Kinetic Theory of Gases MCQ Questions & Answers in Heat and Thermodynamics | Physics

Given
$$\eqalign{ & {C_P} - {C_V} = 5000\,J/mol{e^ \circ }C\,.......\left( {\text{i}} \right) \cr & \frac{{{C_P}}}{{{C_V}}} = 1.6\,......\left( {{\text{ii}}} \right) \cr} $$
From Equation (i) & (ii),
$$\eqalign{ & \Rightarrow \frac{{{C_P}}}{{{C_V}}} - \frac{{{C_V}}}{{{C_V}}} = \frac{{5000}}{{{C_V}}} \Rightarrow 1.6 - 1 = \frac{{5000}}{{{C_V}}} \cr & \Rightarrow {C_V} = \frac{{5000}}{{0.6}} = 8.33 \times {10^3} \cr & {\text{Hence}}\,\,{C_P} = 1.6,{C_V} = 1.6 \times 8.33 \times {10^3} \cr & = 1.33 \times {10^4} \cr} $$

From graph, $${T^2}V = {\text{const}}.\,\,......\left( {\text{i}} \right)$$
As we know that $$T{V^{\gamma - 1}} = {\text{const}}$$
$$ \Rightarrow V{T^{\frac{1}{{\gamma - 1}}}} = {\text{cons}}{\text{.}}\,......\left( {{\text{ii}}} \right)$$
On comparing (1) and (2), we get
$$\eqalign{ & \Rightarrow \gamma = \frac{3}{2} \cr & {\text{Also}}\,\,{v_{rms}} = \sqrt {\frac{{3P}}{\rho }} \,{\text{and}}\,{v_{{\text{sound}}}} = \sqrt {\frac{{P\gamma }}{\rho }} \cr} $$
$$ \Rightarrow \frac{{{v_{rms}}}}{{{v_{{\text{sound}}}}}} = \sqrt {\frac{3}{\gamma }} = \sqrt 2 $$

From $$PV = nRT$$
$${P_A} = \frac{{{\rho _A}{M_A}}}{{RT}}\,\,{\text{and}}\,\,{P_B} = \frac{{{\rho _B}{M_B}}}{{RT}}$$
From question,
$$\eqalign{ & \frac{{{P_A}}}{{{P_B}}} = \frac{{{\rho _A}}}{{{\rho _B}}}\frac{{{M_A}}}{{{M_B}}} = 2\frac{{{M_A}}}{{{M_B}}} = \frac{3}{2} \cr & {\text{So,}}\,\,\frac{{{M_A}}}{{{M_B}}} = \frac{3}{4} \cr} $$

$${v_{rms}} \propto \frac{1}{{\sqrt M }} \Rightarrow {\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}$$
also in mixture temperature of each gas will be same, hence kinetic energy also remains same.

Figure shows the particles each moving with same speed $$v$$ but the different directions. Consider any two particles having angle $$\theta $$ between directions of their velocities.
Then, $$\overrightarrow {{v_{rel}}} = \overrightarrow {{v_B}} - \overrightarrow {{v_A}} $$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.,}}\,\,{v_{rel}} = \sqrt {{v^2} + {v^2} - 2vv\cos \theta } \cr & \Rightarrow {v_{rel}} = \sqrt {2{v^2}\left( {1 - \cos \theta } \right)} = 2v\sin \left( {\frac{\theta }{2}} \right) \cr} $$
So averaging $${v_{rel}}$$ over all pairs
$$\eqalign{ & {{\bar v}_{rel}} = \frac{{\int_0^{2\pi } {{v_{rel}}} d\theta }}{{\int_0^{2\pi } d \theta }} = \frac{{\int_0^{2\pi } 2 v\sin \left( {\frac{\theta }{2}} \right)}}{{\int_0^{2\pi } d \theta }} \cr & = \frac{{2v \times 2\left[ { - \cos \left( {\frac{\theta }{2}} \right.} \right]_0^{2\pi }}}{{2\pi }} \cr & \Rightarrow {{\bar v}_{rel}} = \left( {\frac{{4v}}{\pi }} \right) > v\left[ {{\text{as}}\,\frac{4}{\pi } > 1} \right] \cr} $$

Total Kinetic Energy $$ = U = \frac{f}{2}nRT$$
In case of $${H_2}$$ degree of freedom is greatest and number of moles $$n$$ is highest.
So this is the case of maximum kinetic energy.

For an ideal gas $$PV = nRT$$
⇒ Co - efficient of volume expansion
$${\left( {\frac{{\Delta V}}{{\Delta T}}} \right)_P} = \frac{{nR}}{P} = {\text{Constant}}$$
Note : Average translation K.E. for $${O_2}$$ is $$\frac{3}{2}kT$$
(Three degrees of freedom for translational motion).
Now decrease in pressure increases the volume.
⇒ It increases mean free path of the molecules. Also average K.E. does not depend on the gas, so molecules of each component of mixture of gases have same average translational energy.

$$\eqalign{ & {C_P} = \frac{7}{2}R;{C_V} = {C_P} - R = \frac{7}{2}R - R = \frac{5}{2}R \cr & \frac{{{C_P}}}{{{C_V}}} = \frac{{\frac{7}{2}R}}{{\frac{5}{2}R}} = \frac{7}{5} \cr} $$

As the temperature increases, the average velocity increases. So, the collisions are faster.

The internal energy of a monatomic gas is due to linear motion only and that of the diatomic gas is due to rolling (Linear rotatory) motion.