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Q.

A gaseous mixture enclosed in a vessel consists of one 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 moles of the gas B if $γ$ for the gaseous mixture is $(\frac{19}{13})$ .

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Detailed Solution

AS FOR IDEAL GAS $C_{p} - C_{u} = R$ and $γ = (\frac{C_{P}}{C_{υ}})$
SO $C_{υ} = \frac{R}{(γ - 1)}$
$\begin{array}{l} ∴ {(C_{ν})}_{1} = \frac{R}{(\frac{5}{3}) - 1} = \frac{3}{2} R \\ {(C_{ν})}_{2} = \frac{R}{(\frac{7}{5}) - 1} = \frac{5}{2} R \end{array}$
Now from conservation of energy,
$\begin{array}{l} i . e, Δ U = Δ U_{1} + Δ U_{1} \\ (μ_{1} + μ_{2}) {(C_{ν})}_{m i x} Δ T = [μ_{1} {(C_{υ})}_{1} + μ_{2} {(C_{υ})}_{2}] Δ T \end{array}$

$i . e, {(C_{υ})}_{m i x} = \frac{μ_{1} {(C_{υ})}_{1} + μ_{2} {(C_{υ})}_{2}}{μ_{1} + μ_{2}}$

$W e h a v e \frac{13}{6} R = \frac{1 \times \frac{3}{2} R + μ_{2} \frac{5}{2} R}{1 + μ_{2}} = \frac{(3 + 5 μ_{2}) R}{2 (1 + μ_{2})}$

$\Rightarrow \begin{array}{l} 13 + 13 μ_{2} = 9 + 15 μ_{2} \Rightarrow μ_{2} = 2 m o l e s \end{array}$

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