A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is (are)

According to question no. of mole of hydrogen = no. of mole of helium =1

Total internal energy, u

$=\frac{f_1}{2}nRT+\frac{f_2}{d}mRT\Rightarrow u=\frac{3}{2}RT+\frac{5}{2}RT=4RT$

$\therefore$Average internal energy per mole $=\frac{u}{2n}=\frac{4RT}{2}=2RT$

We know that $V_{sound}=\sqrt{\frac{\gamma RT}{M}}$

$\frac{n_1+n_2}{{\gamma }_{mix}-1}=\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}\Rightarrow \frac{2}{{\gamma }_{mix}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$

$\frac{2}{{\gamma }_{mix}-1}=\frac{3}{2}+\frac{5}{2}=4\Rightarrow {\gamma }_{mix}-1=\frac{1}{2}\therefore {\gamma }_{mix}=\frac{3}{2}$

$\frac{{\left(V_s\right)}_{mix}}{{\left(V_s\right)}_{He}}=\sqrt{\frac{{\gamma }_{mix}}{M_{mix}}\times \frac{M_{He}}{{\gamma }_{He}}}=\sqrt{\frac{\frac{3}{2}\times 4}{3\times \frac{5}{3}}}=\sqrt{\frac{6}{5}}$

$\left[\therefore M_{mix}=\frac{1\times 2+1\times 4}{2}=3\right]$

We know that $V_{rms}=\sqrt{\frac{3RT}{M}}$

$\therefore \frac{{\left(V_{rms}\right)}_{He}}{{\left(V_{rms}\right)}_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{H_E}}}=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}$