$2$ mole ideal He gas and $3$ mole ideal $H_{2}$ gas at constant volume find out $C_{v}$ of mixture

\frac{21 R}{10}

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\frac{11 R}{10}

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\frac{21 R}{5}

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\frac{11 R}{5}

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Solution

The correct option is C $\frac{21 R}{10}$
The given mixture of two moles of monoatomic gas mixed with three mole of diatomic gas can be given as
$C = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$

$= \frac{2 \times \frac{3}{2} R + 3 \times \frac{5}{2} R}{2 + 3}$

$= \frac{3 R + 15 \frac{R}{2}}{5} = (\frac{21 R}{10})$

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\frac{C_{p}}{C_{v}} = 5 / 3

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(γ = \frac{7}{5})

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H_{2}

(v = 5 / 3)

(γ = 7 / 5)

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γ

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