Question

A gaseous mixture contains 1 mole of a gas $(\gamma =5/3)$ and another gas B (with $\gamma =7/5)$ at a temperature T. Assuming that gases do not react with each other and behave ideally and $\gamma$ for mixture is $19/13$. The fraction of gas B in the mixture is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $1$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{2}{3}$

As we know,

$C_p-C_v=R$

$\frac{C_p}{C_v}-1=\frac{R}{C_v}$

$C_v=\frac{R}{\gamma -1}$

A: $\gamma =\frac{5}{3}$

(for monoatomic) $C_v=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R$

B: $\gamma =\frac{7}{5}$

(for diatomic) $C_v=\frac{R}{(\frac{7}{5}-1)}=\frac{5}{2}R$

assume mixture contains 1 mole of A and n mole of B

So for mixture

$C_{v_{mix}}=\frac{1\times \frac{3}{2}R+n\times \frac{5}{2}R}{1+n}=\frac{(3+5n)R}{2(1+n)}$

$\frac{R}{\frac{19}{13}-1}=\frac{(3+5n)R}{2(1+n)}$

$\frac{13}{6}=\frac{3+5n}{2(1+n)}$

$n=2$

So, fraction of B $=\frac{2}{1+2}=\frac{2}{3}$