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

Answer: 2

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 $\frac{3}{2}+\frac{5n_2}{2}=\frac{13\left(1+n_2\right)}{6}$
or $3\left(3+5n_2\right)=13\left(1+n_2\right)$
$9+15n_2=13+13n_2$
$n_2=2$ gram mole.