A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be:

Take R = 8.32 J mole1 k1.

The average kinetic energy (per molecule)of any (ideal) gas (be it monoatomic like Argon, diatomic like oxygen or polyatomic) is always equal to $\frac{3}{2}{K}_{B}T.$ If depends only on temperature, and is independent of the nature of the gas. So, Answer is 1 : 1.

Degree of freedom, f = 8

$\therefore C_V=\frac{8}{2}R=4Rand{C}_P=4R+R=5R$

So, H = $\Delta U+W$

at constant pressure, W = $P\Delta v$

150 = $nR\Delta T$

So,  $H=n{C}_V\Delta T=n5R\Delta T=5\left(nR\Delta T\right)$

= 5 × 150

= 750 J

$\overline{x}$ (mean free path) = $\frac{1}{\sqrt[]{2}\pi d^2{n}_v}$  

$n_v=\frac{n{N}_A}{v},$ n → No. of moles in volume v N_A ® Avogadro’s Number

$\frac{n}{v}=\frac{P}{R_T}$

$\overline{x}\propto v\propto \frac{1}{\rho (density}\&\overline{x}\alpha TatconstantP)$

^{$\rho \uparrow \iff \overline{x}\downarrow |\overline{x}\uparrow \iff T\uparrow$}                           

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is  $=\frac{1}{2}{k}_{B}T$ (for Mono atomic gases)