Two perfect gases at absolute temperature $T_{1} and T_{2}$ are mixed. There is no loss of energy. Find the temperature of mixture if masses of molecules are $m_{1} and m_{2}$ and the number of molecules in the gases are $n_{1} and n_{2}$ respectively

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$T = n_{1} T_{1} + n_{2} T_{2}$

$T = \frac{T_{1}}{n_{1}} + \frac{T_{2}}{n_{2}}$

$T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

$T = \frac{n_{1}^{2} T_{1} + n_{2}^{2} T_{2}}{n_{1}^{2} + n_{2}^{2}}$

answer is C.

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Detailed Solution

For non-reactive gases
$* n_{mix} = n_{1} + n_{2} \dots (1)^{*} u_{mix} = u_{1} + u_{2} \dots (2)$
Consider $n_{mix} = n_{1} + n_{2}$
$\Rightarrow n_{1} = \frac{P_{1} V_{1}}{{RT}_{1}}; n_{2} = \frac{P_{2} V_{2}}{{RT}_{2}} \Rightarrow n_{m \dot{i} x} = \frac{P_{mi \dot{x}} V_{mix}}{{RT}_{mix}}$
Also $u_{m \dot{ix}} = u_{1} + u_{2}$
$\begin{array}{l} u_{1} = n_{1} f_{1} {RT}_{1}; u_{2} = n_{2} f_{2} {RT}_{2} \\ u_{mix} = n_{mix} f_{mix} {RT}_{mix} \end{array}$
Only for gases where $f_{1} = f_{2} = f_{mix}$
$T_{mix} = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$