Two monoatomic ideal gas at temperature $T_{1}$ and $T_{2}$ are mixed. There is no loss of energy. If the masses of molecules of the two gases are $m_{1}$ and $m_{2}$ and number of their molecules are $n_{1}$ and $n_{2}$ respectively. The temperature of the mixture will be:

\frac{T_{1} + T_{2}}{n_{1} + n_{2}}

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\frac{T_{1}}{n_{1}} + \frac{T_{2}}{n_{2}}

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\frac{n_{2} T_{1} + n_{1} T_{2}}{n_{1} + n_{2}}

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\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}

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Solution

The correct option is C $\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$
Net P-V work done=0 as no net change in volume takes place.
Also net heat absorbed= 0
Thus by first law, net internal energy change=0
$\Rightarrow n_{1} C_{v_{1}} Δ T_{1} + n_{2} C_{v_{2}} Δ T_{2} = 0$

Both gases are monoatomic $\Rightarrow C_{v_{1}} = C_{v_{2}} = \frac{3}{2} R$
Let final temperature of mixture at equilibrium be T
Then $\frac{3}{2} R (n_{1} (T - T_{1}) + n_{2} (T - T_{2})) = 0 \Rightarrow T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

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