Question

A gaseous mixture enclosed in a vessel contains $1\, g$ 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 $\gamma = \frac{C_{P}}{C_{V}}$
$\therefore \quad C_{V} = \frac{R}{\gamma - 1}$
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
$C_{V} = \frac{n_{A}\left(C_{V}\right)_{A} + n_{B} \left(C_{V}\right)_{B}}{n_{A} + n_{B}}$
$\therefore \quad \frac{13}{6} \,R = \frac{1\times \frac{3}{2}R+n \times\frac{5}{2}R}{1+n} = \frac{\left(3+5n\right)\,R}{2\left(1+n\right)}$
$\therefore \quad n = 2$