An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁and temperature T₁. The other chamber has volume V₂ and contains same ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

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$\frac{T_{1} T_{2} (P_{1} V_{1} {+ P}_{2} V_{2})}{P_{1} V_{1} T_{2} {+ P}_{2} V_{2} T_{1}}$

$\frac{P_{1} V_{1} T_{1} {+ P}_{2} V_{2} T_{2}}{P_{1} V_{1} {+ P}_{2} V_{2}}$

$\frac{P_{1} V_{1} T_{2} {+ P}_{2} V_{2} T_{1}}{P_{1} V_{1} {+ P}_{2} V_{2}}$

$\frac{T_{1} T_{2} (P_{1} V_{1} {+ P}_{2} V_{2})}{P_{1} V_{1} T_{1} {+ P}_{2} V_{2} T_{2}}$

answer is A.

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Detailed Solution

Here Q = 0 and W = 0 .

Therefore from first law of thermodynamics $ΔU= Q+W= 0$

Internal energy of first vessel + Internal energy of second vessel = Internal energy of combined vessel

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

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

For first vessel $n_{1} = \frac{P_{1} V_{1}}{{RT}_{1}}$ and for second vessel $n_{2} = \frac{P_{2} V_{2}}{{RT}_{2}}$

$∴ T = \frac{\frac{P_{1} V_{1}}{{RT}_{1}} {× T}_{1} + \frac{P_{2} V_{2}}{{RT}_{2}} {× T}_{2}}{\frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}}}$

$= \frac{T_{1} T_{2} (P_{1} V_{1} {+P}_{2} V_{2})}{P_{1} V_{1} T_{2} {+P}_{2} V_{2} T_{1}}$

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