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An insulated container of gas has two ch...

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

A

`(T_1 T_2 (P_1 V_1 + P_2 V_2))/(P_1 V_1 T_2 + P_2 V_2 T_1)`

B

`(P_1 V_1 T_1 + P_2 V_2 T_2)/(P_1 V_1 + P_2 V_2)`

C

`(P_2 V_2 T_2 + P_2 V_2 T_1)/(P_1 V_1 + P_2 V_2)`

D

`(T_1 T_2 (P_1 V_1 + P_2 V_2))/(P_1 V_1 T_1 + P_2 V_2 T_2)`

Text Solution

Verified by Experts

The correct Answer is:
A
As no work is done and the system is thermally insulated from the surrounding, the sum of the internal energy of the gas in the two partitions is constant, i.e., `U=U_1+U_2`
If both gases have the same degrees of freedom, then
`U=(f(n_1+n_2)RT)/(2) implies U_1= (fn_1RT_1)/(2),U_2= (fn_2RT_2)/(2)`
Solving, we get : `T= ((p_1v_1+p_2v_2)T_1T_2)/(p_1v_1T_2+p_2v_2T_1)`.
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