Question

The value of $\frac{C_p}{C_v}$ for the mixture of 2 mols of oxygen and 5 moles of ozone is:

• A. 1.34
• B. 1.41
• C. 1.51
• D. 1.67

Answer 1

The correct option is A 1.34

We have $C_p=(1+\frac{f}{2})R$ and $C_v=\frac{f}{2}R$. For a diatomic gas degree of freedom f is 5 and for triatomic gas it is 7.

Thus we get $C_p$ for the mixture as:

$C_p=(\frac{n_1}{n_1+n_2}\left)\right(1+\frac{5}{2})R+(1+\frac{7}{2})R=\frac{n_1}{n_1+n_2}8R$

and

$C_v=(\frac{n_1}{n_1+n_2}\left)\right(\frac{5}{2}R+\frac{7}{2}R)=\frac{n_1}{n_1+n_2}6R$.

Thus the ratio $\frac{C_p}{C_v}$ is given as $\frac{8}{6}=1.34$