Two perfect gases at absolute temperatures $T_{1} a n d T_{2}$ are mixed. There is no loss of energy in this process. If $n_{1} a n d n_{2}$ are the respective number of molecules of the gases, the temperature of the mixture will be

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

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

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

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

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Solution

The correct option is A
$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Average kinetic energy per molecule of a perfect gas $= \frac{3}{2} k T$
$∴$ Average KE of molecules of the first gas $\frac{3}{2} n_{1} k T_{1}$
Average KE of molecules of the second gas = $\frac{3}{2} n_{2} k T_{2}$
$∴$ Total KE of the molecules of the two gases before they are mixed is
$K = \frac{3}{2} n_{1} k T_{1} + \frac{3}{2} n_{2} k T_{2} = \frac{3}{2} (n_{1} T_{1} + n_{2} T_{2}) k (1)$
If T is the temperature of the mixture, the kinetic energy of the molecules $(n_{1} + n_{2})$ in the mixture is
$K^{'} = \frac{3}{2} (n_{1} + n_{2}) k T (2)$
Since there is no loss of energy K=K'. Equating (1) and (2) we get
$n_{1} T_{1} + n_{2} T_{2} = (n_{1} + n_{2}) T o r T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

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