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 othe 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

Options

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)`

Answer

A

Explanation

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)`.