As given molecules are of hydrogen.

$\therefore$ Volume occupied by 1 mole

$$\begin{aligned} & =1 \mathrm{~mole} \text { of the gas at NTP } \\ & =22400 \mathrm{~mL}=22400 \mathrm{cc} \end{aligned}$$

$\therefore$ Number of molecules in 1 cc of hydrogen

$$=\frac{6.023 \times 10^{23}}{22400}=2.688 \times 10^{19}$$

$\mathrm{H}_2$ is a diatomic gas, having a total of 5 degrees of freedom (3 translational + 2 rotational)

$\therefore$ Total degrees of freedom possessed by all the molecules

$$\begin{aligned} & =5 \times 2.688 \times 10^{19} \\ & =1.344 \times 10^{20} \end{aligned}$$

According to kinetic interpretation of temperature, absolute temperature of a given sample of a gas is proportional to the total translational kinetic energy of its molecules.

Hence, any change in absolute temperature of a gas will contribute to corresponding change in translational KE and vice-versa.

Assuming $n=$ number of moles.

Given, $m=$ molar mass of the gas.

When, the container stops, its total KE is transferred to gas molecules in the form of translational $K E$, thereby increasing the absolute temperature.

If $\Delta T=$ change in absolute temperature.

$$\text { Then, } \mathrm{KE} \text { of molecules due to velocity } \mathrm{v}_0, \mathrm{KE}=\frac{1}{2}(m n) v_0^2\quad \text{... (i)}$$

$$\text { Increase in translational } \mathrm{KE}=n \frac{3}{2} R(\Delta T)\quad \text{... (ii)}$$

$$\begin{aligned} &\text { According to kinetic theory Eqs. (i) and (ii) are equal }\\ &\begin{aligned} \Rightarrow \quad & \frac{1}{2}(m n) v_0^2 =n \frac{3}{2} R(\Delta T) \\ \Rightarrow \quad & (m n) v_0^2 =n 3 R(\Delta T) \\ \Rightarrow \quad & \Delta T =\frac{(m n) v_0^2}{3 n R} \\ \Rightarrow \quad & \Delta T =\frac{m v_0^2}{3 R} \end{aligned} \end{aligned}$$

The moon has small gravitational force (pull) and hence, the escape velocity is small. The value of escape velocity for the moon is $4.6 \mathrm{~km} / \mathrm{s}$.

As the moon is in the proximity of the earth as seen from the sun, the moon has the same amount of heat per unit area as that of the earth.

The air molecules have large range of speeds. Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number of molecules have speed greater than escape velocity and they escape.

Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again a significant number of molecules escape as their speed exceed escape speed. Hence, over a long time the moon has lost most of its atmosphere.

$$\begin{aligned} \text { At } 300 \mathrm{~K}, v_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}} & =\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{7.3 \times 10^{-26}}}=1.7 \mathrm{~km} / \mathrm{s} \\ v_{\text {es }} \text { for moon } & =4.6 \mathrm{~km} / \mathrm{s} \quad \quad\left[\mathrm{v}_{\mathrm{es}}=\text { escape velocity }\right] \end{aligned}$$

(b) As the molecules move higher; their potential energy increases and hence, kinetic energy decreases and hence, temperature reduces.

At greater height more volume is available and gas expands and hence, some cooling takes place.

$$\begin{aligned} &\text { (a) We know that }\\ &v_{\mathrm{rms}}^2=\frac{\sum_\limits i n_i v_i^2}{\sum n_i} \end{aligned}$$

This is the rms speed for all molecules collectively.

$$\begin{aligned} \text{Now,}\quad v_{\mathrm{rms}} & =\left(\frac{\sum_\limits i n_i v_i^2}{\sum n_i}\right)^{\frac{1}{2}} \\ & =\sqrt{\frac{n_1 v_1^2+n_2 v_2^2+n_3 v_3^2+\ldots \ldots \ldots+n_n v_n^2}{n_1+n_2+n_3+\ldots \ldots+n_n}} \\ & =\sqrt{\frac{n_1 v_1^2+n_2 v_2^2+n_3 v_3^2+n_4 v_4^2+n_5 v_5^2}{n_1+n_2+n_3+n_4+n_5}} \\ & =\sqrt{\frac{\left.10 \times(200)^2+20 \times(400)^2+40 \times 600\right)^2+20 \times(800)^2+10 \times(1000)^2}{100}} \\ & =\sqrt{1000 \times(4+32+144+128+100)} \\ & =\sqrt{408 \times 1000} \approx 639 \mathrm{~m} / \mathrm{s} \end{aligned}$$

$$\begin{aligned} &\text { Now, according to kinetic theory of gasses }\\ &\begin{aligned} \frac{1}{2} m v_{\mathrm{ms}}^2 & =\frac{3}{2} k_B T \quad\left[\begin{array}{l} K_B=\text { Boltzmann constant } \\ m=\text { mass of gaseous molecules } \end{array}\right] \\ \therefore \quad T=\frac{1}{3} \frac{m v_{\mathrm{ms}}^2}{k_B} & =\frac{1}{3} \times \frac{3.0 \times 10^{-26} \times 4.08 \times 10^5}{1.38 \times 10^{-23}} \\ & =2.96 \times 10^2 \mathrm{~K}=296 \mathrm{~K} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { (b) If all the molecules with speed } 1000 \mathrm{~m} / \mathrm{s} \text { escape, then }\\ &\begin{aligned} v_{\mathrm{rms}}^2 & =\frac{10 \times(200)^2+20 \times(400)^2+40 \times(600)^2+20 \times(800)^2}{90} \\ & =\frac{10 \times 100^2 \times(1 \times 4+2 \times 16+4 \times 36+2 \times 64)}{90} \\ & =10000 \times \frac{308}{9}=342 \times 1000 \mathrm{~m}^2 / \mathrm{s}^2 \\ v_{\mathrm{rms}} & =584 \mathrm{~m} / \mathrm{s} \\ \text { Again } \quad T & =\frac{1}{3} \frac{\mathrm{mv} v_{\mathrm{ms}}^2}{k} \\ & =\frac{1}{3} \times \frac{3 \times 10^{-26} \times 3.42 \times 10^5}{1.38 \times 10^{-23}} \\ & =2.478 \times 10^2 \\ & =247.8 \approx 248 \mathrm{~K} \end{aligned} \end{aligned}$$

The situation can be considered as the time of relaxation, based on kinetic theory of gases. Mean free path is the distance between two successive collisions, which we will consider here as the distance travelled by the plane before it just avoids the collision safe radius is equivalent to radius of the atom.

Hence, the required time

$$\begin{aligned} t=\frac{l}{v}, l=\text { mean free path } & =\frac{1}{\sqrt{2} \pi d^2 n}, n=\text { number density }=\frac{N}{V} \\ n & =\frac{\text { Number of aeroplanes }(\mathrm{N})}{\text { Volume }(\mathrm{V})} \\ & =\frac{10}{20 \times 20 \times 1.5}=0.0167 \mathrm{~km}^{-3} \\ t & =\frac{1}{\sqrt{2} \pi d^2(\mathrm{~N} / \mathrm{V})} \times \frac{1}{\mathrm{~V}} \quad \quad[\mathrm{~V}=\text { velocity of aeroplane }] \end{aligned}$$

By putting the given data,

$$\begin{aligned} t & =\frac{1}{\sqrt{2} \times 3.14 \times(20)^2 \times 0.0167 \times 10^{-6} \times 150} \\ & =\frac{10^6}{1776.25 \times 2.505} \\ & =\frac{10^6}{4449.5}=224.74 \mathrm{~h} \\ & \approx 225 \mathrm{~h} \end{aligned}$$