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Q.

A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with $γ = \frac{5}{3}$ and some amount of gas B with $γ = \frac{7}{5}$ at a temperature T. The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if $γ$ for the gaseous mixture is $\frac{19}{13}$ .

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answer is 2.

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Detailed Solution

As for an ideal gas $C_{P} - C_{V} = R$ and $γ = \frac{C_{P}}{C_{V}}$
$C_{V} = \frac{R}{γ - 1} {(C_{V})}_{1} = \frac{R}{\frac{5}{3} - 1} = \frac{3}{2} R {(C_{V})}_{2} = \frac{R}{\frac{7}{5} - 1} = \frac{5}{2} R {(C_{V})}_{m i x} = \frac{R}{\frac{19}{13} - 1} = \frac{13}{6} R$

Now from conservation of energy $Δ U = Δ U_{1} + Δ U_{2}$
$(n_{1} + n_{2}) {(C_{V})}_{m i x} Δ T = [n_{1} C_{V_{1}} + n_{2} C_{V_{2}}] Δ T$
By solving we get $n_{2} = 2$ .

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