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

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 = 1.01325×0Pa
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we  get

PV = nRT

HC VERMA kinetic theory of gases exercise IMG 1

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 cmof ideal gas at STP = ​6.022×1023 
Now,
Number of molecules in 1 cm3 of ideal gas at STP

HC VERMA kinetic theory of gases exercise IMG 2

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 m​3
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

HC VERMA kinetic theory of gases exercise IMG 3

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  cm3 of  O2 = 1 mol  O2

HC VERMA kinetic theory of gases exercise IMG 4

Question-5 :-

Equal masses of air are sealed in two vessels, one of volume V0 and 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 V0 and 2V0 be P1, T1 and P2 , T2, respectively.
Since the two vessels have the same mass of gasn1 = 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

HC VERMA kinetic theory of gases exercise IMG 5

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, Pmax = 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)
V1V2 = V
Applying the five variable gas equation, we get

HC VERMA kinetic theory of gases exercise IMG 6

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, =m/M=2/2= 1 mole

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

HC VERMA kinetic theory of gases exercise IMG 7

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, = 1.01325×105 Pa   (At STP)
Temperature, T = 273 K    (At STP)
Using the ideal gas equation, we get

PV=nRT . . . (1)

HC VERMA kinetic theory of gases exercise IMG 8

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, P​1 = 0.72  m of Hg
Temperature in Kalka, T2= 35+273 = 308 K
Pressure in Kalka, P​2 = 0.76  m of Hg
Let density of air at Simla and Kalka be ρ1 and ρ2 respectively. Then,

HC VERMA kinetic theory of gases exercise IMG 9

Taking ratios , we get

HC VERMA kinetic theory of gases exercise IMG 10

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 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.

(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,
n= 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

HC VERMA kinetic theory of gases exercise IMG 12

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,

HC VERMA kinetic theory of gases exercise IMG 13

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

HC VERMA kinetic theory of gases exercise IMG 14

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

HC VERMA kinetic theory of gases exercise IMG 15

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,
HC VERMA kinetic theory of gases exercise IMG 16

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,

HC VERMA kinetic theory of gases exercise IMG 17

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  HC VERMA kinetic theory of gases exercise IMG 18

HC VERMA kinetic theory of gases exercise IMG 19

We know,
Momentum = m × Vavg
= 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 HC VERMA kinetic theory of gases exercise IMG 18

Let temperature for H and He respectively be  T​1 and T2, respectively.
For hydrogen:
MH = 2g = 2 × 10-3 kg
For helium:
MHe= 4 g = 4 × 10-3 kg
Now,

HC VERMA kinetic theory of gases exercise IMG 20

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

HC VERMA kinetic theory of gases exercise IMG 21

For escape velocity of Earth :-

Let r be the radius of Earth

HC VERMA kinetic theory of gases exercise IMG 22

Multiplying numerator and denominator by R, we get

HC VERMA kinetic theory of gases exercise IMG 23


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 ,

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. img 1

Molar mass of H2 = MH = 2×10-3 kg

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

HC VERMA kinetic theory of gases exercise IMG 25

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 (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 M​1 and M2, respectively.
Let mass of gas in the left and right chamber be m1 and m2, respectively. Then,

HC VERMA kinetic theory of gases exercise IMG 26

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

HC VERMA kinetic theory of gases exercise IMG 27

The time between two collisions is given by

HC VERMA kinetic theory of gases exercise IMG 28

Number  of  collisions  in  1  s

HC VERMA kinetic theory of gases exercise IMG 29

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

HC VERMA kinetic theory of gases exercise IMG 30

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

HC VERMA kinetic theory of gases exercise IMG 31

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,

HC VERMA kinetic theory of gases exercise IMG 32

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  HC VERMA kinetic theory of gases exercise IMG 33

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

HC VERMA kinetic theory of gases exercise IMG 34

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

HC VERMA kinetic theory of gases exercise IMG 35

Mass of leaked gas = 32 × 0.005 = 0.16 g

HC VERMA kinetic theory of gases exercise IMG 38

⇒ R3 = 2.2 × 10-3 m

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