Numerical Problems-1 - Chapter 9 Kinetic theory of Gases 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A fresh air is composed of nitrogen $\mathrm{N}_2(78 \%)$ and oxygen $\mathrm{O}_2(21 \%)$. Find the rms speed of $\mathrm{N}_2$ and $\mathrm{O}_2$ at $20^{\circ} \mathrm{C}$.
Answer:
Absolute temperature $\mathrm{T}=20^{\circ} \mathrm{C}+273=293 \mathrm{~K}$
Gas constant $\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
For, Nitrogen $\left(\mathrm{N}_2\right)$,
Molar mass $(M)=28$ ger $\mathrm{mol}=28 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}$

$
\begin{aligned}
\therefore \quad v_{r m s} & =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \times 8.32 \times 293}{28 \times 10^{-3}}}=\sqrt{\frac{7313.28}{28 \times 10^{-3}}} \\
\left(v_{r m s}\right)_{\mathrm{N}_2} & =511 \mathrm{~ms}^{-1}
\end{aligned}
$
For, Oxygen $\left(\mathrm{O}_2\right)$,
Molar mass $(\mathrm{M})=32 \mathrm{~g}$ per mol $=32 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}$
$
\begin{aligned}
\therefore \quad v_{r m s} & =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \times 8.32 \times 293}{32 \times 10^{-3}}}=\sqrt{\frac{7313.28}{32 \times 10^{-3}}} \\
\left(v_{r m s}\right)_{\mathrm{O}_2} & =478 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 2.
If the rms speed of methane gas in the Jupiter's atmosphere is $471.8 \mathrm{~m} \mathrm{~s}^{-1}$, show that the surface temperature of Jupiter is sub-zero.
Answer:
RMS speed of methane gas $\left(\mathrm{v}_{\mathrm{rms}}\right)=471.8 \mathrm{~ms}^{-1}$
Molar mass of methane gas $(\mathrm{M})=16.04 \mathrm{~g}$ per mol
$
\mathrm{M}=16.04 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}
$
Gas constant $\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
$
\begin{aligned}
v_{r m s} & =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\
\left(v_{r m s}\right)^2 & =\frac{3 \mathrm{RT}}{\mathrm{M}} \\
\mathrm{T} & =\frac{\left(v_{r m s}\right)^2 \times \mathrm{M}}{3 \mathrm{R}}=\frac{(471.8)^2 \times 16.04 \times 10^{-3}}{3 \times 8.31} \\
& =\frac{3.57 \times 10^6 \times 10^{-3}}{24.93}=0.143 \times 10^3 \\
\mathrm{~T} & =143 \mathrm{~K}-273 \\
\mathrm{~T} & =-130^{\circ} \mathrm{C}
\end{aligned}
$
Question 3.
Calculate the temperature at which the rms velocity of a gas triples its value at S.T.P. (Standard temperature $\mathrm{T}_1=273 \mathrm{~K}$ )
Answer:

At STP temperature $\mathrm{T}_1=273 \mathrm{~K}$
RMS velocity of a gas, $\left(v_{r m s}\right)_1=v$
New RMS velocity of a gas, at temperature $\left(\mathrm{T}_2\right)$
$
\left(v_{r m s}\right)_2=3 v
$
New temperature $\mathrm{T}_2=$ ?
$
\begin{aligned}
v_{r m s} & =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\
\frac{\left(v_{r m s}\right)_1}{\left(v_{r m s}\right)_2} & =\sqrt{\frac{\mathrm{T}}{\mathrm{T}_2}} \\
\mathrm{~T}_2 & =\left(\frac{\left(v_{r m s}\right)_2}{\left(v_{r m s}\right)_1}\right)^2 \times \mathrm{T}_1=\left(\frac{3 v}{v}\right)^2 \times 273 \\
\mathrm{~T}_2 & =9 \times 273=2457 \mathrm{~K}
\end{aligned}
$
Question 4.
A gas is at temperature $80^{\circ} \mathrm{C}$ and pressure $5 \times 10^{-10} \mathrm{Nm}^{-2}$. What is the number of molecules per $\mathrm{m}^3$ if Boltzmann's constant is $1.38 \times 10^{-23} \mathrm{JK}^{-1}$.
Answer:
Temperature of a gas $(\mathrm{T})=80^{\circ} \mathrm{C}+273=353 \mathrm{~K}$
Pressure of a gas $(\mathrm{P})=5 \times 10^{-10} \mathrm{Nm}^{-2}$
Boltzmann's constant $\left(\mathrm{K}_{\mathrm{B}}\right)=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}$
$
V=1 m^3
$
Number of molecules, $n=\frac{\mathrm{PV}}{k \mathrm{~T}}=\frac{5 \times 10^{-10} \times 1}{1.38 \times 10^{-23} \times 353}$
$
\begin{aligned}
& =\frac{5 \times 10^{-10}}{487.14 \times 10^{-23}}=0.01026 \times 10^{13} \\
& n=1.02 \times 10^{11}
\end{aligned}
$
Question 5.
From kinetic theory of gases, show that Moon cannot have an atmosphere (Assume $\mathrm{k}=1.38 \times 10^{-}$ $23 \mathrm{~J} \mathrm{~K}^{-1}$ Temperature $\left.\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}\right)$

Answer:
At absolute temperature $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$
Boltzmann's Constant $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}$
$m=3.33 \times 10^{-27} \mathrm{~kg}$ (mass of hydrogen molecule)
$v_{r m s}=\sqrt{\frac{3 k_{\mathrm{B}} \mathrm{T}}{m}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{3.33 \times 10^{-27}}}$
$=\sqrt{\frac{1130.22 \times 10^{-23}}{3.33 \times 10^{-27}}}=1842.11$
$v_{r m s}=1.84 \times 10^3 \mathrm{~ms}^{-1}$
Question 6.
If $10^{20}$ oxygen molecules per second strike $4 \mathrm{~cm}^2$ of wall at an angle of $30^{\circ}$ with the normal when moving at a speed of $2 \times 10^3 \mathrm{~ms}^{-1}$, find the pressure exerted on the wall.
(mass of 1 atom $=2.67 \times 10^{-26} \mathrm{~kg}$ ).
Answer:
Mass of $\mathrm{O}_2$ atom $=16 \times$ mass of 1 atom
$=16 \times 10^{20} \times 2.67 \times 10^{-26} ; \mathrm{m}=42.72 \times 10^{-6} \mathrm{~kg}$
Momentum of the $\mathrm{O}_2$ molecule $(\mathrm{P})=m v=42.72 \times 10^{-6} \times 2 \times 10^3$
$
\mathrm{P}=85.44 \times 10^{-3} \mathrm{~kg} \mathrm{~ms}^{-1}
$
Momentum normal to the wall at angle $30^{\circ}$
Question 7.
During an adiabatic process, the pressure of a mixture of monoatomic and diatomic gases is found to be proportional to the cube of the temperature. Find the value of $\gamma=\left(C_p / C_V\right)$
Answer:

In adiabatic process, $\quad \mathrm{T}^\gamma \mathrm{P}^{1-\gamma}=$ constant
$
\begin{aligned}
& \mathrm{P} \propto \mathrm{T}^3 \\
& \mathrm{PT}^{-3}=\text { constant } \\
& \mathrm{PT}^{\frac{\gamma}{1-\gamma}}=\text { constant } \\
& \mathrm{PT}^{\frac{\gamma}{1-\gamma}}=\mathrm{PT}^{-3} \\
&
\end{aligned}
$
Comparing the powers, $\frac{\gamma}{1-\gamma}=-3=-3+3 \gamma$
$
2 \gamma=3 ; \gamma=\frac{3}{2}
$
Question 8.
Calculate the mean free path of air molecules at STP. The diameter of $\mathrm{N}_2$ and $\mathrm{O}_2$ is about $3 \times 10^{-10}$ $\mathrm{m}$. Answer:
$
\mathrm{P}=1 \mathrm{~atm}=1.01 \times 10^5 \mathrm{~Pa}, k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}, \mathrm{~T}=273 \mathrm{~K}
$
From ideal gas law, $n=\frac{\mathrm{P}}{k \mathrm{~T}}$
$
\begin{aligned}
& n=\frac{1.01 \times 10^5}{1.38 \times 10^{-23} \times 273}=\frac{1.01 \times 10^5 \times 10^{23}}{376.74}=2.68 \times 10^{-3} \times 10^{28} \\
& n=2.68 \times 10^{25} \text { molecules } / \mathrm{m}^3
\end{aligned}
$
Mean free path of the air molecule,
$
\begin{aligned}
\lambda & =\frac{1}{\sqrt{2} \pi n d^2}=\frac{1}{1.414 \times 3.14 \times 2.68 \times 10^{25} \times\left(3 \times 10^{-10}\right)^2} \\
& =\frac{1}{1.0709 \times 10^{-18} \times 10^{25}}=0.9338 \times 10^{-7} \\
\lambda & =9.3 \times 10^{-8} \mathrm{~m}
\end{aligned}
$
Question 9.
A gas made of a mixture of 2 moles of oxygen and 4 moles of argon at temperature $T$. Calculate the energy of the gas in terms of RT. Neglect the vibrational modes.
Answer:
For two moles of diatomic nitrogen with no vibrational mode,
$
U_1=2 \times \frac{5}{2} \mathrm{RT}=5 \mathrm{RT}
$
For four mole of monatomic argon,
$
\mathrm{U}_2=4 \times \frac{3}{2} \mathrm{RT}=6 \mathrm{RT}
$
Total energy of the gas, $\mathrm{U}=\mathrm{U}_1+\mathrm{U}_2=5 \mathrm{RT}+6 \mathrm{RT}$ $\mathrm{U}=11 \mathrm{RT}$
Question 10.
Estimate the total number of air molecules in a room of capacity $25 \mathrm{~m}^3$ at a temperature of $27^{\circ} \mathrm{C}$.

Answer:
$
\mathrm{T}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}, k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}, \mathrm{~V}=25 \mathrm{~m}^3
$
As Boltzmann's Constant, $k_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}} \Rightarrow \mathrm{R}=k_{\mathrm{B}} \mathrm{N}$
Now, $\quad \mathrm{PV}=n \mathrm{RT}=n k_{\mathrm{B}} \mathrm{NT}$
The number of molecules in the room,
$
\begin{aligned}
n \mathrm{~N} & =\frac{\mathrm{PV}}{k_{\mathrm{B}} \mathrm{T}}=\frac{1.013 \times 10^5 \times 25}{1.38 \times 10^{-23} \times 300} \\
& =\frac{25.325 \times 10^5}{414 \times 10^{-23}}=0.06117 \times 10^{28} \\
n \mathrm{~N} & =6.117 \times 10^{26} \text { molecules }
\end{aligned}
$

Multiple Choice Questions
I. Choose the correct answer from the following:
Question 1.
Oxygen and hydrogen gases are at the same temperature the ratio of the average $K . E$ of an oxygen molecule and that of a hydrogen molecule is
(a) 16
(b) 4
(c) 1
(d) $\frac{1}{4}$
Answer:
(c) 1
Solution:
$\mathrm{P}_1 V_1=\mathrm{P}_2 V_2$, if $T$ is constant.
Question 2 .
According to the kinetic theory of gases,
(a) the pressure of a gas is proportional to the rms speed of the molecules.
(b) the rms speed of the molecules of a gas is proportional to the absolute temperature.
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute
temperature.
(d) the pressure of a gas is proportional to the square root of the rms speed of the molecules.

Answer:
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
Solution:
The rms speed of the molecules of a gas is proportional to the square root of the absolute temperature. .

Question 3.
Pressure exerted by a perfect gas is equal to
(a) mean K.E. per unit volume.
(b) half of mean K.E. per unit volume.
(c) one-third of mean K.E. per unit volume.
(d) two-third of mean K.E. per unit volume.
Answer:
(d) two-third of mean K.E. per unit volume.
Solution: $P=\frac{2}{3} \mathrm{~K}$.E. (average $\mathrm{K}$.E. of the gas per unit volume).
Question 4.
The temperature of an ideal gas is increased from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The root mean square speed of its molecules becomes.
(a) 3 times
(b) double
(c) 4 times
(d) 6 times
Answer:
(b) double
1200
Solution:
From RMS speed $v_{r m s}=\sqrt{\frac{3 k \mathrm{~T}}{\mathrm{~m}}}$
$
\begin{aligned}
\frac{v_2}{v_1} & =\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{273+927}{273+27}}=\sqrt{\frac{1200}{300}}=2 \\
v_2 & =2 v_1
\end{aligned}
$
Question 5.
Two gases are enclosed in a container at constant temperature. One of the gases, which is diatomic, has relative molecular mass eight times the other, which is monoatomic. The ratio pf the rms speed of the molecules of the monoatomic gas to that of the molecules of the diatomic gas is
(a) 8
(b) 4
(c) $2 \sqrt{2}$
(d) 2

Answer:
(c) $2 \sqrt{2}$
Solution:
The $r m s$ speed is independent of the atomicity of the gas. $v=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$
$
\frac{v_{\text {mono }}}{v_{d i}}=\sqrt{\frac{\mathrm{M}_{d i}}{\mathrm{M}_{\text {mono }}}}=\sqrt{8}=2 \sqrt{2}
$
Question 6.
If the absolute temperature of a gas is increased 3 times the rms velocity of the molecules will be
(a) 3 times
(b) 9 times
(c) 73 times
(d) 76 times
Answer:
(c) 73 times
Question 7.
At a given temperature which of the following gases possesses maximum rms velocity of molecules?
(a) $\mathrm{H}_2$
(b) $\mathrm{O}_2$
(c) $\mathrm{N}_2$
(d) $\mathrm{CO}_2$
Answer:
(a) $\mathrm{H}_2$
Solution:
$v_{r m s} \propto \frac{1}{\mathrm{M}}$ Since $\mathrm{M}$ is minimum for $\mathrm{H}_2, v_{r m s}$ is maximum for it.
Question 8.
Two vessels have equal volumes. One of them contains hydrogen at one atmosphere and the other helium at two atmospheres. If both the samples are at the same temperature, the rms velocity of the hydrogen molecules is .....
(a) equal to that of the helium molecules
(b) twice that of the helium molecules
(c) half that of the helium molecules
(d) $\sqrt{2}$ times that of the helium molecules
Answer:
(d) $\sqrt{2}$ times that of the helium molecules

Solution:
$
\frac{v_{\mathrm{H}_2}}{v_{\mathrm{He}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{He}}}{\mathrm{M}_{\mathrm{H}_2}}}=\sqrt{2}
$
Question 9.
A gas is enclosed in a container which is then placed on a fast moving train. The temperature of the gas.$\ldots$
(a) rises
(b) remains unchanged
(c) falls
(d) becomes unsteady
Answer:
(c) falls
Question 10.
The mean translational K.E. of a perfect gas molecule at absolute temperature $\mathrm{T}$ is ( $\mathrm{K}$ is Boltzmann constant)
(a) $\frac{1}{2} k \mathrm{~T}$
(b) $k \mathrm{~T}$
(c) $\frac{3}{2} k \mathrm{~T}$
(d) $\frac{5}{2} k \mathrm{~T}$
Answer:
(c) $\frac{3}{2} k \mathrm{~T}$
Question 11.
Ajar has mixture of hydrogen and Oxygen gases in the ratio $1: 5$. The ratio of mean kinetic energies of hydrogen and Oxygen molecules is .......
(a) $1: 5$
(b) $5: 1$
(c) $1: 1$
(d) $1: 25$
Answer:
(c) $1: 1$
Solution:
The mean kinetic energy depends only on the temperature.
Question 12.
The pressure exerted on the walls of the container by a gas is due to the fact that the gas molecules
(a) lose their K.E
(b) Stick to the walls
(c) are accelerated towards the walls
(d) change their momenta due to collision with the walls.
Answer:

(d) change their momenta due to collision with the walls.
Question 13.
Pressure exerted by a gas is
(a) independent of the density of the gas
(b) inversely proportional to the density of the gas
(c) directly proportional to the density of the gas
(d) directly proportional to the square of the density of the gas.
Answer:
(c) directly proportional to the density of the gas
Question 14.
Four molecules have speed $2 \mathrm{~km} / \mathrm{s}, 3 \mathrm{~km} / \mathrm{s}, 4 \mathrm{~km} / \mathrm{s}$ and $5 \mathrm{~km} / \mathrm{s}$. The rms speed of these molecules in $\mathrm{km} / \mathrm{s}$ is ......
(a) $\sqrt{\frac{27}{2}}$
(b) $\sqrt{27}$
(c) $2 \sqrt{27}$
$(d) \sqrt{54}$
Answer:
(a) $\sqrt{\frac{27}{2}}$
Solution:
$
v_{r m s}=\sqrt{\frac{v_1^2+v_2^2+\ldots .+v_n^2}{n}}=\sqrt{\frac{2^2+3^2+4^2+5^2}{4}}=\sqrt{\frac{27}{2}}
$
Question 15.
A real gas behaves as an ideal gas at ......
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(a) low pressure and high temperature

Question 16.
The kinetic theory of gases breaks down most at
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(b) high pressure and low temperature
Question 17.
Two different ideal gases are enclosed in two different vessels at the same pressure. If $\rho_1$ and $\rho_2$ are their densities and $v_1$ and $v_2$ their rms speeds, respectively then is equal to
(a) $\frac{\rho_1^2}{\rho_2^2}$
(b) $\frac{\rho_2^2}{\rho_1^2}$
(c) $\sqrt{\frac{\rho_1}{\rho_2}}$
(d) $\sqrt{\frac{\rho_2}{\rho_1}}$
Answer:
(a) $\sqrt{\frac{\rho_2}{\rho_1}}$
Solution:
(d) $\sqrt{\frac{\rho_2}{\rho_1}}$
Question 18 .
A cylinder of capacity 20 litres is filled with hydrogen gas. The total average K.E. of translatory motion of its molecules is $1.5 \times 10^5 \mathrm{~J}$. The pressure of hydrogen in the cylinder is
(a) $2 \times 10^6 \mathrm{Nm}^{-2}$
(b) $3 \times 10^6 \mathrm{Nm}^{-2}$
(c) $4 \times 10^6 \mathrm{Nm}^{-2}$
(d) $5 \times 10^6 \mathrm{Nm}^{-2}$
(d) $5 \times 10^6 \mathrm{Nm}^{-2}$
Solution:
$
P=\frac{2}{3} \frac{E}{V}=\frac{2}{3}\left(\frac{1.5 \times 10^5}{20 \times 10^{-3}}\right)=5 \times 10^6 \mathrm{Nm}^{-2}
$

Question 19.
The molecular weights of oxygen and hydrogen are 32 and 2 , respectively. The rms velocities of their molecules at a given temperatures, will be in the ratio
(a) $4: 1$
(b) $1: 4$
(e) $1: 16$
(6) $16: 1$
Answer:
(b) $1: 4$
Solution:
$
\frac{\left(v_{r m s}\right)_{\mathrm{O}_2}}{\left(v_{r m s}\right)_{\mathrm{H}_2}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{O}_2}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}
$
Question 20.
The average energy of a molecules of a monoatomic gas at temperature $\mathrm{T}$ is ( $\mathrm{K}$ Boltzmann constant)
(a) $\frac{1}{2} \mathrm{kT}$
(b) $\mathrm{kT}$
(c) $\frac{3}{2} \mathrm{kT}$
(d) $\frac{5}{2} \mathrm{kT}$
Answer:
(c) $\frac{3}{2} \mathrm{kT}$
Question 21.
The temperature at which the molecules of nitrogen will have the same rms velocity as the molecules of oxygen at $127^{\circ} \mathrm{C}$ is
(a) $77^{\circ} \mathrm{C}$
(b) $350^{\circ} \mathrm{C}$
(c) $273^{\circ} \mathrm{C}$
(d) $457^{\circ} \mathrm{C}$
Answer:

(a) $77^{\circ} \mathrm{C}$
Solution:
Let the required temperature be $T$. Then
$
\frac{3 \mathrm{RT}}{28}=\frac{3 \mathrm{R}(273+127)}{32} \Rightarrow \mathrm{T}=350 \mathrm{~K}=77^{\circ} \mathrm{C}
$
Question 22.
The temperature of a gas is raised from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The root mean square speed of its molecules
(a) becomes $\sqrt{\frac{927}{27}}$ times the earlier value
(b) gets halved
(c) remains the same
(d) gets doubled
Answer:
(d) gets doubled
Solution:
$
\frac{v_2}{v_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{273+927}{273+27}}=2
$
Question 23.
The temperature at which the $\mathrm{K} . \mathrm{E}$ of a gas molecules is double its value at $27^{\circ} \mathrm{C}$ is
(a) $54^{\circ} \mathrm{C}$
(b) $300 \mathrm{~K}$
(c) $327^{\circ} \mathrm{C}$
(d) $108^{\circ} \mathrm{C}$
Answer:
(c) $327^{\circ} \mathrm{C}$
Solution:
$
\mathrm{E}=\frac{3}{2} k \mathrm{~T} ; \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{E}_2}{\mathrm{E}_1} \Rightarrow \mathrm{T}_2=\left(\frac{\mathrm{E}_2}{\mathrm{E}_1}\right) \mathrm{T}_1=\left(\frac{2}{1}\right) \times(273+27)=600 \mathrm{~K}=327^{\circ} \mathrm{C}
$
Question 24.
The temperature of an ideal gas is increased from $120 \mathrm{~K}$ to $480 \mathrm{~K}$. If at $120 \mathrm{~K}$ the root mean square velocity of the gas molecules is $\mathrm{v}$, at $480 \mathrm{~K}$ it becomes
(a) $4 \mathrm{v}$
(b) $2 \mathrm{v}$
(c) $\frac{v}{2}$

(d) $\frac{v}{4}$
Answer:
(b) $2 \mathrm{v}$
Solution:
$
v_{r m s}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}
$
Let the required velocity be $v^{\prime}$. Then $\frac{v^{\prime}}{v}=\sqrt{\frac{480}{120}}=2 ; v^{\prime}=2 v$
Question 25.
The average translational K.E. of $\mathrm{O}_2$ (molar mass 32) molecules at a particular temperature is 0.048 $\mathrm{eV}$. The translational K.E. of $\mathrm{N}_2$ (molar mass 28) molecules in $\mathrm{eV}$ at the same temperature is
(a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
Answer:
(c) 0.048
Solution:
Average translational K.E. of a gas molecule $=\frac{3}{2} \mathrm{kT}$.
It is independent of molecular mass.
Question 26.
$
\left(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)
$
The K.E. of one mole of a gas at normal temperature and pressure is
(a) $0.56 \times 10^4 \mathrm{~J}$
(b) $1.3 \times .10^2 \mathrm{~J}$
(c) $2.7 \times 10^2 \mathrm{~J}$
(d) $3.4 \times 10^3 \mathrm{~J}$
Answer:
(d) $3.4 \times 10^3 \mathrm{~J}$
Solution:
$
\text { K.E. }=\frac{3}{2} \mathrm{RT}=\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^3 \mathrm{~J}
$
Question 27.
The average K.E. of a hydrogen gas molecule at STP will be (Boltzmann constant $\mathrm{K}_{\mathrm{B}}=1.38 \times 10^{-}$ $\left.23 \mathrm{JK}^{-1}\right)$
(a) $0.186 \times 10^{-28} \mathrm{~J}$

(b) $0.372 \times 10^{-20} \mathrm{~J}$
(c) $0.56 \times 10^{-20} \mathrm{~J}$
(d) $5.6 \times 10^{-20} \mathrm{~J}$
Answer:
(c) $0.56 \times 10^{-20} \mathrm{~J}$
Solution:
$
\mathrm{E}=\frac{3}{2} k \mathrm{~T}=\frac{3}{2} \times 1.38 \times 10^{-23} \times 273=0.56 \times 10^{-20} \mathrm{~J}
$
Question 28.
The rms speed of the particles of fume of mass $5 \times 10^{-17} \mathrm{~kg}$ executing Brownian motion in air at STP is
(a) $1.5 \mathrm{~ms}^{-1}$
(b) $3.0 \mathrm{~ms}^{-1}$
(c) $1.5 \mathrm{~cm} \mathrm{~s}^{-1}$
(d) $3.0 \mathrm{~cm} \mathrm{~s}^{-1}$
Answer:
(c) $1.5 \mathrm{~cm} \mathrm{~s}^{-1}$
Solution:
$
\begin{aligned}
& v_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \times 8.3 \times 273}{5 \times 10^{-17} \times 6.02 \times 10^{23}}}=15 \times 10^{-3} \mathrm{~ms}^{-1} \\
& v_{r m s}=1.5 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
$
Question 29.
To what temperature should the hydrogen at room temperature $\left(27^{\circ} \mathrm{C}\right)$ be heated at constant pressure so that the RMS velocity of its molecules becomes double its previous value?
(a) $1200^{\circ} \mathrm{C}$
(b) $927^{\circ} \mathrm{C}$
(c) $600^{\circ} \mathrm{C}$

(d) $108^{\circ} \mathrm{C}$
Answer:
(b) $927^{\circ} \mathrm{C}$
Solution:
$
v_{r m s}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \frac{v_{r m s}}{v_{r m s}}=\sqrt{\frac{\mathrm{T}^{\prime}}{\mathrm{T}}} \Rightarrow \frac{273+t^{\prime}}{273+27}=(2)^2=4 ; t^{\prime}=927^{\circ} \mathrm{C}
$
Question 30
A vessel contains oxygen at $400 \mathrm{~K}$. Another similar vessel contains an equal mass of hydrogen at $300 \mathrm{~K}$. The ratio of the rms speeds of molecules of hydrogen and oxygen is
(a) $\frac{4}{3}$
(b) $\frac{3}{4}$
(c) $3 \sqrt{2}$
(d) $2 \sqrt{3}$
Answer:
(d) $2 \sqrt{3}$
Solution:
$
v_{r m s}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} ; \frac{\left(v_{r m s}\right)_{\mathrm{H}_2}}{\left(v_{r m s}\right)_{\mathrm{O}_2}}=\sqrt{\frac{300}{2} \times \frac{32}{400}}=2 \sqrt{3}
$
Question 31 .
A chamber contains a mixture of helium gas $(\mathrm{He})$ and hydrogen gas $\left(\mathrm{H}_2\right)$. The ratio of the rootmean-square speeds of the molecules of $\mathrm{He}$ and $\mathrm{H}_2$ is .......
(a) 2
(b) $\sqrt{2}$
(c) $\frac{1}{\sqrt{2}}$
(d) $\frac{1}{2}$
Answer:
(c) $\frac{1}{\sqrt{2}}$

Solution:
$
v_{r m s}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} ; \frac{v_{\mathrm{He}}}{v_{\mathrm{H}_2}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{He}}}}=\frac{1}{\sqrt{2}}
$
Question 32 .
On colliding with the walls in a closed container, the ideal gas molecules.
(a) transfer momentum to the walls
(b) lose momentum completely
(c) move with smaller speeds
(d) perform Brownian motion.
Answer:
(a) transfer momentum to the walls
Question 33.
The speeds of 5 molecules of a gas (in arbitrary units) are as follows: 2, 3, 4, 5, 6 The root mean square speed for these molecule is
(a) 2.91
(b) 3.52
(c) 4.00
(d) 4.24
Answer:
(d) 4.24
Solution:
$
v_{r m s}=\sqrt{\frac{v_1^2+v_2^2+v_3^2+\ldots+v_n^2}{n}} \Rightarrow 4.24 \text { units }
$
Question 34 .
At absolute zero temperature, the K.E. of the molecules becomes

(a) zero
(b) maximum
(c) minimum
(d) none of these
Answer:
(a) zero
Question 35.
If the rms speed of the molecules of a gas is $1000 \mathrm{~ms}^{-1}$ the average speed of the molecule is
(a) $1000 \mathrm{~ms}^{-1}$
(b) $922 \mathrm{~ms}^{-1}$
(c) $780 \mathrm{~ms}^{-1}$
(d) $849 \mathrm{~ms}^{-1}$
Answer:
(b) $922 \mathrm{~ms}^{-1}$
$
\begin{aligned}
v_{r m s} & =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} ; v_{a v}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}} \\
\frac{v_{a v}}{v_{r m s}} & =\sqrt{\frac{8}{3 \pi}}=0.9216 \\
v_{a v} & =0.9216 \times v_{r m s}=921.6 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 36.
The gas having average molecular speed four times that of $\mathrm{SO}_2$ (molecular mass 64) is
(a) $\mathrm{He}$ (molecular mass 4)
(b) $\mathrm{O}_2$ (molecular mass 32)
(c) $\mathrm{H}_2$ (molecular mass 2)
(d) $\mathrm{CH}_4$ (molecular mass 16)
Answer:
(a) $\mathrm{He}$ (molecular mass 4)
Solution:
Average speed $\propto \frac{1}{\sqrt{\text { Molecular mass }}}$

2 - Marks Questions
Question 1.
State Avogadro's law.
Answer:
It states that equal volumes of all gases under similar conditions of temperature and pressure, contain equal number of molecules.
Question 2.
Define root mean square speed $\left(\mathrm{v}_{\mathrm{rms}}\right)$. Write down its equations.
Answer:
Root mean square speed is defined as the square root of the mean of the square of speeds of all molecules. It is denoted by $v_{r m s}=\sqrt{\overline{v^2}}$
Mean square speed $\overline{v^2}=\frac{3 k \mathrm{~T}}{m}$
$
\therefore v_{r m s}=\sqrt{\frac{3 k \mathrm{~T}}{m}}=1.73 \sqrt{\frac{k \mathrm{~T}}{m}}
$
Question 3.
Define Avogadro's number.
Answer:
It is defined as the number of particles present in one mole of the substance. It is denoted by $\mathrm{N}_{\mathrm{A}}$
$
\mathrm{N}_{\mathrm{A}}=6.0225 \times 10^{23} \mathrm{~mol}^{-1}
$
Question 4.
Define Average speed. Write it equation.
Answer:
Average speed is defined as the mean (or) average of all the speeds of molecules.

$
\bar{v}=\frac{v_1+v_2+v_3+\ldots+v_n}{n}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}=\sqrt{\frac{8 k \mathrm{~T}}{\pi m}} \Rightarrow \bar{v}=1.60 \sqrt{\frac{k \mathrm{~T}}{m}}
$
Question 5.
Define most probable speed of the gas. Write its expressions.
Answer:
It is defined as the speed acquired by most of the molecules of the gas.
$
\mathrm{V}_{m p}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{2 k \mathrm{~T}}{m}}=1.41 \sqrt{\frac{k \mathrm{~T}}{m}}
$
Question 6.
Write down the comparison of $v_{r m s}, \bar{v}$ and $v_{m p}$.
Answer:
Comparison of $v_{r m s}, \bar{v}$ and $v_{m p}$ : Among the speeds $v_{r m s}$ is the largest and $v_{m p}$ is the least
$
v_{r m s}>\bar{v}>v_{m p}
$
Ratio-wise, $v_{r m s}: \bar{v}: v_{m p}=\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}=1.732: 1.6: 1.414$

Question 7.
What is the reason "No hydrogen in Earth's atmosphere"?
Answer:
As the root mean square speed of hydrogen is much less than that of nitrogen, it easily escapes from the earth's atmosphere.
In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters many disastrous consequences.
Question 8 .
What is Brownian motion?
Answer:
The motion of the particles in a random and zig-zag mannar in a fluid is called Brownian motion.
Question 9.
What does the universal gas constant $R$ signify? Give its value.
Answer:
The universal gas constant $\mathrm{R}$ signifies the workdone by (or on) a gas per mole per kelvin. Its value is $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Question 10.
What is Boltzmann's constant? Give its value.
Answer:
Boltzmann's constant is defined as the gas constant per molecule.
$
k_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}_{\mathrm{A}}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}
$
Question 11.
When do the real gases obey more correctly the gas equation: $\mathrm{PV}=\mathrm{nRT}$ ?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas. At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.