We have to convert the given temperatures in kelvin .

If pressure of a given mass of the gas is kept constant, then

$$\begin{array}{ll} & V \propto T \\ \Rightarrow & \frac{V}{T}=\text { constant } \quad \left[\begin{array}{l} V=\text { Volume of gas } \\ T=\text { Temperature of gas } \end{array}\right]\\ \Rightarrow & \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \Rightarrow & V_2=V_1\left(\frac{T_2}{T_1}\right) \\ & T_1=273+27=300 \mathrm{~K} \\ & T_2=273+327=600 \mathrm{~K} \\ \text { But } & V_1=100 \mathrm{cc} \\ & V_2=V_1\left(\frac{600}{300}\right) \\ & V_2=2 V_1 \\ \therefore \quad & V_2=2 \times 100=200 \mathrm{cc} \end{array}$$

We know that for a given mass of a gas

$$v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$$

where, $R$ is gas constant $T$ is temperature in kelvin $M$ is molar mass of the gas.

Clearly, $$\quad v_{\mathrm{rms}} \propto \sqrt{T}$$

$$\begin{aligned} &\text { As } R, M \text { are constants, }\\ &\frac{\left(v_{\mathrm{rms}}\right)_1}{\left(v_{\mathrm{rms}}\right)_2}=\sqrt{\frac{T_1}{T_2}} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Given, }\quad \left(v_{\text {rms }}\right)_1 & =100 \mathrm{~m} / \mathrm{s} \\ T_1=27^{\circ} \mathrm{C} & =27+273=300 \mathrm{~K} \\ T_2=127^{\circ} \mathrm{C} & =127+273=400 \mathrm{~K} \end{aligned} \end{aligned}$$

$$\therefore$$ From Eq. (i)

$$\begin{aligned} & \frac{100}{\left(v_{\mathrm{rms}}\right)_2}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2} \\ \Rightarrow \quad & \left(v_{\mathrm{rms}}\right)_2=\frac{2 \times 100}{\sqrt{3}}=\frac{200}{\sqrt{3}} \mathrm{~m} / \mathrm{s} \end{aligned}$$

For n-molecules, we know that

$$v_{\mathrm{rms}}=\sqrt{\frac{v_1^2+v_2^2+v_3^2+\ldots \ldots+v_n^2}{n}} \quad\left[\begin{array}{c} \left.v_{\mathrm{rms}}=\begin{array}{l} \text { root mean } \\ \text { square velocity } \end{array}\right] \end{array}\right.$$

where $v_1, v_2, v_3 \ldots \ldots \ldots v_n$ are individual velocities of $n$-molecules of the gas. For two molecules,

$$v_{\text {ms }}=\sqrt{\frac{v_1^2+v_2^2}{2}} \quad\left[v_1, v_2, v_3, \ldots \ldots \ldots v_n \text { are individual velocity }\right]$$

Given, $$\quad v_1=9 \times 10^6 \mathrm{~m} / \mathrm{s}$$

and $$\quad v_2=1 \times 10^6 \mathrm{~m} / \mathrm{s}$$

$$\begin{aligned} \therefore \quad v_{r m s} & =\sqrt{\frac{\left(9 \times 10^6\right)^2+\left(1 \times 10^6\right)^2}{2}} \\ & =\sqrt{\frac{81 \times 10^{12}+1 \times 10^{12}}{2}} \\ & =\sqrt{\frac{(81+1) \times 10^{12}}{2}} \\ & =\sqrt{\frac{82 \times 10^{12}}{2}} \\ & =\sqrt{41} \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}$$

$\mathrm{O}_2$ 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 $$l_1 \propto \frac{1}{d_1^2} \text { and } l_2 \propto \frac{1}{d_2^2}$$

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