Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? R = 8.3 J/mol K

We first recall the general result for an ideal gas: if a molecule has $$f$$ degrees of freedom, then its molar specific heat at constant volume is given by the formula $$C_v = \dfrac{f}{2}\,R,$$ because the internal energy of one mole is $$U = \dfrac{f}{2}\,RT$$ and differentiating with respect to temperature at constant volume gives $$\left(\dfrac{\partial U}{\partial T}\right)_V = \dfrac{f}{2}\,R.$$

For a monatomic gas such as helium, there are only the three translational degrees of freedom, so $$f = 3.$$ Therefore, for each mole of helium, we have

$$C_{v,\text{He}} = \dfrac{3}{2}\,R.$$

A hydrogen molecule ($$\text{H}_2$$) is di-atomic and is stated to be rigid in the question, which means it possesses three translational and two rotational degrees of freedom, giving $$f = 5.$$ Hence, for each mole of hydrogen,

$$C_{v,\text{H}_2} = \dfrac{5}{2}\,R.$$

Now we form the mixture. The numbers of moles are:

Helium: $$n_{\text{He}} = 2,$$   Hydrogen: $$n_{\text{H}_2} = 3.$$

The total heat capacity of the mixture at constant volume is the sum of the individual contributions:

$$C_V^{\text{(total)}} \;=\; n_{\text{He}}\,C_{v,\text{He}} \;+\; n_{\text{H}_2}\,C_{v,\text{H}_2}.$$

Substituting the values just obtained, we get

$$\begin{aligned} C_V^{\text{(total)}} & = 2\left(\dfrac{3}{2}\,R\right) + 3\left(\dfrac{5}{2}\,R\right) \\ & = 2 \times \dfrac{3}{2}\,R + 3 \times \dfrac{5}{2}\,R \\ & = 3R + \dfrac{15}{2}\,R \\ & = 3R + 7.5R \\ & = 10.5R. \end{aligned}$$

The mixture contains a total of

$$n_{\text{total}} = n_{\text{He}} + n_{\text{H}_2} = 2 + 3 = 5 \text{ moles}.$$

The required quantity is the molar specific heat of the mixture, defined as

$$C_v^{\text{(mix)}} = \dfrac{C_V^{\text{(total)}}}{n_{\text{total}}}.$$

Substituting the numbers:

$$\begin{aligned} C_v^{\text{(mix)}} & = \dfrac{10.5R}{5} \\ & = 2.1R. \end{aligned}$$

We are given $$R = 8.3 \text{ J mol}^{-1}\text{K}^{-1},$$ so

$$\begin{aligned} C_v^{\text{(mix)}} & = 2.1 \times 8.3 \\ & = 17.43 \text{ J mol}^{-1}\text{K}^{-1}. \end{aligned}$$

This numerical value matches closest with the option $$17.4 \text{ J/mol K}.$$

Hence, the correct answer is Option A.