mol^-1

Answer-19 :-

we know ,

Molar mass of H₂ = M_H = 2×10^-3kg

Molar mass of N₂ = M_N = 28×10^-3kg
Now,

Question-20 :-

Figure (Figure 24- E2) shows a vessel partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part.

(Figure 24- E2)

Answer-20 :-

Let the temperature of gas in both the chambers be T.
Let the molar mass of gas in the left chamber and right chamber be M₁and M₂, respectively.
Let mass of gas in the left and right chamber be m₁ and m₂, respectively. Then,

Question-21 :-

Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38 × 10⁻⁵ cm.

Use R = 8.31 JK^-1 mol^-1

Answer-21 :-

Here,

λ = 1.38 × 10^-8 m

T = 273 K
M = 2 × 10^-3 kg

Average speed of the H molecules is given by

The time between two collisions is given by

Number of collisions in 1 s

Question-22 :-

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?

Use R = 8.31 JK^-1 mol^-1

Answer-22 :-

Here,

P = 10⁵ Pa

T = 300 K

For H₂

M = 2×10^-3 kg

(a) Mean speed is given by

Let us consider a cubic volume of 1m³.

V = 1 m³

Momentum of 1 molecule normal to the striking surface before collision = mu sin 45°

Momentum of 1 molecule normal to the striking surface after collision = – mu sin 45°

Change in momentum of the molecule = 2mu sin 45° = √2mu

Change in momentum of n molecules = 2mnu sin 45° = √2mnu

Let Δt be the time taken in changing the momentum.

Force per unit area due to one molecule =√2mu/Δt = √2mu/Δt

Observed pressure due to collision by n molecules = √2mnu/Δt = 10⁵

Question-23 :- (Kinetic Theory of Gases Exercise HC Verma )

Air is pumped into an automobile tyre’s tube up to a pressure of 200 kPa in the morning when the air temperature is 20°C. During the day the temperature rises to 40°C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature.

Answer-23 :-

Here,

P₁= 2 × 10⁵ Pa

P₂ = ?

T₁ = 293 K

T₂ = 313 K

V₂ = V₁ + 0.02 V₁ = V₁ (1.02)

Now,

Question-24 :-

Oxygen is filled in a closed metal jar of volume 1.0 × 10⁻³ m³ at a pressure of 1.5 × 10⁵Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 10⁵ Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.

Answer-24 :-

Here,

V₁ = 1.0 × 10^-3 m³

T₁ = 400K

P₁ = 1.5 × 10⁵ Pa

P₂ = 1.0 × 10⁵ Pa

T₂ = 300

M = 32 g

Number of moles in the jar before

Volume of the gas when pressure becomes equal to external pressure is given by

Net volume of leaked gas = V₂ – V₁

= 1.125 × 10^-3 – 1.0 × 10^-3

= 1.25 × 10^-4 m³

Let n₂ be the number of moles of leaked gas. Applying equation of state on this amount of gas, we get

Mass of leaked gas = 32 × 0.005 = 0.16 g

An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0 × 10⁵ Pa and density of water = 1000 kg m⁻³.

Answer-25 :-

Here,

h = 3.3 m
P₁ = P_o + ρgh
⇒ P₁ = 1.0 × 10⁵ + 1000 × 9.8 × 3.3
⇒ P₁ = 1.32 × 10⁵ Pa
P₂ = 1.0 × 10⁵Pa
Since temperature remains the same, applying Boyle’s law we get
P₁ V₁ = P₂ V₂