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=fn2RT22Solving we get T=P1V1+P2V2T1T2P1V1T2+P2V2T1