Question

A gaseous mixture enclosed in a vessel consists of one $\text{gram mole}$ of a gas $A$ with $\gamma =\frac{5}{3}$ and some amount of gas $B$ with $\gamma =\frac{7}{5}$ at a temperature $T$. The gases $A$ and $B$ do not react with each other and are assumed to be ideal. Find the number of moles of the gas $B$ if $\gamma$ for the gaseous mixture is $\left(\frac{19}{13}\right)$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

As for an ideal gas, $C_P-C_V=R$ and $\gamma =\left(\frac{C_P}{C_V}\right)$
So, $C_V=\frac{R}{(\gamma -1)}$
$\therefore (C_V{)}_1=\frac{R}{\left(\frac{5}{3}\right)-1}=\frac{3}{2}R$;
& $(C_V{)}_2=\frac{R}{\left(\frac{7}{5}\right)-1}=\frac{5}{2}R$
Also, $(C_V{)}_{\text{mix}}=\frac{R}{\left(\frac{19}{13}\right)-1}=\frac{13}{6}R$

Since both the gases do not react,
$\Delta U_{mix}=\Delta U_1+\Delta U_2$
i.e. $(n_1+n_2)(C_V{)}_{\text{mix}}\Delta T=n_1(C_V{)}_1+n_2(C_V{)}_2\Delta T$
$\Rightarrow (C_V{)}_{\text{mix}}=\frac{n_1(C_V{)}_1+n_2(C_V{)}_2}{n_1+n_2}$
Substituting the data,
$\frac{13}{6}R=\frac{1\times \frac{3}{2}R+n_2\frac{5}{2}R}{1+n_2}=\frac{(3+5n_2)R}{2(1+n_2)}$
or $13+13n_2=9+15n_2$
or $n_2=2$ is the no. of moles of gas B.