HC Verma ii Solutions for Class 12 Science Physics Chapter 24 - Kinetic Theory Of Gases

Question 1:

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.  

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.    

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. 

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. 

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. 

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. 

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.  

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.   

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. 

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.

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.

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.   

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.

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.

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.

Hence, correct answer is (d).

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 ​forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) 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.

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.

Question 6:

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

Question 7:

The straight line T₁ has greater slope than T_​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 8:

Root mean squared velocity is given by
$$\begin{gathered} v_{rms}=\sqrt{\frac{3RT}{M}} \\ \Rightarrow {\left(v_{rms}\right)}^2=\frac{3RT}{M} \\ \Rightarrow {\left(v_{rms}\right)}^2 \alpha T \end{gathered}$$

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.

Question 10:

(b) 2000 ms⁻¹

Given,
Molecular mass of hydrogen, M_H = 2
Molecular mass of oxygen, M_o = 32
RMS speed is given by,

 $$\begin{gathered} v_{rms=}\sqrt{\frac{3RT}{M}} \\ \Rightarrow \sqrt{\frac{\mathit{3}RT}{M_O}}=500 \end{gathered}$$
Now,
$$\begin{gathered} \Rightarrow \frac{v_{Orms}}{v_{Hrms}}=\frac{\sqrt{{\displaystyle \frac{3RT}{M_O}}}}{\sqrt{{\displaystyle \frac{3RT}{M_H}}}} \\ \Rightarrow \frac{v_{O}rms}{v_{Hrms}}=\frac{\sqrt{{\displaystyle \frac{3RT}{32}}}}{\sqrt{{\displaystyle \frac{3RT}{2}}}} \\ \Rightarrow \frac{v_{Orms}}{{v_H}_{rms}}=\frac{1}{4} \\ \Rightarrow \frac{500}{v_{Hrms}}=\frac{1}{4} \\ \Rightarrow v_{Hrms}=4\times 500=2000 {\mathrm{ms}}^{-1} \end{gathered}$$

Hence, the correct answer is (b).

Question 11:

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

$$\begin{gathered} PV=nRT \\ \Rightarrow P=\frac{nRT}{V} \\ \Rightarrow 2\times {10}^5=\frac{nRT}{V} ...\left(1\right) \\ \mathrm{When} \mathrm{half} \mathrm{of} \mathrm{the} \mathrm{gas} \mathrm{is} \mathrm{removed}, \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{left} \mathrm{behind} =\frac{\mathrm{n}}{2} \\ \mathrm{Let} \mathrm{the} \mathrm{pressure} \mathrm{be} \mathrm{P}\text{'}. \\ \mathrm{P}\text{'}=\frac{\mathrm{n}}{2}\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}} \\ \mathrm{Now}, \\ \mathrm{P}\text{'}=\frac{1}{2}\times 2\times {10}^5={10}^5 \left[\mathrm{From} \mathrm{eq}. \left(1\right)\right] \end{gathered}$$
=100 kPa

Thus, (a) is the correct answer.

Question 12:

$$\begin{gathered} \mathrm{Given}: \mathrm{v}=\sqrt{\frac{3\mathrm{RT}}{32}} \\ \mathrm{Let} \mathrm{the} \mathrm{new} \mathrm{rms} \mathrm{speed} \mathrm{be} \mathrm{v}\text{'}. \\ \mathrm{Molecule} \mathrm{dissociate}, M=16 \\ v\text{'}=\sqrt{\frac{3R\left(2T\right)}{16}} \\ =\sqrt{\frac{3R\left(4T\right)}{32}} \\ =2\sqrt{\frac{3RT}{32}}=2v \end{gathered}$$

Thus, (c) is the correct answer.

Question 13:

$$\begin{gathered} \mathrm{Here}, \\ PV=nRT ...\left(1\right) \\ \mathrm{Also}, \\ k=\frac{R}{N} \\ \Rightarrow R=kN ...\left(2\right) \\ \mathrm{Now}, \\ PV=nkNT \left[\mathrm{From} \mathrm{eq}. \left(1\right) \mathrm{and} \mathrm{eq}. \left(2\right)\right] \\ \Rightarrow nN=\frac{PV}{kT} \\ nN = \mathrm{Number} \mathrm{of} \mathrm{molecules} \\ \frac{\mathrm{P}\mathrm{V}}{\mathrm{k}\mathrm{T}}=\mathrm{Number} \mathrm{of} \mathrm{molecules} \end{gathered}$$

Thus, (d) 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
$$\begin{gathered} \Rightarrow P=\frac{nR}{V}T \\ \mathrm{Given}: P \alpha T \\ \mathrm{This} \mathrm{means} \frac{\mathrm{n}\mathrm{R}}{\mathrm{V}} \mathrm{is} \mathrm{a} \mathrm{constant}. \mathrm{So}, \mathrm{V} \mathrm{is} \mathrm{also} \mathrm{a} \mathrm{constant}. \end{gathered}$$

Constant V implies the process is isochoric.

Thus, (c) 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

Question 1:

$$\begin{gathered} \mathrm{In} \mathrm{an} \mathrm{ideal} \mathrm{gas}, \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{state} \mathrm{is} \mathrm{given} \mathrm{by} \\ PV=nRT \\ \Rightarrow PV=n{N}_A\frac{R}{N_A}T \\ \Rightarrow PV=n{N}_{A}kT \\ \Rightarrow \frac{1}{n{N}_A}=\frac{kT}{PV} \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{gas} \mathrm{M}, \mathrm{we} \mathrm{get} \\ \frac{M}{n{N}_A}=\frac{MkT}{PV} \\ \mathrm{Now}, n{N}_A \mathrm{gives} \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{molecules} \mathrm{of} \mathrm{the} \mathrm{gas}. \\ \mathrm{Also}, \frac{M}{n{N}_A} \mathrm{gives} \mathrm{the} \mathrm{mass} \mathrm{of} \mathrm{a} \mathrm{single} \mathrm{molecule}. \\ \mathrm{Hence},\frac{\mathrm{M}\mathrm{k}\mathrm{T}}{\mathrm{P}\mathrm{V}} \mathrm{is} \mathrm{the} \mathrm{mass} \mathrm{of} \mathrm{a} \mathrm{single} \mathrm{molecule} \mathrm{of} \mathrm{the} \mathrm{gas}, \\ \mathrm{Molecular} \mathrm{mass} \mathrm{is} \mathrm{a} \mathrm{property} \mathrm{of} \mathrm{the} \mathrm{gas}. \end{gathered}$$

Thus, (d) is the correct answer.

Question 2:

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.  
Pressure, P = 1.01325$\times$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=$\frac{8.314\times 273}{1.01325\times {10}^5}=0.0224 {\mathrm{m}}^3$

   

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$\times$10²³
 Number of molecules in 22.4​$\times$10³ cm³ of ideal gas at STP = ​6.022$\times$10²³

Now,
Number of molecules in 1 cm³ of ideal gas at STP = $\frac{6.022\times {10}^{23}}{22.4\times {10}^3}=2.688\times {10}^{19}$

Question 4:

Given:
Volume of ideal gas, V = 1 cm³= 10^-6 m^​3
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10⁻⁸ m of Hg
Density of ideal gas, ρ = 13600 kgm^-3
Pressure $\left(P\right)$ is given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get
          
$$\begin{gathered} n=\frac{PV}{RT} \\ \Rightarrow n=\frac{\rho gh\times \mathrm{V}}{RT} \\ \Rightarrow n=\frac{13600\times 9.8\times {10}^{-8}\times {10}^{-6}}{8.31\times 273} \\ \Rightarrow n=5.87\times {10}^{-13} \end{gathered}$$

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

Question 5:

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

$$\begin{gathered} 22.4\times {10}^3 {\mathrm{cm}}^3 \mathrm{of} {\mathrm{O}}_2=1 \mathrm{mol} {\mathrm{O}}_2 \\ 1 {\mathrm{cm}}^3 \mathrm{of} {\mathrm{O}}_2 = \frac{1}{22.4\times {10}^3 } \mathrm{mol} {\mathrm{O}}_2 \\ 1 \mathrm{mol} \mathrm{of} {\mathrm{O}}_2 = 32 \mathrm{g} \\ \frac{1}{22.4\times {10}^3 }\mathrm{mol} \mathrm{of} {\mathrm{O}}_2 =\frac{32}{22.4\times {10}^3 }=1.43\times {10}^{-3} \mathrm{g} \\ = 1.43 \mathrm{mg} \end{gathered}$$

Question 6:

Let the pressure and temperature for the vessels of volume V₀ and 2V₀ be P_​1, T₁ and P₂ , T₂, respectively.
Since the two vessels have the same mass of gas, n₁ = n₂ = n.
$$\begin{gathered} T_1=300 \mathrm{K} \\ {T}_2 = 600 \mathrm{K} \\ \mathrm{Using} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{state} \mathrm{for} \mathrm{perfect} \mathrm{gas}, \mathrm{we} \mathrm{get} \\ PV=nRT \\ \mathrm{For} \mathrm{the} \mathrm{vessel} \mathrm{of} \mathrm{volume} V_{\mathrm{o}}: \\ {P}_1{V}_o=nR{T}_1 ...\left(1\right) \\ \mathrm{For} \mathrm{the} \mathrm{vessel} \mathrm{of} \mathrm{volume} 2V_{\mathrm{o}}: \\ {P}_2\left(2V_{\mathrm{o}}\right)=nR{T}_2 ...\left(2\right) \\ \mathrm{Dividing} \mathrm{eq}. \left(2\right) \mathrm{by} \mathrm{eq}. \left(1\right), \mathrm{we} \mathrm{get} \\ \frac{2P_2}{P_{\mathit{1}}}=\frac{T_{\mathit{2}}}{T_{\mathit{1}}}=\frac{600}{300}=2 \\ \Rightarrow \frac{P_{\mathit{2}}}{P_{\mathit{1}}}=1 \\ \Rightarrow P_{\mathit{2}}\mathit{:}{P}_{\mathit{1}}=1:1 \end{gathered}$$

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
$PV=nRT$

$$\begin{gathered} PV = nRT \\ \Rightarrow n =\frac{PV}{RT} \\ \Rightarrow n=\frac{\rho ghV}{RT} \\ \Rightarrow n =\frac{{10}^{-6}\times 13600\times 10\times 250\times {10}^{-6} }{8.314\times 300} \\ \mathrm{Now}, \mathrm{number} \mathrm{of} \mathrm{molecules} =nN \\ =\frac{{10}^{-6}\times 13600\times 10\times 250\times {10}^{-6} }{8.314\times 300}\times 6\times {10}^{23} \\ =8\times {10}^{15} \end{gathered}$$

Question 8:

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

$$\begin{gathered} \frac{P_{1}V}{T_1}=\frac{P_{2}V}{T_2} (\because {\mathrm{V}}_1={\mathrm{V}}_2) \\ \Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \Rightarrow T_2=\frac{P_2\times T_1}{P_1} \\ \Rightarrow T_2=\frac{1.0\times {10}^6\times 300}{8.0\times {10}^5}=375 \mathrm{K} \end{gathered}$$

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
$$\begin{gathered} \Rightarrow P=\frac{nRT}{V} \\ \Rightarrow P=\frac{1\times 8.3\times 300}{0.02} \end{gathered}$$
$\Rightarrow$P = 1.24 × 10⁵ Pa

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 $\times$ 10⁵ Pa   (At STP)
Temperature, T = 273 K    (At STP)

Using the ideal gas equation, we get

$$\begin{gathered} PV = nRT ...\left(1\right) \\ n = \frac{m}{M} ...\left(2\right) \\ \therefore PV = \frac{m}{M}RT \\ \Rightarrow M=\frac{m}{V}\frac{RT}{P} \\ \Rightarrow M=\rho \frac{RT}{P} \\ \Rightarrow M=1.25\times \frac{8.31\times 273}{{10}^5} \\ \Rightarrow M=2.83\times {10}^{-2} \\ =28.3 \mathrm{g}-{\mathrm{mol}}^{-1} \end{gathered}$$

Question 11:

Here,
Temperature in Simla, T₁= 15 + 273 = 288 K       
Pressure in Simla, P_​1 = 0.72  m of Hg 
Temperature in Kalka, T₂= 35+273 = 308 K
Pressure in Kalka, P_​2 = 0.76  m of Hg 
Let density of air at Simla and Kalka be $\rho$₁ and $\rho$₂, respectively. Then,
$$\begin{gathered} PV=\frac{\mathit{m}}{\mathit{M}}RT \\ \Rightarrow \frac{\mathit{m}}{\mathit{V}}\mathit{=}\frac{\mathit{P}\mathit{M}}{\mathit{R}\mathit{T}} \\ \Rightarrow \rho \mathit{=}\frac{\mathit{P}\mathit{M}}{\mathit{R}\mathit{T}} \end{gathered}$$

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

$$\begin{gathered} \frac{{\rho }_{\mathit{1}}}{{\rho }_2}=\frac{P_1}{T_1}\times \frac{T_2}{P_2} \\ \Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{0.72}{288}\times \frac{308}{0.76} \\ \Rightarrow \frac{{\rho }_2}{{\rho }_{\mathit{1}}}=0.987 \end{gathered}$$

Question 12:

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.
$$\begin{gathered} \mathrm{For} \mathrm{first} \mathrm{part}: \\ \mathrm{Applying} \mathrm{equation} \mathrm{of} \mathrm{state}, \mathrm{we} \mathrm{get} \\ {P}_{1}V=nRT ...\left(1\right) \\ \mathrm{For} second \mathrm{part}: \\ \mathrm{Applying} \mathrm{equation} \mathrm{of} \mathrm{state}, \mathrm{we} \mathrm{get} \\ {P}_2\left(3V\right)=nRT ...\left(2\right) \\ \mathrm{Dividing} \mathrm{eq}. \left(1\right) \mathrm{by} \mathrm{eq}. \left(2\right), \mathrm{we} \mathrm{get} \\ \frac{P_{1}V}{P_2\left(3V\right)}=1 \\ \Rightarrow \frac{P_1}{P_2}=\frac{3}{1} \\ \Rightarrow P_1:P_2=3:1 \end{gathered}$$

Question 13:

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, M_​0 = 2 g/mol=0.002 kg /mol
We know,
$$\begin{gathered} C=\sqrt{\frac{3RT}{M_0}} \\ \Rightarrow C=\sqrt{\frac{3\times 8.3\times 300}{0.002}} \\ \Rightarrow C=1932.6 {\mathrm{ms}}^{-1} \end{gathered}$$

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

$$\begin{gathered} \sqrt{\frac{3\times 8.3T}{0.002}}=2\times 1932.6 \\ \Rightarrow T=1200 \mathrm{K} \end{gathered}$$

Question 14:

Here,
V = 10^-3 m³
Density = 0.177 kgm^-3
P = 10⁵pa   
$$\begin{gathered} C=\sqrt{\frac{3P}{\rho }}=\sqrt{\frac{3\times {10}^5}{0.177}} \\ =1301.9 {\mathrm{ms}}^{-1} \end{gathered}$$

Question 15:

We know from kinetic theory of gases that the average translational energy per molecule is $\frac{3}{2}kT$.
Now,
E_avg= 0.040 eV = $0.040\times 1.6\times {10}^{-19}=6.4\times {10}^{-21}J$ 
$$\begin{gathered} 6.40\times {10}^{-21}=\frac{3}{2}\times 1.38\times {10}^{-23}\times T \\ \Rightarrow T=\frac{2}{3}\times \frac{6.40\times {10}^{-21}}{1.38\times {10}^{-23}}=309.2 K \end{gathered}$$

Question 16:

$$\begin{gathered} \mathrm{Here}, \\ {\mathrm{V}}_{\mathrm{avg}}=\frac{\sqrt{8\mathrm{RT}}}{\sqrt{\mathrm{\pi M}}}=\frac{\sqrt{8\times 8.83\times 300}}{\sqrt{3.14\times 0.032}} \\ =445.25 \mathrm{m}/\mathrm{s} \\ \mathrm{We} \mathrm{know}, \\ \mathrm{T}=\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{6400000\times 2}{445.25} \\ =\frac{28747.83 h}{3600}=7.985 h=8 h \end{gathered}$$

Question 17:

Here,
​m = 6.64 × 10⁻²⁷ kg
T = 273 K

$$\begin{gathered} \mathrm{Average} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{He} \mathrm{atom} \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mathrm{V}}_{\mathrm{avg}}=\sqrt{\frac{8kT}{\pi m}} \\ =\sqrt{\frac{8\times 1.38\times {10}^{-23}\times 273}{3.14\times 6.64\times {10}^{-27}}} \\ =1202.31 \end{gathered}$$

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

Question 18:

Mean velocity is given by

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

Let temperature for H and He respectively be  T_​1 and T₂, respectively.
For hydrogen:
M_H = 2g = 2$\times$10^-3 kg
For helium:
M_He= 4 g = 4$\times$10^-3 kg

Now,
A/q$$\begin{gathered} \sqrt{\frac{8R{T}_1}{\pi M_H}}=\sqrt{\frac{8R{T}_2}{\pi M_{He}}} \\ \Rightarrow \sqrt{\frac{8R{T}_1}{2\times {10}^{-3}\pi }}=\sqrt{\frac{8R{T}_2}{\pi \times 4\times {10}^{-3}}} \\ \Rightarrow \sqrt{\frac{T_1}{2}}=\sqrt{\frac{T_2}{4}} \\ \Rightarrow \frac{T_1}{T_2}=\frac{1}{2} \\ \Rightarrow T_1:T_2=1:2 \end{gathered}$$

Question 19:

Mean speed of the molecule is given by

$$\begin{gathered} \sqrt{\frac{8\mathrm{RT}}{\mathrm{\pi M}}} \\ \mathrm{For} \mathrm{H} \mathrm{molecule}, M =2\times {10}^{-3}kg \\ =\sqrt{\frac{4\mathrm{RT}\times {10}^3}{\pi }} \end{gathered}$$

For escape velocity of Earth:

 $$\begin{gathered} \mathrm{Let} r\mathit{ }\mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{Earth}. \\ v=\sqrt{\frac{2GM}{r}} \\ \mathrm{Multiplying} \mathrm{neumerator} \mathrm{and} \mathrm{denominator} \mathrm{by} \mathrm{R}, \mathrm{we} \mathrm{get} \\ {v}_c=\sqrt{\frac{GM}{r^2}2r} \\ g=\frac{GM}{r^2} \\ {v}_c=\sqrt{2gr} \end{gathered}$$

$$\begin{gathered} \sqrt{\frac{4\mathrm{RT}\times {10}^3}{\pi }}=\sqrt{2gr} \\ \Rightarrow \frac{2\times 8.314\times T\times {10}^3}{3.142}=9.8\times 6.37\times {10}^6 \\ \Rightarrow T\approx 11800 K \end{gathered}$$

Question 20:

We know,

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

Molar mass of H₂ = M_H = 2$\times$10^-3 kg

Molar mass of N₂ = M_N = 28$\times$10^-3 kg
Now,
$$\begin{gathered} <V{>}_H=\sqrt{\frac{8RT}{\pi M_H}} \\ <V{>}_N=\sqrt{\frac{8RT}{\pi M_N}} \\ \frac{<V{>}_H}{<V{>}_N}=\sqrt{\frac{M_N}{M_H}}=\sqrt{\frac{28}{2}}=\sqrt{14}=3.74 \end{gathered}$$

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

$$\begin{gathered} A/Q \\ {v}_{rms}=\sqrt{\frac{3kT}{m_1}}=\sqrt{\frac{8kT}{\mathrm{\pi }{m}_2}} \\ \Rightarrow 3m_1 = \frac{8m_2}{3.14} \\ \Rightarrow \frac{m_2}{m_1}=1.18 \end{gathered}$$