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An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁and contains ideal gas at pressure P₁ and temperature T₁ . The other chamber has volume V₂ and contains 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

(1) Here Q = 0 and W = 0. Therefore from first law of thermodynamics ∆U = Q + W = 0

$∴$ Internal energy of the system with partition = Internal energy of the system without partition

$n_{1} C_{v} T_{1} + n_{2} C_{v} T_{2} = (n_{1} + n_{2}) C_{v} T ∴ T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

$B u t n_{1} = \frac{P_{1} V_{1}}{R T_{1}} a n d n_{2} = \frac{P_{2} V_{2}}{R T_{2}} ∴ T = \frac{\frac{P_{1} V_{1}}{R T_{1}} \times T_{1} + \frac{P_{2} V_{2}}{R T_{2}} \times T_{2}}{\frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}}}$

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

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