HC Verma ii Solutions for Class 12 Science Physics Chapter 24 Kinetic Theory Of Gases are provided here with simple step-by-step explanations. These solutions for Kinetic Theory Of Gases are extremely popular among class 12 Science students for Physics Kinetic Theory Of Gases Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma ii Book of class 12 Science Physics Chapter 24 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s HC Verma ii Solutions. All HC Verma ii Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Question 1:

No, kinetic energy of the molecules does not increase. This is because velocity of the molecules does not increase with respect to the walls of the gas cylinder, when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas's kinetic energy and there will be a rise in the temperature.

#### Question 2:

No, kinetic energy of the molecules does not increase. This is because velocity of the molecules does not increase with respect to the walls of the gas cylinder, when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas's kinetic energy and there will be a rise in the temperature.

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. Boiling point of a liquid depends on the pressure above its surface. Higher the pressure above the liquid, higher will be its boiling point.

When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate this fall in pressure, more liquid undergoes phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.

#### Question 3:

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. Boiling point of a liquid depends on the pressure above its surface. Higher the pressure above the liquid, higher will be its boiling point.

When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate this fall in pressure, more liquid undergoes phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.

No, the gas won't obey ideal gas equation due to the following reasons:

1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder becaue liquid state does not obey the ideal gas equation.

#### Question 4:

No, the gas won't obey ideal gas equation due to the following reasons:

1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder becaue liquid state does not obey the ideal gas equation.

(a) Temperature is defined as the average kinetic energy of he partciles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities.

(b) No, we cannot define temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules.

#### Question 5:

(a) Temperature is defined as the average kinetic energy of he partciles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities.

(b) No, we cannot define temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules.

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, temperature will be the same for all the molecules.

#### Question 6:

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, temperature will be the same for all the molecules.

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:

1. As per the Kinetic theory, neutrons do not interact with each othe. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.

2. Neutrons are smaller than hydrogen. This fulfils another kinetic theory postulate that gas molecules should be points and should have negligible size.

#### Question 7:

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:

1. As per the Kinetic theory, neutrons do not interact with each othe. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.

2. Neutrons are smaller than hydrogen. This fulfils another kinetic theory postulate that gas molecules should be points and should have negligible size.

No, the pressure on gas won't increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.

#### Question 8:

No, the pressure on gas won't increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because pressure of a gas is formed due to the molecule's collision with the walls of the container.

#### Question 9:

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because pressure of a gas is formed due to the molecule's collision with the walls of the container.

Two postulates of kinetic theory will not be valid in this case. These are given below:

1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time.

#### Question 10:

Two postulates of kinetic theory will not be valid in this case. These are given below:

1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time.

If the gas is ideal, there will be no temperature change. Moreover, Charles's law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles's law.

#### Question 11:

If the gas is ideal, there will be no temperature change. Moreover, Charles's law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles's law.

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at 1000C,. Thus, cooking process gets faster.

#### Question 12:

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at 1000C,. Thus, cooking process gets faster.

If the molecules are not allowed to collide with each other, they will have long mean free paths and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at very low temperature.

#### Question 13:

If the molecules are not allowed to collide with each other, they will have long mean free paths and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at very low temperature.

Yes, it is possible to boil water at 300C by reducing the external pressure. A liquid boils when its vapour pressure equals external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.

No, the flask containing water boiling at 300C will not be hot.

#### Question 14:

Yes, it is possible to boil water at 300C by reducing the external pressure. A liquid boils when its vapour pressure equals external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.

No, the flask containing water boiling at 300C will not be hot.

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body.
As a result of it, temperature of our body falls down due to loss of heat and we feel cold.

#### Question 1:

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body.
As a result of it, temperature of our body falls down due to loss of heat and we feel cold.

(d) Kinetic energy.

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

#### Question 2:

(d) Kinetic energy.

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies.
At high temperature, molecules move very fast. So, they tend to collide elastically and â€‹forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) is the correct answer.

#### Question 3:

At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies.
At high temperature, molecules move very fast. So, they tend to collide elastically and â€‹forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) is the correct answer.

According to the kinetic theory, molecules show straight line in motion (translational).  So, the kinetic energy is essentially transitional.

Thus, (a) is the correct answer.

#### Question 4:

According to the kinetic theory, molecules show straight line in motion (translational).  So, the kinetic energy is essentially transitional.

Thus, (a) is the correct answer.

Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature.

Thus, (d) is the correct answer.

#### Question 5:

Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature.

Thus, (d) is the correct answer.

The rms speed of a gas is given by â€‹$\sqrt{\frac{3RT}{{M}_{o}}}$.
Since hydrogen has the lowest Mo compared to other molecules, it will have the highest rms speed.
Thus, (a) is the correct answer.

#### Question 6:

The rms speed of a gas is given by â€‹$\sqrt{\frac{3RT}{{M}_{o}}}$.
Since hydrogen has the lowest Mo compared to other molecules, it will have the highest rms speed.
Thus, (a) is the correct answer.

The straight line T1 has greater slope than Tâ€‹2. This means $\frac{P}{\rho }$ ratio is greater for T1 than T2. Now, rms velocity of a gas is given by $\sqrt{\frac{3P}{\rho }}$. This means rms velocity of gas with T1 molecules is greater than T2 molecules. Again, gas with higher temperature has higher rms velocity.

So, T1 > T2.

Thus, (a) is the correct answer.

#### Question 7:

The straight line T1 has greater slope than Tâ€‹2. This means $\frac{P}{\rho }$ ratio is greater for T1 than T2. Now, rms velocity of a gas is given by $\sqrt{\frac{3P}{\rho }}$. This means rms velocity of gas with T1 molecules is greater than T2 molecules. Again, gas with higher temperature has higher rms velocity.

So, T1 > T2.

Thus, (a) is the correct answer.

Root mean squared velocity is given by

Thus, (c) is the correct answer.

#### Question 8:

Root mean squared velocity is given by

Thus, (c) is the correct answer.

Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Thus, (c) is the correct answer.

#### Question 9:

Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Thus, (c) is the correct answer.

(b) 2000 ms−1

Given,
Molecular mass of hydrogen, MH = 2
Molecular mass of oxygen, Mo = 32
RMS speed is given by,

Now,

Hence, the correct answer is (b).

#### Question 10:

(b) 2000 ms−1

Given,
Molecular mass of hydrogen, MH = 2
Molecular mass of oxygen, Mo = 32
RMS speed is given by,

Now,

Hence, the correct answer is (b).

Let the number of moles in the gas be n.
Applying equation of state, we get

=100 kPa

Thus, (a) is the correct answer.

#### Question 11:

Let the number of moles in the gas be n.
Applying equation of state, we get

=100 kPa

Thus, (a) is the correct answer.

Thus, (c) is the correct answer.

#### Question 12:

Thus, (c) is the correct answer.

Thus, (d) is the correct answer.

#### Question 13:

Thus, (d) is the correct answer.

According to the graph, P is directly proportional to T.

Applying the equation of state, we get

PV = nRT

Constant V implies the process is isochoric.

Thus, (c) is the correct answer.

#### Question 14:

According to the graph, P is directly proportional to T.

Applying the equation of state, we get

PV = nRT

Constant V implies the process is isochoric.

Thus, (c) is the correct answer.

As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.

Thus, (b) is the correct answer.

#### Question 15:

As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.

Thus, (b) is the correct answer.

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid.

Thus, (a) is the correct answer.

#### Question 16:

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid.

Thus, (a) is the correct answer.

As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Thus, (d) is the correct answer.

#### Question 17:

As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Thus, (d) is the correct answer.

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume,  both the vessels will have same pressure.

Thus, (a) is the correct answer.

#### Question 1:

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume,  both the vessels will have same pressure.

Thus, (a) is the correct answer.

According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision.

Thus, (c) and (d) are correct answers.

#### Question 2:

According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision.

Thus, (c) and (d) are correct answers.

The average speed of molecules is given by $\sqrt{\frac{8kT}{\pi m}}$. We observe that greater the mass, lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with smaller average speed.

Thus, (b) is the correct answer.

#### Question 3:

The average speed of molecules is given by $\sqrt{\frac{8kT}{\pi m}}$. We observe that greater the mass, lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with smaller average speed.

Thus, (b) is the correct answer.

The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Thus, (b) is the correct answer.

#### Question 4:

The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Thus, (b) is the correct answer.

Pressure of an ideal gas is given by PV = $\frac{1}{3}mn{u}^{2}$. We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change.

Thus, (c) is the correct answer.

#### Question 5:

Pressure of an ideal gas is given by PV = $\frac{1}{3}mn{u}^{2}$. We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change.

Thus, (c) is the correct answer.

Average momentum of a gas sample is zero, so it does not depend upon any of these parameters.

Thus, (d) is the correct answer.

#### Question 6:

Average momentum of a gas sample is zero, so it does not depend upon any of these parameters.

Thus, (d) is the correct answer.

(a) The kinetic energy of 1 mole
(c) The number of molecules in 1 mole

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases.
Number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey Avogadro's law.
Thus, (a) and (c) are correct answers.

#### Question 7:

(a) The kinetic energy of 1 mole
(c) The number of molecules in 1 mole

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases.
Number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey Avogadro's law.
Thus, (a) and (c) are correct answers.

Thus, (d) is the correct answer.

#### Question 1:

Thus, (d) is the correct answer.

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325$×$105 Pa
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we  get

PV = nRT

V = $\frac{RT}{P}$

V=

#### Question 2:

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325$×$105 Pa
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we  get

PV = nRT

V = $\frac{RT}{P}$

V=

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022$×$1023
Number of molecules in 22.4â€‹$×$103 cmof ideal gas at STP = â€‹6.022$×$1023

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

#### Question 3:

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022$×$1023
Number of molecules in 22.4â€‹$×$103 cmof ideal gas at STP = â€‹6.022$×$1023

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

Given:
Volume of ideal gas, V = 1 cm3= 10-6 mâ€‹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 $\left(P\right)$ is given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get

$n=\frac{PV}{RT}\phantom{\rule{0ex}{0ex}}⇒n=\frac{\rho gh×\mathrm{V}}{RT}\phantom{\rule{0ex}{0ex}}⇒n=\frac{13600×9.8×{10}^{-8}×{10}^{-6}}{8.31×273}\phantom{\rule{0ex}{0ex}}⇒n=5.87×{10}^{-13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

#### Question 4:

Given:
Volume of ideal gas, V = 1 cm3= 10-6 mâ€‹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 $\left(P\right)$ is given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get

$n=\frac{PV}{RT}\phantom{\rule{0ex}{0ex}}⇒n=\frac{\rho gh×\mathrm{V}}{RT}\phantom{\rule{0ex}{0ex}}⇒n=\frac{13600×9.8×{10}^{-8}×{10}^{-6}}{8.31×273}\phantom{\rule{0ex}{0ex}}⇒n=5.87×{10}^{-13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

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

#### Question 5:

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

Let the pressure and temperature for the vessels of volume V0 and 2V0 be Pâ€‹1, T1 and P2 , T2, respectively.
Since the two vessels have the same mass of gas, n1 = n2 = n.

#### Question 6:

Let the pressure and temperature for the vessels of volume V0 and 2V0 be Pâ€‹1, T1 and P2 , T2, respectively.
Since the two vessels have the same mass of gas, n1 = n2 = n.

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, $\rho$ =  13600 kgm−3
Avogadro constant, N = 6 × 1023 mol−1
Pressure $\left(P\right)$ is given by
P = $\rho gh$

Using the ideal gas equation, we get
$PV=nRT$

#### Question 7:

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, $\rho$ =  13600 kgm−3
Avogadro constant, N = 6 × 1023 mol−1
Pressure $\left(P\right)$ is given by
P = $\rho gh$

Using the ideal gas equation, we get
$PV=nRT$

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)
V1= V2 = V

Applying the five variable gas equation, we get

#### Question 8:

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)
V1= V2 = V

Applying the five variable gas equation, we get

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 = $\frac{m}{M}=\frac{2}{2}$= 1 mole

Rydberg's constant, R = 8.3 J/Kmol

From the ideal gas equation, we get

PV = nRT
$⇒P=\frac{nRT}{V}\phantom{\rule{0ex}{0ex}}⇒P=\frac{1×8.3×300}{0.02}$
$⇒$P = 1.24 × 105 Pa

#### Question 9:

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 = $\frac{m}{M}=\frac{2}{2}$= 1 mole

Rydberg's constant, R = 8.3 J/Kmol

From the ideal gas equation, we get

PV = nRT
$⇒P=\frac{nRT}{V}\phantom{\rule{0ex}{0ex}}⇒P=\frac{1×8.3×300}{0.02}$
$⇒$P = 1.24 × 105 Pa

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas, $\rho$= 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

#### Question 10:

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas, $\rho$= 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

Here,
Temperature in Simla, T1= 15 + 273 = 288 K
Pressure in Simla, Pâ€‹1 = 0.72  m of Hg
Temperature in Kalka, T2= 35+273 = 308 K
Pressure in Kalka, Pâ€‹2 = 0.76  m of Hg
Let density of air
at Simla and Kalka be $\rho$1 and $\rho$2, respectively. Then,
$PV=\frac{\mathit{m}}{\mathit{M}}RT\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{m}}{\mathit{V}}\mathit{=}\frac{\mathit{P}\mathit{M}}{\mathit{R}\mathit{T}}\phantom{\rule{0ex}{0ex}}⇒\rho \mathit{=}\frac{\mathit{P}\mathit{M}}{\mathit{R}\mathit{T}}$

Thus,

${\rho }_{1}=\frac{{P}_{1}M}{\mathrm{R}{T}_{1}}$
${\rho }_{2}=\frac{{P}_{2}M}{\mathrm{R}{T}_{2}}$

Taking ratios, we get

$\frac{{\rho }_{\mathit{1}}}{{\rho }_{2}}=\frac{{P}_{1}}{{T}_{1}}×\frac{{T}_{2}}{{P}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{\rho }_{1}}{{\rho }_{2}}=\frac{0.72}{288}×\frac{308}{0.76}\phantom{\rule{0ex}{0ex}}⇒\frac{{\rho }_{2}}{{\rho }_{\mathit{1}}}=0.987$

#### Question 11:

Here,
Temperature in Simla, T1= 15 + 273 = 288 K
Pressure in Simla, Pâ€‹1 = 0.72  m of Hg
Temperature in Kalka, T2= 35+273 = 308 K
Pressure in Kalka, Pâ€‹2 = 0.76  m of Hg
Let density of air
at Simla and Kalka be $\rho$1 and $\rho$2, respectively. Then,
$PV=\frac{\mathit{m}}{\mathit{M}}RT\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{m}}{\mathit{V}}\mathit{=}\frac{\mathit{P}\mathit{M}}{\mathit{R}\mathit{T}}\phantom{\rule{0ex}{0ex}}⇒\rho \mathit{=}\frac{\mathit{P}\mathit{M}}{\mathit{R}\mathit{T}}$

Thus,

${\rho }_{1}=\frac{{P}_{1}M}{\mathrm{R}{T}_{1}}$
${\rho }_{2}=\frac{{P}_{2}M}{\mathrm{R}{T}_{2}}$

Taking ratios, we get

$\frac{{\rho }_{\mathit{1}}}{{\rho }_{2}}=\frac{{P}_{1}}{{T}_{1}}×\frac{{T}_{2}}{{P}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{\rho }_{1}}{{\rho }_{2}}=\frac{0.72}{288}×\frac{308}{0.76}\phantom{\rule{0ex}{0ex}}⇒\frac{{\rho }_{2}}{{\rho }_{\mathit{1}}}=0.987$

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

â€‹Volume of the first part = V
Volume of the second part =â€‹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.

#### Question 12:

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

â€‹Volume of the first part = V
Volume of the second part =â€‹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.

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, Mâ€‹0 = 2 g/mol=0.002 kg /mol
We know,

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

#### Question 13:

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, Mâ€‹0 = 2 g/mol=0.002 kg /mol
We know,

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

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

#### Question 14:

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

We know from kinetic theory of gases that the average translational energy per molecule is $\frac{3}{2}kT$.
Now,
Eavg= 0.040 eV = $0.040×1.6×{10}^{-19}=6.4×{10}^{-21}J$

#### Question 15:

We know from kinetic theory of gases that the average translational energy per molecule is $\frac{3}{2}kT$.
Now,
Eavg= 0.040 eV = $0.040×1.6×{10}^{-19}=6.4×{10}^{-21}J$

#### Question 16:

Here,
â€‹m = 6.64 × 10−27 kg
T = 273 K

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

#### Question 17:

Here,
â€‹m = 6.64 × 10−27 kg
T = 273 K

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

Mean velocity is given by

${V}_{avg}=\sqrt{\frac{8RT}{\pi M}}$

Let temperature for H and He respectively be  Tâ€‹1 and T2, respectively.
For hydrogen:
MH = 2g = 2$×$10-3 kg
For helium:
MHe= 4 g = 4$×$10-3 kg

Now,
A/q$\sqrt{\frac{8R{T}_{1}}{\pi {M}_{H}}}=\sqrt{\frac{8R{T}_{2}}{\pi {M}_{He}}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{\frac{8R{T}_{1}}{2×{10}^{-3}\pi }}=\sqrt{\frac{8R{T}_{2}}{\pi ×4×{10}^{-3}}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{\frac{{T}_{1}}{2}}=\sqrt{\frac{{T}_{2}}{4}}\phantom{\rule{0ex}{0ex}}⇒\frac{{T}_{1}}{{T}_{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{T}_{1}:{T}_{2}=1:2$

#### Question 18:

Mean velocity is given by

${V}_{avg}=\sqrt{\frac{8RT}{\pi M}}$

Let temperature for H and He respectively be  Tâ€‹1 and T2, respectively.
For hydrogen:
MH = 2g = 2$×$10-3 kg
For helium:
MHe= 4 g = 4$×$10-3 kg

Now,
A/q$\sqrt{\frac{8R{T}_{1}}{\pi {M}_{H}}}=\sqrt{\frac{8R{T}_{2}}{\pi {M}_{He}}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{\frac{8R{T}_{1}}{2×{10}^{-3}\pi }}=\sqrt{\frac{8R{T}_{2}}{\pi ×4×{10}^{-3}}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{\frac{{T}_{1}}{2}}=\sqrt{\frac{{T}_{2}}{4}}\phantom{\rule{0ex}{0ex}}⇒\frac{{T}_{1}}{{T}_{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{T}_{1}:{T}_{2}=1:2$

Mean speed of the molecule is given by

For escape velocity of Earth:

#### Question 19:

Mean speed of the molecule is given by

For escape velocity of Earth:

We know,

${V}_{avg}=\sqrt{\frac{8RT}{\pi M}}$

Molar mass of H2 = MH = 2$×$10-3 kg

Molar mass of N2 = MN = 28$×$10-3 kg
Now,
$}_{H}=\sqrt{\frac{8RT}{\pi {M}_{H}}}\phantom{\rule{0ex}{0ex}}}_{N}=\sqrt{\frac{8RT}{\pi {M}_{N}}}\phantom{\rule{0ex}{0ex}}\frac{}_{H}}{}_{N}}=\sqrt{\frac{{M}_{N}}{{M}_{H}}}=\sqrt{\frac{28}{2}}=\sqrt{14}=3.74$

#### Question 20:

We know,

${V}_{avg}=\sqrt{\frac{8RT}{\pi M}}$

Molar mass of H2 = MH = 2$×$10-3 kg

Molar mass of N2 = MN = 28$×$10-3 kg
Now,
$}_{H}=\sqrt{\frac{8RT}{\pi {M}_{H}}}\phantom{\rule{0ex}{0ex}}}_{N}=\sqrt{\frac{8RT}{\pi {M}_{N}}}\phantom{\rule{0ex}{0ex}}\frac{}_{H}}{}_{N}}=\sqrt{\frac{{M}_{N}}{{M}_{H}}}=\sqrt{\frac{28}{2}}=\sqrt{14}=3.74$

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 Mâ€‹1 and M2, respectively.
Let mass of gas in the left and right chamber be m1 and m2, respectively. Then,

#### Question 21:

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 Mâ€‹1 and M2, respectively.
Let mass of gas in the left and right chamber be m1 and m2, respectively. Then,

Here,

T = 273 K
M =

Average speed of the H molecules is given  by

The time between two collisions is given by

#### Question 22:

Here,

T = 273 K
M =

Average speed of the H molecules is given  by

The time between two collisions is given by

Here,
P = 105 Pa
T = 300 K
For H2
M = 2×10-3 kg

Here,
P = 105 Pa
T = 300 K
For H2
M = 2×10-3 kg

#### Question 36:

Let the initial pressure of the chambers A and B be PA1 and PB1, respectively.
Let the final pressure of chambers A and B be PA2 and PB2, respectively.

#### Question 37:

Let the initial pressure of the chambers A and B be PA1 and PB1, respectively.
Let the final pressure of chambers A and B be PA2 and PB2, respectively.

#### Question 46:

Here,
Relative humidity = 100%

#### Question 47:

Here,
Relative humidity = 100%

#### Question 51:

We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 650 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 480 of methyl alcohol.

#### Question 52:

We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 650 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 480 of methyl alcohol.

Here,
T = 980F

We drop perpendicular corresponding to a temperature of 36.70C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg.

#### Question 53:

Here,
T = 980F

We drop perpendicular corresponding to a temperature of 36.70C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg.

#### Question 62:

(a) Relative humidity is given by

⇒ VP = 0.4 × 1.6 × 103

Evaporation occurs as long as the atmosphere is not saturated.

Net pressure change = 1.6 × 103 − 0.4 × 1.6 × 103
= (1.6 − 0.4 × 1.6) 103
= 0.96 × 103

Let the mass of water evaporated be m. Then,

(b) At 20°C, SVP = 2.4 KPa
At 15°C, SVP = 1.6 KPa

Net pressure change = (2.4 − 1.6) × 103 Pa
= 0.8 × 103 Pa

Mass of water evaporated is given by

$m=\frac{m\text{'}×8.3×293}{18}\phantom{\rule{0ex}{0ex}}⇒m\text{'}=\frac{0.8×50×18×{10}^{3}}{8.3×293}$

= 296.06 ≈ 296 g

View NCERT Solutions for all chapters of Class 12