$$\begin{aligned} &\text { Given, number of moles of helium }=5\\ &T=7^{\circ} \mathrm{C}=7+273=280 \mathrm{~K} \end{aligned}$$

$$\begin{aligned} &\text { (a) Hence, number of atoms (He is monoatomic) }\\ &\begin{aligned} & =\text { Number of moles } \times \text { Avogadro's number } \\ & =5 \times 6.023 \times 10^{23} \\ & =30.015 \times 10^{23} \\ & =3.0 \times 10^{24} \text { atoms } \end{aligned} \end{aligned}$$

$$\text { (b) Now, average kinetic energy per molecule }=\frac{3}{2} k_B T$$

Here, $k_B=$ Boltzmann constant. $$\quad$$ (It has only 3 degrees of freedom)

$$\begin{aligned} &\therefore \text { Total energy of all the atoms }\\ &\begin{aligned} & =\text { Total internal energy } \\ & =\frac{3}{2} k_b T \times \text { number of atoms } \\ & =\frac{3}{2} \times 1.38 \times 10^{-23} \times 280 \times 3.0 \times 10^{24} \\ & =1.74 \times 10^4 \mathrm{~J} \end{aligned} \end{aligned}$$

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.

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

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

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