A gaseous mixture enclosed in a vessel consists of one gm mole of a gas A With $(γ = 5 / 3)$ and another B with $(γ = 7 / 5)$ at a temperature T. The gases A and B do not React with each other and assumed to be ideal. Find the number of gram moles of the gas B if $γ$ of the gaseous mixture is 19/13.

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Detailed Solution

For two gases, we can write
$\frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1} = \frac{n_{1} + n_{2}}{γ_{m e a n} - 1}$
Here, $\begin{matrix} n_{1} = 1 a n d n_{2} = ? \\ γ_{1} = 5 / 3 a n d γ_{2} = 7 / 5 \end{matrix}$
Substituting these values in above equation, we get
$\frac{1}{\frac{5}{3} - 1} + \frac{n_{2}}{\frac{7}{5} - 1} = \frac{1 + n_{2}}{\frac{19}{13} - 1}$
Or $\frac{3}{2} + \frac{5 n_{2}}{2} = \frac{13 (1 + n_{2})}{6}$
Or $3 (3 + 5 n_{2}) = 13 (1 + n_{2})$
$9 + 15 n_{2} = 13 + 13 n_{2}$
$n_{2} = 2$ gram mole.