Question

One mole of an ideal monoatonic gas $(\gamma =\frac{5}{3})$ is mixed with one mole of a diatomic gas $(\gamma =\frac{7}{5})$.
($\gamma$ denotes the ratio of specific heat at constant pressure, to that at constant volume) find $\gamma$ for the mixture.

Answer 1

For mixture of gases,

$\frac{n_1+n_2}{{\gamma }_m-1}=\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}$

Given that,

$n_1=1$

${\gamma }_1=\frac{5}{3}$

$n_2=1$

${\gamma }_2=\frac{7}{5}$

${\gamma }_m\left(\text{For mixture}\right)=?$

$\therefore \frac{1+1}{{\gamma }_m-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$

$\Rightarrow \frac{2}{{\gamma }_m-1}=\frac{3}{2}+\frac{5}{2}$

$\Rightarrow \frac{2}{{\gamma }_m-1}=\frac{8}{2}$

$\Rightarrow 2=4({\gamma }_m-1)$

$\Rightarrow 2=4{\gamma }_m-4$

$\Rightarrow 4{\gamma }_m=4+2$

$\Rightarrow {\gamma }_m=\frac{6}{4}=\frac{3}{2}$

Hence, $\gamma$ for the mixture will be $\frac{3}{2}$.