Ideal Gas Mixture Energy and Sound Speed Ratios

To determine the correct statement(s), let us analyze each option step-by-step using kinetic theory of gases.

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1. Analysis of Option (A): Average Energy per Mole

For an ideal gas, the total internal energy is the sum of the internal energies of its constituent gases: $U = U_{\text{H}_2} + U_{\text{He}}$

• Hydrogen ($\text{H}_2$) is a rigid diatomic gas at room temperature, having $f_{\text{H}_2} = 5$ degrees of freedom. For $n_{\text{H}_2} = 1\text{ mole}$: $U_{\text{H}_2} = n_{\text{H}_2} \cdot \frac{f_{\text{H}_2}}{2} R T = 1 \cdot \frac{5}{2} R T = \frac{5}{2} R T$

• Helium ($\text{He}$) is a monatomic gas, having $f_{\text{He}} = 3$ degrees of freedom. For $n_{\text{He}} = 1\text{ mole}$: $U_{\text{He}} = n_{\text{He}} \cdot \frac{f_{\text{He}}}{2} R T = 1 \cdot \frac{3}{2} R T = \frac{3}{2} R T$

The total internal energy of the mixture is: $U = \frac{5}{2} R T + \frac{3}{2} R T = 4 R T$

The total number of moles in the mixture is: $n_{\text{total}} = n_{\text{H}_2} + n_{\text{He}} = 1 + 1 = 2\text{ moles}$

Therefore, the average internal energy per mole of the gas mixture is: $U_{\text{avg}} = \frac{U}{n_{\text{total}}} = \frac{4 R T}{2} = 2 R T$

Thus, Option (A) is correct.

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2. Analysis of Option (B): Ratio of Speed of Sound

The speed of sound in an ideal gas is given by: $v = \sqrt{\frac{\gamma R T}{M}}$ where $\gamma = \frac{C_p}{C_v}$ is the adiabatic index and $M$ is the molar mass of the gas.

For the Gas Mixture:

• Molar mass of the mixture ($M_{\text{mix}}$): $M_{\text{mix}} = \frac{n_{\text{H}_2} M_{\text{H}_2} + n_{\text{He}} M_{\text{He}}}{n_{\text{H}_2} + n_{\text{He}}} = \frac{1 \cdot 2 + 1 \cdot 4}{1 + 1} = \frac{6}{2} = 3\text{ g/mol}$

• Molar heat capacity at constant volume ($C_{v, \text{mix}}$): $C_{v, \text{mix}} = \frac{n_{\text{H}_2} C_{v, \text{H}_2} + n_{\text{He}} C_{v, \text{He}}}{n_{\text{H}_2} + n_{\text{He}}} = \frac{1 \cdot \left(\frac{5}{2} R\right) + 1 \cdot \left(\frac{3}{2} R\right)}{2} = 2 R$

• Molar heat capacity at constant pressure ($C_{p, \text{mix}}$): $C_{p, \text{mix}} = C_{v, \text{mix}} + R = 2 R + R = 3 R$

• Adiabatic index of the mixture ($\gamma_{\text{mix}}$): $\gamma_{\text{mix}} = \frac{C_{p, \text{mix}}}{C_{v, \text{mix}}} = \frac{3 R}{2 R} = \frac{3}{2}$

Speed of sound in the mixture: $v_{\text{mix}} = \sqrt{\frac{\left(\frac{3}{2}\right) R T}{3}} = \sqrt{\frac{R T}{2}}$

For Pure Helium Gas:

• Molar mass $M_{\text{He}} = 4\text{ g/mol}$
• Adiabatic index $\gamma_{\text{He}} = \frac{5}{3}$

Speed of sound in pure helium: $v_{\text{He}} = \sqrt{\frac{\left(\frac{5}{3}\right) R T}{4}} = \sqrt{\frac{5 R T}{12}}$

Ratio of Speed of Sound:

$\frac{v_{\text{mix}}}{v_{\text{He}}} = \frac{\sqrt{\frac{R T}{2}}}{\sqrt{\frac{5 R T}{12}}} = \sqrt{\frac{12}{2 \times 5}} = \sqrt{\frac{6}{5}}$

Thus, Option (B) is correct.

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3. Analysis of Options (C) and (D): Ratio of RMS Speeds

The root-mean-square (rms) speed of an ideal gas molecule is given by: $v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}$

• For Helium atoms ($M_{\text{He}} = 4\text{ g/mol}$): $v_{\text{rms, He}} = \sqrt{\frac{3 R T}{4}}$

• For Hydrogen molecules ($M_{\text{H}_2} = 2\text{ g/mol}$): $v_{\text{rms, } \text{H}_2} = \sqrt{\frac{3 R T}{2}}$

The ratio of the rms speed of helium atoms to that of hydrogen molecules is: $\frac{v_{\text{rms, He}}}{v_{\text{rms, } \text{H}_2}} = \frac{\sqrt{\frac{3 R T}{4}}}{\sqrt{\frac{3 R T}{2}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$

Thus, Option (D) is correct and Option (C) is incorrect.

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Conclusion

The correct options are A, B, and D.