Let n₁ and n₂ moles of two different ideal gases are mixed. If the adiabatic coefficient of the two gases are $γ_{1}$ and $γ_{2}$ respectively, then adiabatic coefficient $γ$ of the mixture is given through the relation___

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$\frac{(n_{1} {+n}_{2}) γ}{γ -1} = \frac{n_{1} γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}$

$n_{1} {+n}_{2} {= n}_{1} γ_{1} {+n}_{1} γ_{2}$

Both B and C

$(n_{1} + n_{2}) (γ - 1) = n_{1} (γ_{1} - 1) + n_{2} (γ_{2} - 1)$

answer is B.

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Detailed Solution

$C_{V} = \frac{R}{γ - 1}, C_{V_{1}} = \frac{R}{γ_{1} - 1} C_{V_{2}} = \frac{R}{γ_{2} - 1}$
But $C_{V_{m}} = \frac{n_{1} c_{V_{1}} + n_{2} c_{v_{2}}}{n_{1} + n_{2}}$
$= \frac{R}{γ - 1} (n_{1} + n_{2}) = \frac{n_{1} R}{γ_{1} - 1} + \frac{n_{2} R}{γ_{2} - 1}$
$\frac{n_{1} + n_{2}}{γ - 1} = \frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}$
But $V = \frac{C_{p}}{C_{v}} = \frac{n_{1} c_{P_{1}} + n_{2} {cp}_{2}}{n_{1} c_{v_{1}} + n_{2} c_{v_{2}}}$
But solves we get $\frac{n_{1} + n_{2}}{γ - 1} γ = \frac{n_{1} γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}$