Consider a mixture of $$n$$ moles of helium gas and $$2n$$ moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its $$\frac{C_P}{C_V}$$ value will be:

We are given a gaseous mixture that contains $$n$$ moles of helium (a mon-atomic gas) and $$2n$$ moles of oxygen (a di-atomic gas that is to be treated as a rigid molecule). We have to evaluate the ratio $$\dfrac{C_P}{C_V}$$ for the whole mixture, treating it as an ideal gas.

First, we recall the basic kinetic-theory relation connecting the molar heat capacity at constant volume with the number of degrees of freedom $$f$$ of a molecule: for any ideal gas,

$$C_V = \dfrac{f}{2}\,R,$$

where $$R$$ is the universal gas constant. Correspondingly,

$$C_P = C_V + R.$$

For helium, which is mon-atomic, there are only the three translational degrees of freedom, so $$f_{\text{He}} = 3$$. Hence

$$C_{V,\text{He}} = \dfrac{3}{2}R, \qquad C_{P,\text{He}} = \dfrac{3}{2}R + R = \dfrac{5}{2}R.$$

For oxygen, which is di-atomic and described as a rigid rotor, the molecule possesses three translational and two rotational degrees of freedom, giving $$f_{\text{O}_2}=5$$. Therefore,

$$C_{V,\text{O}_2} = \dfrac{5}{2}R, \qquad C_{P,\text{O}_2} = \dfrac{5}{2}R + R = \dfrac{7}{2}R.$$

Let us now compute the molar heat capacities of the entire mixture. The total number of moles present is

$$n_{\text{total}} = n + 2n = 3n.$$

The mixture’s molar heat capacity at constant volume is obtained by weighting each component’s heat capacity by the number of moles of that component and then dividing by the total moles:

$$\begin{aligned} C_{V,\text{mix}} &= \dfrac{\,n \,C_{V,\text{He}} \;+\; 2n \,C_{V,\text{O}_2}\,}{\,n_{\text{total}}\,} \\ &= \dfrac{\,n\left(\dfrac{3}{2}R\right) + 2n\left(\dfrac{5}{2}R\right)}{3n}. \end{aligned}$$

Simplifying the numerator inside the fraction:

$$n\left(\dfrac{3}{2}R\right) = \dfrac{3}{2}\,nR,$$

$$2n\left(\dfrac{5}{2}R\right) = 2n \cdot \dfrac{5}{2}R = 5nR.$$

Adding these two contributions gives

$$\dfrac{3}{2}\,nR + 5nR = \left(\dfrac{3}{2} + 5\right)nR = \left(\dfrac{3 + 10}{2}\right)nR = \dfrac{13}{2}\,nR.$$

Dividing by the denominator $$3n$$ produces

$$C_{V,\text{mix}} = \dfrac{\dfrac{13}{2}\,nR}{3n} = \dfrac{13}{6}\,R.$$

Next, the mixture’s molar heat capacity at constant pressure is simply one $$R$$ higher:

$$C_{P,\text{mix}} = C_{V,\text{mix}} + R = \dfrac{13}{6}R + R = \left(\dfrac{13}{6} + \dfrac{6}{6}\right)R = \dfrac{19}{6}\,R.$$

Finally, we take the ratio

$$\dfrac{C_P}{C_V}\Bigg|_{\text{mix}} = \dfrac{\dfrac{19}{6}R}{\dfrac{13}{6}R} = \dfrac{19}{13}.$$

Hence, the correct answer is Option A.