In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400K. The heat capacities per mole of an ideal monatomic gas are CV=23R,CP=25R, and those for an ideal diatomic gas are CV=25R,CP=27R.
Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be
(A)250R
(B)200R
(C)100R
(D)−100R✓ Correct
Answer: (D)
Solution
When the partition is free to move, both gases undergo isobaric (constant pressure) processes. Heat balance:
n1CP1(700−T)=n2CP2(T−400)
2×25R(700−T)=2×27R(T−400)
5(700−T)=7(T−400)
3500−5T=7T−2800
12T=6300, so T=525 K.
Work done by monatomic gas: W1=n1RΔT1=2R(525−700)=−350R.
Work done by diatomic gas: W2=n2RΔT2=2R(525−400)=250R.
Total work: Wnet=W1+W2=−350R+250R=−100R.
JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option
