Question

A gaseous mixture enclosed in a vessel contains $1g$ mole of a gas $A$ (with $\gamma =5/3$) and another gas $B$ (with $\gamma =7/5$) at a temperature $T$. The gases A and $B$ do not react with each other and assumed to be ideal. The number of gram moles of $B$, if $\gamma$ for the gaseous mixture is $19/13$ is

• A. $2$
• B. $12$
• C. $16$
• D. $8$

Answer 1

Answer: (A)

Let the mixture contain $n$ moles of gas $B$
As $C_P-C_V=R$ and $\gamma =\frac{C_P}{C_V}$
$\therefore C_V=\frac{R}{\gamma -1}$
For gas $A$, $C_V=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R$
For gas $B$, $C_V=\frac{R}{\frac{7}{5}-1}=\frac{5}{2}R$
For the mixture, $C_V=\frac{R}{\frac{19}{13}-1}=\frac{13}{6}R$
By conservation of energy, molar specific heat of the mixture is
$C_V=\frac{n_A{\left(C_V\right)}_A+n_B{\left(C_V\right)}_B}{n_A+n_B}$
$\therefore \frac{13}{6}R=\frac{1\times \frac{3}{2}R+n\times \frac{5}{2}R}{1+n}=\frac{\left(3+5n\right)R}{2\left(1+n\right)}$
$\therefore n=2$