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 beA. `(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_(1)V_(1)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))`

An insulated container of gas has two chambers separated by an insulat

1.	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 beA. `(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_(1)V_(1)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» Correct Answer - A As no, work is done and system is thermally insulated from surrounding, it mean sum of internal energy of gas in two partitions is constant i.e, `U = U_(1) +U_(2)` Assume both gases have same degree of freedom, then Assuming both gases have same degree of freedom, then `U = (f(n_(1)+n_(2))RT)/(2)` and `U_(1) = (fn_(1)RT_(1))/(2), U_(2) = (fn_(2)RT_(2))/(2)` `n_(1) = (P_(1)V_(1))/(RT_(1))` and `(P_(2)V_(2))/(RT_(2))` Solving we get `T = ((p_(1)V_(1)+p_(2)V_(2))T_(1)T_(2))/(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))`

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