An insulated container of two diatomic gases has two chambers separated by an insulting partition. One of the chambers has volume $V_{1}$ and contains ideal gas at pressure $p_{1}$ and temperature $T_{1}$ . The other chamber has volume $V_{2}$ and contains ideal gas at pressure $p_{2}$ and temperature $T_{2}$ . If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

\frac{T_{1} T_{2} (p_{1} V_{1} + p_{2} V_{2})}{p_{1} V_{1} T_{1} + p_{2} V_{2} T_{2}}

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\frac{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}{p_{1} V_{1} + p_{2} V_{2}}

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\frac{p_{1} V_{1} T_{1} + p_{2} V_{2} T_{2}}{p_{1} V_{1} + p_{2} V_{2}}

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\frac{T_{1} T_{2} (p_{1} V_{1} + p_{2} V_{2})}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}

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Solution

The correct option is D $\frac{T_{1} T_{2} (p_{1} V_{1} + p_{2} V_{2})}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}$
As no work is done and system is thermally insulated from surrounding, it means sum of internal energy of gas in two partitions is constant i.e., $U = U_{1} + U_{2}$ .

Assuming both gases have same degree of freedom, then
$U = \frac{f (n_{1} + n_{2}) R T}{2} U_{1} = \frac{f (n_{1} R T_{1})}{2}, U_{2} = \frac{f n_{2} R T_{2}}{2}$

$U = \frac{f (n_{1} + n_{2}) R T}{2} = \frac{f (n_{1} R T_{1})}{2} + \frac{f n_{2} R T_{2}}{2}$

$U = \frac{f (n_{1} + n_{2}) R T}{2} = \frac{f (n_{1} R T_{1})}{2} + \frac{f n_{2} R T_{2}}{2}$
$T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$
Substituting
$n_{1} = \frac{p_{1} V_{1}}{R T_{1}}$ , $n_{2} = \frac{p_{2} V_{2}}{R T_{2}}$

we get,
$T = \frac{(p_{1} V_{1} + p_{2} V_{2}) T_{1} T_{2}}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}$

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