# 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 |

-: Select Topics :-

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* = 1.01325×0^{5 }Pa

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 cm^{3} 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 = 6.022×10^{23 }

Number of molecules in 22.4×10^{3} cm^{3 }of ideal gas at STP = 6.022×10^{23 }

Now,

Number of molecules in 1 cm^{3} of ideal gas at STP

**Question-3 :-**

Find the number of molecules in 1 cm^{3} of an ideal gas at 0°C and at a pressure of 10^{−5}mm of mercury.

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

**Answer-3 :-**

Given:

Volume of ideal gas,* V* = 1 cm^{3}^{ }= 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

Number of molecules *= N* × *n*

= 6.023 × 10^{23}×5.874 × 10^{−13}

= 35.384 × 10^{10}

= 3.538 × 10^{11}

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

Calculate the mass of 1 cm^{3} of oxygen kept at STP.

**Answer-4 :-**

We know that 22.4 L of O_{2} contains 1 mol O_{2} at STP. Thus,

22.4×10^{3} cm^{3 }of O_{2 }= 1 mol O_{2}

^{}

**Question-5 :-**

Equal masses of air are sealed in two vessels, one of volume *V*_{0}_{ }and the other of volume 2*V*_{0}. 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 V_{0}_{ }and 2V_{0} be P_{1}, T_{1} and P_{2} , T_{2}, respectively.

Since the two vessels have the same mass of gas*, *n_{1} = n_{2}_{ }= n.

T_{1}=300K

T_{2}=600K

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

For the vessel of volume Vo:

P_{1}Vo=nRT_{1}…(1)

For the vessel of volume 2 Vo:

P_{2}(2Vo)=nRT_{2}…(2)

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

2P_{2}/P1 = T_{2}/T_{1 }= 600/300 = 2

⇒ P_{2}/P_{1 }= 1

⇒ P_{2 }: P_{1 }= 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 × 10^{23} 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 × 10^{23} 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 × 10^{6} Pa. It contains a gas at 8.0 × 10^{5} 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 × 10^{6} Pa

Pressure in the gas cylinder, *P*_{1} = 8.0 × 10^{5} Pa

Temperature in the cylinder,* T*_{1} = 300 K

Let *T _{2}* be the temperature at which the cylinder will break.

Volume is constant. Thus, ………… (Given)

*V*

_{1}=

*V*

_{2}=

*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 m^{3} 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 m^{3}

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 *= 1.01325×10^{5} 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, T_{1}= 15 + 273 = 288 K

Pressure in Simla, P_{1} = 0.72 m of Hg

Temperature in Kalka, T_{2}= 35+273 = 308 K

Pressure in Kalka, P_{2} = 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,

n_{1 }= n_{2}= 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,

T_{1} = T_{2} = T

Let pressure of first and second parts be P_{1} and P_{2}, respectively.

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

P1V=nRT……….(1)

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

P_{2}(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, *M*_{}_{0} = 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 cm^{3} at STP. Calculate the rms speed of the gas molecules.

**Answer-13 :-**

Here,

V = 10^{-3} m^{3}

Density = 0.177 kgm^{-3}

P = 10^{5}pa

**Question-14 :-**

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10^{−19}J). 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 T_{1} and T_{2}, respectively.

For hydrogen:

M_{H} = 2g = 2 × 10^{-3 }kg

For helium:

M_{He}= 4 g = 4 × 10^{-3 }kg

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 H_{2} = M_{H} = 2×10^{-3 }kg

Molar mass of N_{2} = M_{N} = 28×10^{-3 }kg

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 M_{1 }and M_{2}, respectively.

Let mass of gas in the left and right chamber be m_{1} and m_{2}, 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 = 10^{5} Pa

T = 300 K

For H_{2}

M = 2×10^{-3} kg

(a) Mean speed is given by

Let us consider a cubic volume of 1m^{3}.

V = 1 m^{3}

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 = 10 ^{5}**

**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,

P_{1}= 2 × 10^{5} Pa

P_{2} = ?

T_{1} = 293 K

T_{2} = 313 K

V_{2} = V_{1} + 0.02 V_{1} = V_{1} (1.02)

Now,

**Question-24 :-**

Oxygen is filled in a closed metal jar of volume 1.0 × 10^{−3} m^{3} at a pressure of 1.5 × 10^{5}Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 10^{5} 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,

V_{1} = 1.0 × 10^{-3} m^{3}

T_{1} = 400K

P_{1} = 1.5 × 10^{5} Pa

P_{2} = 1.0 × 10^{5} Pa

T_{2} = 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 = V_{2} – V_{1}

= 1.125 × 10^{-3} – 1.0 × 10^{-3}

= 1.25 × 10^{-4} m^{3}

Let n_{2} 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

**Question-25 :-**

^{5}Pa and density of water = 1000 kg m

^{−3}.

**Answer-25 :-**

h = 3.3 m

P

_{1}= P

_{o}+ ρgh

⇒ P

_{1}= 1.0 × 10

^{5}+ 1000 × 9.8 × 3.3

⇒ P

_{1}= 1.32 × 10

^{5}Pa

P

_{2}= 1.0 × 10

^{5 }Pa

Since temperature remains the same, applying Boyle’s law we get

P

_{1}V

_{1}= P

_{2}V

_{2}

_{2}be the new radius. Then,

⇒ R_{3} = 2.2 × 10^{-3} m

**Question-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 m^{3}. One of the tubes gets punctured and the volume of the tube reduces to 0.0005 m^{3}. 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}

**Answer-26 :-**

Here,

P_{1} = 2 × 10^{5} pa

V_{1} = 0.002 m^{3}

V_{2} = 0.0005 m^{3}

T_{1} = T_{2} = 300 K

Number of moles initially ,

⇒ n_{1} = 0.16

Applying equation of state, we get

P_{2} V_{2} = n_{2} RT

Assuming the final pressure becomes equal to the atmospheric pressure, we get

P_{2} = 1.0 × 10^{5} pa

⇒ n_{2} = 0.02

Number of leaked moles= n_{2} – n_{1}

= 0.16 -0.02

= 0.14

**Question-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}

**Answer-27 :-**

Here ,

m = 0.040g

M = 4g

n = 0.040/4=0.01

T_{1} = (100 + 273) K = 373 K

He is a monoatomic gas. Thus,

C_{v }= 3 × ( 1/2 R )

⇒ C_{v} = 1.5 × 8.3 = 12.45

Let the initial internal energy be U_{1} .

Let the final internal energy be U_{2} .

U_{2} -U_{1} = n C_{v} ( T_{2} -T_{1} )

⇒ 0.01 × 12.45( T_{2} – 373) = 12

⇒ T_{2} = 469 K

The temperature in °C can be obtained as follows: 469 – 273 = 196° C

**Question-28 :-**

During an experiment, an ideal gas is found to obey an additional law pV^{2} = 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}

**Answer-28 :-**

Applying equation of state of an ideal gas, we get

PV = nRT

⇒ P = nRT/V . . . 1

Taking differentials, we get

⇒ PdV + VdP = nRdT . . . 2

Applying the additional law, we get

PV^{2} = c

V^{2} dP + 2VPdV = 0

⇒ VdP + 2PdV = 0 . . . 3

Subtracting eq. (3) from eq. (2) , we get

PdV = -nRdT

⇒ dV = −nR/P dT

Now ,

**Question-29 :- ****(Kinetic Theory of Gases Exercise HC Verma )**

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 m^{3}. Find the pressure of the mixture.

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

**Answer-29 :-**

Here ,

V = 0 .166 m^{3}

T = 300 K

Mass of O_{2} = 1.60 g

M_{O} = 32 g

n_{O} = 1.60/32 = 0.05

Mass of N_{2} = 2.80 g

MN=28g

nN = 2.80/28 = 0.1

Partial pressure of O_{2 }is given by

Total pressure is sum of the partial pressures.

⇒ P = P_{N} + P_{O} = 750 + 1500 = 2250 Pa

**Question-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.

**Answer-30 :-**

Here,

h = 1 m

P_{1} = 0.75 mHg = 0.75 ρg Pa

ρ = 13500 kg/m^{3}

Let h be the height of the mercury above the piston.

P_{2} = P_{1} + hρg

Let the CSA be A.

V_{1} = Ah = A

V_{2} = (1-h)A

Applying Boyle’s law, we get

P_{1} V_{1} = P_{2} V_{2}

⇒ 0.75 ρgA = P_{2} (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

**Question-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 Ρ/T = 1 / 2 (P_{A}T_{A}+P_{B}T_{B}) when equilibrium is achieved.

**Answer-31 :-**

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

Total Pressure is the sum of the partial pressures . It is given by

P = P’_{A + }P’_{B}

**Question-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}

**Answer-32 :-**

(a) Here ,

V_{1}=5×10^{-5}m^{3}

P_{1}=10^{5}Pa

T_{1}=273K

M = 28.8 g

** **

⇒ m = 0.0635 g

(b) Here,

V_{1}=5×10^{-5}m^{3}

P_{1}=10^{5}Pa

P_{2}=10^{5}Pa

T_{1}=273K

T_{2}=373K

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

P=10^{5}Pa

V_{1}=5×10^{-5}m^{3}

Applying equation of state, we get

PV = nRT

**Question-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.

**Answer-33 :-**

Let the CSA of the tube be A .

Initial volume of air , V_{1} = 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 P_{1.}

Now , Pressure of the mercury and trapped air balances the atmospheric pressure . Thus ,

P1 + 0.1ρg = 0.75ρg

⇒ P_{1 }= 0.65ρg

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]

P_{2 }= 0.75ρg + 0.1ρg = 0.85ρg

Applying the Boyle’s law when the temperature remains constant , we get

P_{1} V_{1 }= P_{2}V_{2 }

Let the new height of the trapped air be x .

⇒ 0.65ρg0.2A = 0.85ρgxA

⇒ x = 0.15 m = 15 cm

**Question-34 :- ****(Kinetic Theory of Gases Exercise HC Verma )**

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.

**Answer-34 :-**

Let CSA of the tube be A.

on the colder side :

P_{1 }= 0.76 m Hg

T_{1} = 300K

V_{1} = V

T_{2 }= 273K

V_{2} = Ax

on the hotter side :

P_{1}= 0.76 m Hg

T_{1 }= 300K

V_{1}‘ = V

T_{2}‘ = 400K

V_{2}‘ = Ay

In equilibrium , the pressures on both side will balance each other .

⇒ P_{2}‘ = P_{2}

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

**Question-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.

**Answer-35 :-**

Case I Atmospheric pressure + pressure due to mercury column

Case II Atmospheric pressure + Component of the pressure due to mercury column

**l = 48 cm**

**Kinetic Theory of Gases Exercise HC Verma** Questions Solutions

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

(Page-36)

**Question-36 :-**

Figure **(Figure 24-E4)** shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in the parts *A* and *B* are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time.

**(Figure 24-E4)**

**Answer-36 :-**

The middle wall is weakly conducting. Thus after a long time the temperature of both the parts will equalise. The final position of the separating wall be at distance x from the left end. So it is at a distance 30 – x from the right end Putting combined gas equation of one side of the separating wall,Read more on Sarthaks.com – https://www.sarthaks.com/63030/figure-24-shows-cylindrical-tube-of-length-cm-which-partitioned-tight-fitting-separator

** ⇒ 30 – x = 2x**

**⇒ 3x = 30**

**⇒ x = 10**

**⇒ 10 cm**

The separator will be at a distance 10 cm from left end.

**Question-37 :-**

A vessel of volume V_{0} contains an ideal gas at pressure p_{0} and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.

**Answer-37 :-**

Let P be the pressure and n be the number of moles of gas inside the vessel at any given time t .

Suppose a small amount of gas of dn moles is pumped out and the decrease in pressure is dP .

Applying equation of state to the gas inside the vessel , we get

The pressure of the gas taken out is equal to the inner pressure .

Applying equation of state , we get

( P – d P ) d V = d n RT

⇒ P dV = dn RT ……….(2)

From eq. (1) and eq. (2) , we get

⇒ dV = rdt

⇒ dV = -rdt ……..(3) [Since pressures decreases , rate is negative]

Now ,

(a)

Integrating the equation P = P_{0} to P = P and time t = 0 to t = t , we get

**Question-38 :-**

One mole of an ideal gas undergoes a process

where pο and Vο are constants . Find the temperature of the gas when V=Vο .

**Answer-38 :-**

Given :

Multiplying both sides by V, we get

**Question-39 :-**

Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows, doors, etc.

**Answer-39 :-**

We Know internal energy at a particular temperature

U=nCvT

Air in home is are chiefly diatomic molecules , so

Now by eqn. of state

Now pressure P is Constant also V of the room = constant

Thus ,

U = Constant

**Question-40 :-**

Figure (Figure 24-E5) shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate dN/dt.

**Answer-40 :-**

Here ,

P_{1}=10^{5}Pa

A=π(0.05)^{2}

L = 0.2 m

V = AL = 0.0016 m^{3}

T_{1 }= 300 K

T_{2 }= 600K

µ = 0.20

Applying 5 variable equation of state , we get

Net pressure , P = P_{2} – T_{1} = 2 × 10^{5}– 10^{5} = 10^{5}

Total force acting on the stopper = PA = 10^{5} × π × 0.05)2

Applying law of friction , we get

**Question-41 :- ****(Kinetic Theory of Gases Exercise HC Verma )**

Figure (Figure 24-E6) shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T_{0} and its pressure is p_{0} which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to 2T_{0 }?

**Answer-41 :-**

**(a)** Since pressure from outside and inside the cylinder is the same, there is no net pressure acting on the pistons. So, tension will be zero.

**(b)**

**T _{1 }= Tο**

**T _{2 }=2Tο**

**P _{2 }= Pο = 10^{5}Pa**

**CSA = A**

Let the Pistons be L distance apart .

V = AL

Applying five variable gas equation , we get

Net Force acting outside = **2Pο – Pο = Pο**

Force acting on a piston F = PοA

By the free body diagram , we get

**F – T = 0**

**T = PοA**

**Question-42 :-**

Figure (Figure 24-E7) shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h_{0} and pressure 2p_{0} where p_{0} is the atmospheric pressure. There is a hole in the wall of the tank at a depth h_{1} below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h_{2} of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.

**Answer-42 :-**

** **

(b) K.E. of the water = Pressure energy of the water at that layer

**Question-43 :-**

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm^{2} and weight 1 kg in figure. The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process.

**Answer-43 :-**

Atmospheric pressure inside the cylindrical vessel, Pο=10^{5}Pa

**A = 10 cm ^{2} = 10 × 10^{-4} m^{2}**

Pressure due to the weight of the piston =** mg/A =1 × 9.81/0 × 10 ^{-4}**

**P _{1 } = 10^{5 }+ 9.8 × 10^{3}**

**V _{1}=0.2 ×10 × 10^{-4 }= 2 × 10^{-4}**

After evacuation , external pressure above the piston = 0

**P _{2}** = 0 + 9.8× 10

^{3}

Now,

**P _{1}**

**V**=

_{1}**P**

_{2}**V**

_{2}Let L be the final length of the gas column . Then,

**V2 = 10 × 10 ^{-4}L**

**⇒(10 ^{5}+ 9.8× 10^{3}) × 0.2 × 10 × 10^{-4 }= 9.8 × 10^{3 }× 10 ×10^{-4 }L**

**L = 2.2 m**

**Question-44 :-**

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm^{2} and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder.

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

**Answer-44 :-**

Here ,

V=50 m^{3}

T=273+15=288K

RH=40%

(a) Let x be the amount of water that evaporates.

P = RH × SVP

⇒ P = 0.4 × 1600 = 640

water will evaporate until VP becomes equal to SVP.

(b) T = 288 + 5 = 293K

SVP = 2400 Pa

Difference in pressure = 2400 – 1600 = 800 Pa

Let x be the amount of water that evaporates.

**Question-45 :-**

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

**Answer-45 :-**

Here ,

**P _{1}**=0.76 m Hg

**P _{2}**=P

**P _{1}**=273K

**T _{2}**=335K

Let each of the bulbs have n1 moles initially .

Let the number of moles left in second bulb after its pressure reached P be n2.

Applying equation of state , we get

Number of moles left in the second bulb after the temperature rose =

**n _{3}** = its own

**n**moles + the it received from the first

_{1}⇒ P = 0.8375

⇒ P = 84 cm of Hg

**Kinetic Theory of Gases Exercise HC Verma** Questions Solutions

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

#### (Page-37)

**Question-46 :-**

The weather report reads, “Temperature 20°C : Relative humidity 100%”. What is the dew point?

**Answer-46 :-**

Relative humidity = 100%

⇒ Vapour pressure of air = SVP at the same temperature

So , the air is saturated at 20ºC . So , dew point is 20º C .

**Question-47 :-**

The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C − 3.2 kPa.

**Answer-47 :-**

Here ,

T = 298 K

RH = 60%

=0.6

Saturated vapour pressure=3.2×10^{3} Pa

⇒vapour pressure of water vapour

(VP) = 0.6×3.2×10^{3 }= 1.92×10^{3} Pa

If the water vapour is completely removed from the air , then net pressure = 1.04×10^{5}-1.92×10^{3}

= 1.02 × 10^{5}Pa

=102 kPa

**Question-48 :- ****(Kinetic Theory of Gases Exercise HC Verma )**

The temperature and the dew point in an open room are 20°C and 10°C. If the room temperature drops to 15°C, what will be the new dew point?

**Answer-48 :-**

Here ,

Temperature = 20ºC

Dew Point = 10ºC

Air becomes saturated at 10ºC . But if the room temperature is lowered to 15ºC , the Dew point will still be at 10ºC .

**Question-49 :-**

Pure water vapour is trapped in a vessel of volume 10 cm^{3}. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

**Answer-49 :-**

Here ,

RH = 40%

condensation occurs when VP = Pο .

⇒ **P _{1}**=0.4Pο

⇒ **P _{2}**=Pο

Since the process is isothermal , applying Boyle’s law we get

Thus water vapour condenses at volume 4.3 cm^{3}

**Question-50 :-**

A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced in the tube and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapour pressure at the room temperature is 1.0 cm.

**Answer-50 :-**

Here ,

Atmospheric pressure , P = 0.76 m Hg

pressure due to water vapour inside , P′ = 0.754 mHg

Vapour pressure = P – P′ = 0.76 – 0.754 = 0.006 mHg

SVH = 0.01 mHg

^{}

**Question-51 :-**

Using figure, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm.

^{}

^{(figure 24-E9)}

**Answer-51 :-**

^{}

We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 65^{0} of methyl alcohol.

For 0.5 atm pressure, corresponding pressure in mm Hg will be 375 mm. We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 375 mm. This gives the boiling point 48^{0}of methyl alcohol.

**Question-52 :-**

The human body has an average temperature of 98°F. Assume that the vapour pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure for the vapour pressures.

**Answer-52 :-**

Here ,

T=98ºF

When we convert the temperature to °C , we get

We drop perpendicular corresponding to a temperature of 36.7^{0}C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg.

**Question-53 :-**

A glass contains some water at room temperature 20°C. Refrigerated water is added to it slowly. when the temperature of the glass reaches 10°C, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20°C and at 8.9 mm of mercury it is 10°C.

**Answer-53 :-**

Here ,

Dew point = 10º C [∵ Dew appears at 10ºC]

At boiling point, SVP equals atmospheric pressure.

At 20º C , SVP = 17.5 mmHg

At dew point , SVP = 8.9 mmHg

**Question-54 :-**

50 m^{3} of saturated vapour is cooled down from 30°C to 20°C. Find the mass of the water condensed. The absolute humidity of saturated water vapour is 30 g m^{−3} at 30°C and 16 g m^{−3} at 20°C.

**Answer-54 :-**

We Know that 1 m^{3} of air contains 30 g of water vapour at 30ºC.

So, amount of water vapour in 50 m^{3} of air at 30ºC = (30×50) g = 1500 g

Also , 1 m^{3 }of air contains 16 g of water vapour at 20∘C.

Amount of water vapour in 50 m^{3} of air at 20ºC. = (16×50) g = 800 g

Amount of water vapour condensed = (1500 – 800) g = 700 g

**Question-55 :-**

A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If the saturation vapour pressure at the atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value.

**Answer-55 :-**

Atmospheric pressure = 76 cm Hg

SVP = 0.80 cm Hg

When water is introduced into the barometer, water evaporates.

Thus, it exerts its vapour pressure over the mercury meniscus.

As more and more water evaporates, the vapour pressure increases that forces down the mercury level further.

Finally, when the volume is saturated with the vapour at the atmospheric temperature, the highest vapour pressure, i.e. SVP is observed and the fall of mercury level reaches its minimum. Thus,

Net pressure acting on the column = 76 – 0.80 cmHg

Net length of Hg column at SVG = 75.2 cm

**Question-56 :-**

50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27°C. The water level in the jar is same as the level outside. The saturation vapour pressure at 27°C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.

**Answer-56 :-**

Here ,

Atmospheric pressure, Pο = 99.4 × 10^{3} Pa

SVP at 27ºC Pω = 3.4 × 10^{3} Pa

T = 300K

V = 50 × 10^{-6} m^{3}

Now,

Pressure inside the jar = Pressure outside the jar [∵ Level of water is same inside and outside of the jar]

Pressure outside the jar = Atmospheric pressure

Pressure inside the jar = VP of oxygen + SVP of water at 27ºC

⇒ Pο = P + Pω

⇒ P = Pο – Pω = 99.4 × 10^{3} – 3.4 × 10^{3} = 96 × 10^{3}

Applying equation of state , we get

PV = nRT

⇒ 96 × 10^{3 }× 50 × 10^{-6}= n×8.3×300

**Question-57 :- ****(Kinetic Theory of Gases Exercise HC Verma )**

A faulty barometer contains certain amount of air and saturated water vapour. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74.0 cm of mercury. Saturation vapour pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir.

**Answer-57 :-**

Given :

Let the CSA be A.

**case 1 : **

V_{1 }= (x – 74) A

SVP = 1 cm Hg

Atmospheric pressure, Pο = 76 cm Hg

Mercury column height = 74.0 cm

Let P be the air pressure above the barometer. Then,

Atmospheric pressure = SVP + Air pressure above the barometer mercury level + Mercury column height

⇒ 1+P+74 = 76

⇒ P = 1 cm

**case 2 : **

Atmospheric pressure, Pο′ = 74.0 cm Hg

Let P‘ be the air pressure. Then,

P′ + 72.10 + 1 = 76

⇒ P′ = 0.9

V_{2 }=( x – 72.1)A

Applying Boyle’s law, we get

PV_{1 }=P′V_{2 }

⇒ 1 × (x – 74)A = 0.9 × (x – 72.1) A

⇒ x = 91.1 cm

Length of the tube = 91.1 cm

**Question-58 :-**

On a winter day, the outside temperature is 0°C and relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room? The saturation vapour pressure at 0°C is 4.6 mm of mercury and at 20°C it is 18 mm of mercury.

**Answer-58 :-**

Given :

SVP at 0ºC = 4.6 mm Hg

RH = 40%

Here , V is constant .

T_{1} = 273K

T_{2}= 293K

Applying equation of state, we get

SVP at the same temperature = 18 mmHg

**Question-59 :-**

The temperature and humidity of air are 27°C and 50% on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate it. The saturation vapour pressure at 27°C = 3600 Pa.

**Answer-59 :-**

Here ,

SVP = 3600 Pa

T = 273 + 27 = 300K

V = 1 m^{3}

M = 18 g for water

RH = 50%

Let m_{1} be the mass of water present in the 50% humid air.

PV = nRT

Required pressure for saturation = 3600 Pa

Let m_{2} be the amount of water required for saturation .

Total excess water vapour that has to be added = m_{2}–m_{1}

⇒ 36 – 13 = 13 g

**Question-60 :-**

The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m^{3}. The saturation vapour pressure at 300 K 3.3 kPa. Calculate the mass of the water vapour present in the room.

**Answer-60 :-**

Here,

T = 300K

SVP = 3300 Pa at 300K

RH = 20%

⇒ P/SV P = 0.2

⇒ P = 0.2 × SVP = 0.2 × 3300 = 660

V = 50 m^{3}

M = 18 g

Now ,

PV = nRT

⇒ m = 238.55 g≈238 g

**Question-61 :- (Kinetic Theory of Gases Exercise HC Verma )**

The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m^{3}. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa.

**Answer-61 :-**

Here ,

M = 18g for water

m = 500g

V = 50 m^{3}

T = 300K

SVP = 3300 Pa

RH = 20%

V P/SV P = 0.2

⇒ VP = V_{1} = 0.2 × 3300 = 660 Pa

Partial pressure P_{1} For evaporated water is given by

**Question-62:-**

A bucket full of water is placed in a room at 15°C with initial relative humidity 40%. The volume of the room is 50 m3. (a) How much water will evaporate? (b) If the room temperature is increased by 5°C, how much more water will evaporate? The saturation vapour pressure of water at 15°C and 20°C are 1.6 kPa and 2.4 kPa respectively.

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

**Answer-62 :-**

(a) Relative humidity is given by

Evaporation occurs as long as the atmosphere is not saturated.

Net pressure change = 1.6 × 10^{3} – 0.4 × 1.6 × 10^{3}

= (1.6 – 0.4 × 1.6) 10^{3}

= 0.96 × 10^{3}

Let the mass of water evaporated be m. Then,

=361.45 ≈ 361 g

(b) At 20ºC , SVP = 2.4 KPa

At 15ºC , SVP = 1.6 KPa

Net pressure change = (2.4 – 1.6) × 10^{3 }Pa

= 0.8 × 10^{3 }Pa

Mass of water evaporated is given by

—: End of Kinetic Theory of Gases Exercise **HC Verma** **Solutions** Concept of Physics:–

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