Animated Solution for Physics - Thermodynamics: 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 statements is/are

Imagine a container holding a mixture of two distinct gases in thermal equilibrium at a temperature $T$. We have exactly one mole of hydrogen ($H_2$) and one mole of helium ($He$).

Hydrogen is a diatomic gas, meaning its molecules look like tiny dumbbells. At normal temperatures, it has 3 translational and 2 rotational degrees of freedom, giving it a total of $f_1=5$ degrees of freedom. Helium, on the other hand, is a noble, monoatomic gas. Its atoms are solitary spheres with only 3 translational degrees of freedom, so $f_2=3$.

Let's start by finding the average energy per mole of this mixture. The total internal energy $U$ of a gas mixture is simply the sum of the internal energies of its components.

Substituting our values:

Since we have a total of $n_1+n_2=2$ moles in the container, the average energy per mole is:

This confirms that the first statement is absolutely correct.

To evaluate the speed of sound, we need two crucial properties of our mixture: the effective molar mass ($M_{mix}$) and the adiabatic index (${\gamma }_{mix}$).

The molar mass of the mixture is the total mass divided by the total number of moles. Hydrogen has a molar mass of $M_1=2$ g/mol, and helium has $M_2=4$ g/mol.

Next, we calculate ${\gamma }_{mix}$. Remember, you cannot just average the $\gamma$ values of the individual gases! You must use the specific heat capacities:

For diatomic hydrogen, $C_{P_1}=\frac{7R}{2}$ and $C_{V_1}=\frac{5R}{2}$. For monoatomic helium, $C_{P_2}=\frac{5R}{2}$ and $C_{V_2}=\frac{3R}{2}$.

Speed of Sound and RMS Speed

The speed of sound in a gas is given by the formula $v=\sqrt{\frac{\gamma RT}{M}}$. We want the ratio of the speed of sound in the mixture to that in pure helium.

For pure helium, ${\gamma }_{He}=\frac{5}{3}$ and $M_{He}=4$ g/mol. Taking the ratio, the $R$ and $T$ terms cancel out beautifully:

This confirms that the second statement is also correct!

Finally, let's check the root mean square (RMS) speeds. The RMS speed is given by $v_{rms}=\sqrt{\frac{3RT}{M}}$. Since both gases are in the same container, they share the same temperature $T$. This means the RMS speed is inversely proportional to the square root of the molar mass ($v_{rms}\propto \frac{1}{\sqrt{M}}$).

This proves that the fourth statement is correct, while the third statement is incorrect. The correct options are (a), (b), and (d).