HC Verma Questions and Solutions: Chapter 24: Kinetic Theory of Gases- 2 | HC Verma Solutions - JEE PDF Download

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

Here, STP means a system having a temperature of 273 K and 1 atm pressure.  
Pressure, P = 1.01325 × 0⁵ Pa    
No of moles, n = 1 mol
Temperature, T = 273 K
Applying the equation of an ideal gas, we  get
PV = nRT

Q.2. Find the number of molecules of an ideal gas in a volume of 1.000 cm³ at STP.

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

Q.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.

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

Q.4. Calculate the mass of 1 cm³ of oxygen kept at STP.

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

Q.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.

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 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 V₀ :

For the vessel of volume 2 V₀:

Dividing eq. (2) by  eq. (1), we  get

Q.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

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

Using the ideal gas equation, we get

Q.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.

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

Q.8. 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.

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

⇒ P = 1.24 × 10⁵ Pa

Q.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.

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

Q.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.

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 ρ₁ and ρ₂ respectively. Then,

Taking ratios , we get
 

Q.11. 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.
Use R=8.314J K^-1 mol^-1

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

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

Dividing  eq. (1) by eq. (2), we get

Q.12. 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

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

Q.13. A sample of 0.177 g of an ideal gas occupies 1000 cm³ at STP. Calculate the rms speed of the gas molecules.

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

Q.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.

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

Q.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

Here,

Q.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.

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

Q.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

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

Q.18. 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

Mean speed of the molecule is givne by

For escape velocity of Earth :-
Let r be the radius of Earth

Multiplying numerator and denominator by R, we get

Q.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

we know,

Molar mass of H₂ = M_H = 2×10^-3 kg
Molar mass of N₂ = M_N = 28×10^-3 kg
Now,

Q.20. Figure 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.

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 M₁ and M₂, respectively.
Let mass of gas in the left and right chamber be m₁ and m₂, respectively. Then,
 

Q.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.

Here,

Average speed of the H molecules is given  by

The time between two collisions is given by

Number of collisions in 1s =

Q.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.

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° =
Change in momentum of n molecules = 2mnu sin 45° =
Let Δt be the time taken in changing the momentum.
Force per unit area due to one molecule
Observed pressure due to collision by n molecules =

Q.23. 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.

Here,

P₂ = ?
T₁ = 293  K
T₂ = 313  K
V₂ = V₁ + 0.02 V₁ = V₁ (1.02)
Now,

Q.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.

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

Q.25. 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.

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 × 10⁵ Pa
Since temperature remains the same, applying Boyle's law we get
P₁ V₁ = P₂ V₂

Let R₂ be the new radius. Then,
 
⇒ R₃ = 2.2 × 10^-3 m

Q.26. Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 m³. One of the tubes gets punctured and the volume of the tube reduces to 0.0005 m³. How many moles of air have leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas.
Use R = 8.3 J K^-1 mol^-1

Here,
P₁ = 2 × 10⁵ pa
V₁ = 0.002 m³
V₂ = 0.0005 m³
T₁ = T₂ = 300 K
Number of moles initially, n₁ =

Applying equation of state, we get
P₂ V₂ = n₂ RT
Assuming the final pressure becomes equal to the atmospheric pressure, we get
P₂ = 1.0 × 10⁵ pa

Number of leaked moles= n₂ - n₁
= 0.16 -0.02
= 0.14

Q.27. 0.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J.
Use R = 8.3 J K^-1 mol^-1.

Here ,
m = 0.040g
M = 4g

T₁ = (100 + 273) K = 373 K
He is a monoatomic gas. Thus,

⇒ C_v = 1.5 × 8.3 = 12.45
Let the initial internal energy be U₁ .
Let the final internal energy be U₂ .
U₂ -U₁ = n C_v ( T₂ -T₁ )
⇒ 0.01 × 12.45( T₂ - 373) = 12
⇒ T₂ = 469  K
The temperature in °C can be obtained as follows: 469 - 273 = 196° C.

Q.28. During an experiment, an ideal gas is found to obey an additional law pV² = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V.
Use R = 8.3 J K^-1 mol^-1

Applying equation of state of an ideal gas, we get
PV = nRT

Taking differentials, we get
⇒ PdV + VdP = nRdT  . . . 2
Applying the additional law, we get
PV² = c
V² dP + 2VPdV = 0
⇒ VdP + 2PdV = 0  . . . 3
Subtracting eq. (3) from eq. (2) , we get
 PdV = -nRdT

Integrating between T₂ and T₁ , we get

Q.29. A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m³. Find the pressure of the mixture.
Use R = 8.3 J K^-1 mol^-1.

Here,
V = 0 .166 m³
T = 300 K
Mass of O₂ = 1.60  g
M_O = 32  g
n_O = 1.60/32 = 0.05
Mass of N₂ = 2.80  g

Total pressure is sum of the partial pressures.
⇒ P = P_N + P_O = 750 + 1500 = 2250 Pa

Q.30. A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.

Here,
h = 1 m
P₁ = 0.75 mHg = 0.75 ρg Pa
ρ = 13500 kg/m³
Let h be the height of the mercury above the piston.
P₂ = P₁ + hρg
Let the CSA be A.
V₁ = Ah = A
V₂ = (1 - h)A
Applying Boyle's law, we get
P₁ V₁ = P₂ V₂
⇒ 0.75 ρgA = P₂ (1 - h)A
⇒ 0.75 ρg = (0.75 ρg + hρg)(1 - h)
⇒ 0.75 = (0.75 + h)(1 - h)
⇒ h = 0.25 m
h = 25 cm

Q.31. Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are p_A, T_A, V in the vessel A and p_B, T_B, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy  when equilibrium is achieved.

Let the partial pressure of the gas in chamber A and B be P'_A and P'_B , respectively.
Applying equation of state for gas A, we get

Similarly, For gas B :

Total Pressure is the sum of the partial pressures . It is given by
P = P'_A + P'_B

Q.32. A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.
Use R = 8.3 J K^-1 mol^-1.

(a) Here ,

⇒ m = 0.0635 g
(b) Here,

M = 28.8 g


Applying equation of state , we get
PV = nRT

Thus, mass of expelled air = 0.017 g
Amount of air in the container = 0.0635 - 0.017 = 0.0465 g
(c) Here,
T = 273K

Applying equation of state, we get
PV = nRT

Q.33. A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.

Let the CSA of the tube be A .
Initial volume of air , V₁ = 20A cm = 0.2A
Length of mercury , h = 0.1 m
Let the pressure of the trapped air when the tube is inverted and vertical be P₁.
Now , Pressure of the mercury and trapped air balances the atmospheric pressure . Thus ,

when the tube is inverted with the closed end down , the pressure acting upon the trapped air is
Atmospheric pressure + Mercury column pressure
Now ,
Pressure of trapped air = Atmospheric Pressure + Mercury column Pressure  [In equilibrium]

Applying the Boyle's law when the temperature remains constant , we get
P₁ V₁ = P₂V₂ 
Let the new height of the trapped air be x .

⇒ x = 0.15 m = 15 cm

Q.34. A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Let  CSA of the tube be A.
on the colder side :
P₁ = 0.76 m Hg
T₁ = 300K
V₁ = V
T₂ = 273K
V₂ = Ax

on the hotter side :
P₁= 0.76 m Hg
T₁ = 300K
V₁' = V
T₂' = 400K
V₂' = Ay

In equilibrium , the pressures on both side will balance each other.
⇒ P₂' = P₂

From the length of the tube , we get
x + y + 0.1=1
⇒ y = 0.9-x

⇒ x = 0.365 m
⇒ x = 36.5 cm

Q.35. An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant.

Here,
Initial pressure = Atmospheric pressure + pressure due to mercury
⇒ P₁ = P₀ + P_Hg
Let the CSA of the tube be A.
P₁ = 0.76 + 0.2 = 0.96 m Hg
T₁ = T₂ = T
V₁ = 0.43 A
If the tube is slanted , then the atmospheric pressure P₀ remains the same . only the P_Hg  changes

Let the length of the air column be l.

⇒ l = 0.48 m
⇒ l = 48 cm