An insulated container of gas has two chambers separated by an insulating 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 contain same 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_{2} + P_{2} V_{2} T_{1}}

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\frac{P_{5} V_{1} T_{1} + P_{2} V_{2} T_{2}}{P_{1} V_{1} + P_{2} V_{2}}

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\frac{P_{1} V_{1} T_{5} + P_{2} V_{2} T_{1}}{P_{1} V_{1} + P_{2} V_{2}}

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\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{4})}{P_{1} V_{1} T_{1} + P_{2} V_{2} T_{6}}

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Solution

The correct option is D $\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}$
Let the final temperature be $T$ and $n_{1}$ & $n_{2}$ be the no. of moles of gas in each container

Using conservation of energy

$(n_{1} +_{2}) C_{V} T = n_{1} C_{V} T_{1} + n_{2} C_{V} T_{2}$

$\Rightarrow (\frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}}) T = \frac{P_{1} V_{1}}{R} + \frac{P_{2} V_{2}}{R}$

$\Rightarrow (\frac{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}{T_{1} T_{2}}) T = P_{1} V_{1} + P_{2} V_{2}$

$\Rightarrow T = \frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}$

Hence, the correct option is $A$

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