$\mathrm{O}_2$ is a diatomic gas having 5 degrees of freedom.

Energy (total internal) per mole of the gas $=\frac{5}{2} R T \quad\left[\begin{array}{l}R=\text { Universal gas constant } \\ T=\text { temperature }\end{array}\right]$

For 2 moles of the gas total internal energy $=2 \times \frac{5}{2} R T=5 R T\quad \text{... (i)}$

Neon $(\mathrm{Ne})$ is a monoatomic gas having 3 degrees of freedom.

$\therefore$ Energy per mole $=\frac{3}{2} R T$

We have 4 moles of Ne .

Hence, Energy $=4 \times \frac{3}{2} R T=6 R T\quad$ ...(ii) [Using Eqs. (i) and (ii)]

$$\begin{array}{rlrl} \therefore \quad & \text { Total energy } & =5 R T+6 R T \\ & =11 R T \end{array}$$

$$\begin{aligned} &\text { Mean free path of a molecule is given by }\\ &l=\frac{1}{\sqrt{2} d^2 n} \end{aligned}$$

where, $n=$ number of molecules/ volume

$d=$ diameter of the molecule

Now, we can write $l \propto \frac{1}{d^2}$

Given, $$\quad d_1=1\mathop A\limits^o, d_2=2\mathop A\limits^o$$

As $$l_1 \propto \frac{1}{d_1^2} \text { and } l_2 \propto \frac{1}{d_2^2}$$

$$\begin{aligned} \Rightarrow \text{So},\quad\frac{l_1}{l_2} & =\left(\frac{d_2}{d_1}\right)^2=\left(\frac{2}{1}\right)^2=\frac{4}{1} \\ \text{Hence, } \quad l_1: l_2 & =4: 1 \end{aligned}$$

Consider the diagram,

Given,

$$\begin{aligned} V_1 & =2.0 \mathrm{~L}, V_2=3.0 \mathrm{~L} \\ \mu_1 & =4.0 \mathrm{~mol}, \mu_2=5.0 \mathrm{~mol} \\ p_1 & =1.00 \mathrm{~atm}, p_2=2.00 \mathrm{~atm} \end{aligned}$$

For chamber $1, p_1, V_1=\mu_1 R T_1$

For chamber 2, $p_2, V_2=\mu_2 R T_2$

When the partition is removed the gases get mixed without any loss of energy. The mixture now attains a common equilibrium pressure and the total volume of the system is sum of the volume of individual chambers $V_1$ and $V_2$.

So, $$\mu=\mu_1+\mu_2, V=V_1+V_2$$

From kinetic theory of gases,

$$\begin{aligned} &\text { For } l \text { mole }\\ &p V=\frac{2}{3} E \quad \text{[E = translational kinetic energy]} \end{aligned}$$

$$\begin{aligned} \text{For $\mu_1$ moles,} \quad & p_1 V_1=\frac{2}{3} \mu_1 E_1 \\ \text{For $\mu_2$ moles, } \quad & p_2 V_2=\frac{2}{3} \mu_2 E_2 \end{aligned}$$

$$\begin{aligned} &\text { Total energy is }\\ &\left(\mu_1 E_1+\mu_2 E_2\right)=\frac{3}{2}\left(p_1 V_1+p_2 V_2\right) \end{aligned}$$

$$\text { From the abvne relation, } \quad p V=\frac{2}{3} E_{\text {total }}=\frac{2}{3} \mu E_{\text {per mole }}$$

$$\begin{aligned} p\left(V_1+V_2\right) & =\frac{2}{3} \times \frac{3}{2}\left(p_1 V_1+p_2 V_2\right) \\ p & =\frac{p_1 V_1+p_2 V_2}{V_1+V_2} \\ & =\left(\frac{1.00 \times 2.0+2.00 \times 3.0}{2.0+3.0}\right) \mathrm{atm} \\ & =\frac{8.0}{5.0}=1.60 \mathrm{~atm} \end{aligned}$$

The average KE will be the same, as conditions of temperature and pressure are the same.

Now as,

$$\begin{aligned} V_{\text {rms }} & =\sqrt{\frac{3 p V}{M}}=\sqrt{\frac{3 R T}{M}} \\ & =\sqrt{\frac{3 R T}{m N}}=\sqrt{\frac{3 k T}{m}} \end{aligned}$$

where, M = molar mass of the gas

m = mass of each molecular of the gas,

R = gas constant

Clearly, $$v_{\mathrm{rms}} \propto \sqrt{\frac{1}{m}}$$

(b) As $k=$ Boltzmann constant

$T=$ absolute temperature (same for all)

But $m_A>m_B>m_C$

$\left(v_{\text {rms }}\right)_A<\left(v_{\text {rms }}\right)_B<\left(v_{\text {rms }}\right)_C$

$\therefore$ or $\left(v_{\mathrm{rms}}\right)_C>\left(v_{\mathrm{rms}}\right)_B>\left(v_{\mathrm{rms}}\right)_A$

Assuming hydrogen molecules as spheres of radius $1 \mathop A\limits^o$.

So, $r=1 \mathop A\limits^o=$ radius.

$$\begin{aligned} \text { The volume of hydrogen molecules } & =\frac{4}{3} \pi r^3 \\ & =\frac{4}{3}(3.14)\left(10^{-10}\right)^3 \\ & \approx 4 \times 10^{-30} \mathrm{~m}^3 \end{aligned}$$

$$\begin{aligned} \text { Number of moles of } \quad \mathrm{H}_2 & =\frac{\text { Mass }}{\text { Molecular mass }} \\ & =\frac{0.5}{2}=0.25 \end{aligned}$$

$$\begin{aligned} & \text { Molecules of } \mathrm{H}_2 \text { present }=\text { Number of moles of } \mathrm{H}_2 \text { present } \times 6.023 \times 10^{23} \\ &=0.25 \times 6.023 \times 10^{23} \\ & \therefore \text { Volume of molecules present }=\text { Molecules number } \times \text { volume of each molecule } \\ &=0.25 \times 6.023 \times 10^{23} \times 4 \times 10^{-30} \\ &=6.023 \times 10^{23} \times 10^{-30} \\ & \approx 6 \times 10^{-7} \mathrm{~m}^3 \quad \text{... (i)} \end{aligned}$$

$$\begin{aligned} &\text { Now, if ideal gas law is considered to be constant. }\\ &\begin{aligned} p_i V_i & =p_f V_f \\ V_f & =\left(\frac{p_i}{p_f}\right) V_i=\left(\frac{1}{100}\right)\left(3 \times 10^{-2}\right)^3 \\ & =\frac{27 \times 10^{-6}}{10^2} \\ & =2.7 \times 10^{-7} \mathrm{~m}^3\quad \text{... (ii)} \end{aligned} \end{aligned}$$

Hence, on compression the volume of the gas is of the order of the molecular volume [form Eq.(i) and Eq.(ii)]. The intermolecular forces will play role and the gas will deviate from ideal gas behaviour.