Question

A gaseous mixture enclosed in a vessel consists of one gm mole of a gas $A$ with $(\gamma=5 / 3)$ and another $B$ with $(\gamma=7 / 5)$ at a temperature $T$. The gases $A$ and $B$ do not react with each other and assumed to be ideal. Find the number of gram moles of the gas if $\gamma$ of the gaseous mixture is $19 / 13$.

Answer 1

For two gases, we can write
$\frac{n_1}{\gamma_1-1}+\frac{n_2}{\gamma_2-1} =\frac{n_1+n_2}{\gamma_{\text {mean }}-1}$
Here, $ n_1 =1 \text { and } n_2=? $
$\gamma_1 =5 / 3 \text { and } \gamma_2=7 / 5$
Substituting these values in above equation, we get
$\frac{1}{\frac{5}{3}-1}+\frac{n_2}{\frac{7}{5}-1}=\frac{1+n_2}{\frac{19}{13}-1}$
or $ \frac{3}{2}+\frac{5 n_2}{2}=\frac{13\left(1+n_2\right)}{6}$
or $ 3\left(3+5 n_2\right)=13\left(1+n_2\right)$
$9+15 n_2=13+13 n_2$
$n_2=2$ gram mole.