Five grams of Helium having rms speed of molecules as 1000 m/s and 24 g of oxygen  having rms speed of 1000 m/s are introduced into a thermally isolated vessel. The  ratio of rms speeds of Helium and oxygen when thermal equilibrium is  attained is  $N\sqrt{2}$ . Find N. Neglect the heat capacity of the vessel.

For Helium,  $1000=\sqrt{\frac{3R{T}_1}{4\times {10}^{-3}}}\Rightarrow 3R{T}_1=4000$
For Oxygen,  $1000=\sqrt{\frac{3R{T}_2}{32\times {10}^{-3}}}\Rightarrow 3R{T}_2=32000$
By internal energy conservation, $U_1+U_2=U_1^{\text{'}}+U_2^{\text{'}}$ 
We know,  $U=\frac{fnRT}{2}$
So,  $\left(\frac{3}{2}\right)\left(\frac{5}{4}\right)R{T}_1+\left(\frac{5}{2}\right)\left(\frac{24}{32}\right)R{T}_2=\left[\left(\frac{3}{2}\right)\left(\frac{5}{4}\right)+\left(\frac{5}{2}\right)\left(\frac{24}{32}\right)\right]RT$
This gives  $T=\frac{T_1+T_2}{2}$
After mixing, RMS speed of Helium is
 $v_1=\sqrt{\frac{3RT}{M_1}}=\sqrt{\frac{3R\left(T_1+T_2\right)}{2M_1}}=\sqrt{\frac{4000+32000}{2\left(4\times {10}^{-3}\right)}}=1500\sqrt{2}m/s$
For Oxygen,
 $$\begin{gathered} v_2=\sqrt{\frac{3R\left(T_1+T_2\right)}{2M_O}}=\sqrt{\frac{4000+32000}{2\left(32\times {10}^{-3}\right)}}=750\sqrt{2}m/s \\ \frac{v_1}{v_2}=\frac{1500\sqrt{2}}{750\sqrt{2}}=2 \end{gathered}$$