Q. 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) $\frac{T_{1}T_{2}(p_{1}V_{1} + p_{2}V_{2})}{p_{1}V_{1}T_{2} + p_{2}V_{2}T_{1}}$
B) $\frac{p_{1}V_{1}T_{1} + p_{2}V_{2}T_{2}}{p_{1}V_{1} + p_{2}V_{2}}$
C) $\frac{p_{1}V_{1}T_{2} + \ p_{2}V_{2}T_{1}}{p_{1}V_{1} + p_{2}V_{2}}$
D) $\frac{T_{1}T_{2}(p_{1}V_{1} + p_{2}V_{2})}{p_{1}V_{1}T_{1} + p_{2}V_{2}T_{2}}$
Explanation
(a) 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 ie, $U = U_{1} + U_{2}$
Assuming both gases have same degree of freedom, then
$U = \frac{f\left( n_{1} + n_{2} \right)RT}{2}$
and $U_{1} = \frac{fn_{1}RT_{1}}{2},\ \ U_{2} = \frac{fn_{2}RT_{2}}{2}$
Solving we get,
$T = \frac{\left( p_{1}V_{1} + p_{2}V_{2} \right)T_{1}T_{2}}{p_{1}V_{1}T_{2} + p_{2}V_{2}T_{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.