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Q.

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 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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a

T1T2P1V1+P2V2P1V1T2+P2V2T1

b

P1V1T1+P2V2T2P1V1+P2V2

c

P1V1T2+P2V2T1P1V1+P2V2

d

T1T2P1V1+P2V2P1V1T1+P2V2T2

answer is A.

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Detailed Solution

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=U1+U2 Assuming both gases have same dof, then U=f(n1+n2)RT2And U1=fn1RT12,U2=fn2RT22​Solving we get T=P1V1+P2V2T1T2P1V1T2+P2V2T1
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