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← JEE Advanced 2014 · Paper 2
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JEE Advanced 2014 · Paper 2, Physics question 11

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Paragraph For Questions 11 & 12

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 400 K. The heat capacities per mole of an ideal monatomic gas are CV=32RC_V=\frac32R, CP=52RC_P=\frac52R, and those for an ideal diatomic gas are CV=52RC_V=\frac52R, CP=72RC_P=\frac72R.

An insulated cylindrical container closed at the top by a movable piston; a horizontal partition divides the gas inside into an upper and a lower compartment.

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Choose one option
Answer and worked solution

Official answer: D

Processes. The partition is fixed, so the lower (monatomic) gas is heated or cooled at constant volume. The upper (diatomic) gas sits under a free piston, so it changes at constant pressure. No heat leaves the container.

Heat lost = heat gained.
2⋅32R(700−T)=2⋅72R(T−400)2\cdot\tfrac32R(700-T)=2\cdot\tfrac72R(T-400)
3(700−T)=7(T−400)  ⟹  10T=4900  ⟹  T=490 K.3(700-T)=7(T-400)\implies10T=4900\implies T=490\text{ K}.

(D)

Topics:First law and internal energySpecific heats of gases
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