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An insulated container of gas has two chambers separated by an insulating partition. One chamber has volume V1 and 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

1. T1T2(P1V1 + P2V2)/(P1V1T2 + P2V2T1)
2. P1V1T1 + P2V2T2)/(P1V1 + P2V2)
3. (P1V1T2 + P2V2T1)/(P1V1 + P2V2)
4. T1T2(P1V1 + P2V2)/(P1V1T1 + P2V2T2)

Correct answer: T1T2(P1V1 + P2V2)/(P1V1T2 + P2V2T1)

Solution

Moles n1 = P1V1/(RT1), n2 = P2V2/(RT2); conserving (n1+n2) Cv T_f = n1 Cv T1 + n2 Cv T2 gives T_f = T1 T2 (P1V1 + P2V2)/(P1V1 T2 + P2V2 T1).

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