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.

When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules remains constant. In this case, as the number of air molecules keep increasing. Hence, this is a case of variable mass. Boyle's law (and even Charle's law) is only applicable in situations, where number of gas molecules remains fixed. Hence, in this case Boyle's law is not applicable.

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

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

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

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

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