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

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_(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

A

Explanation

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))`