(3)

Polyatomic molecule has 3 translational, 3 rotational and vibrational modes.
According to law of equipartition of energy,

$U=(\frac{{\displaystyle 3RT}}{{\displaystyle 2}}+\frac{{\displaystyle 3RT}}{{\displaystyle 2}}+fRT)$

$U=3RT+fRT$

$C_V=\frac{du}{dt}=3R+fR \Rightarrow f=\frac{C_V-3R}{R}=\frac{C_V}{R}-3$

As,  $C_V=\frac{R}{\gamma -1}$

$\therefore$   $f=\frac{R}{(\gamma -1)R}-3\Rightarrow \frac{1}{\gamma -1}-3=\frac{4-3\gamma }{\gamma -1}$

So, the correct option is 3.

(4)

The rms velocity is, $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

where, $R$ is a gas constant, $M$ = molecular mass, $T$ = absolute temperature. So, $A\to Q$

Pressure exerted by ideal gas is $\frac{1}{3}mn{\overline{v}}^2,$ where $m$ is mass of each molecule, $n$ = number of molecules, ${\overline{v}}^2$ = rms speed. So, $B\to P.$

Average kinetic energy of a molecule $\frac{3}{2}{k}_{B}T$

where, $k_B$ = Boltzmann’s constant, $T$ = absolute temperature. So, $C\to S.$

Total internal energy of 1 mole of a diatomic gas, $U=\frac{5}{2}RT$

So, $D\to R$.

(2)

For mono-atomic gas, degree of freedom = 3

Energy associated with each degree of freedom $=\frac{1}{2}{k}_{B}T$

So, energy is $\frac{3}{2}{k}_{B}T.$