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

The correct option is A $2$

Given parameters are:

Here, $n_1=1$

${\gamma }_1=\frac{5}{3}$

${\gamma }_2=\frac{7}{5}$

${\gamma }_{mean}=\frac{19}{13}$

and $n_2=?$

For two gases, we can wire

$\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}=\frac{(n_1+n_2)}{({\gamma }_{mean}-1)}$

Substituting values in the above equation, we get,

$\frac{1}{\frac{5}{3}-1}+\frac{n_2}{\frac{7}{5}-1}=\frac{(1+n_2)}{(\frac{19}{13}-1)}$

$3\times (3+5n_2)=13\times (1+n_2)$

$n_2=2gram/moles$

Option A is correct.