JEE Advanced 2014 · Paper 2 · Physics Q12

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PhysicsSingle correct+3 / -1Question 12 of 60

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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 $C_V=\frac32R$ , $C_P=\frac52R$ , and those for an ideal diatomic gas are $C_V=\frac52R$ , $C_P=\frac72R$ .

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.

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

250R

200R

100R

-100R

Answer and worked solution

Official answer: D

Processes. With the partition free, both gases stay at the (constant) pressure set by the piston, so both change at constant pressure.

Final temperature.
$2\cdot\tfrac52R(700-T)=2\cdot\tfrac72R(T-400)\implies5(700-T)=7(T-400)\implies T=525\text{ K}.$

Work. At constant pressure $W=nR\Delta T$ :
$W=2R(525-700)+2R(525-400)=-350R+250R=-100R.$
(Equivalently: $Q_\text{total}=0$ , so $W=-\Delta U_\text{total}=-[3R(-175)+5R(125)]=-100R$ .)

(D)

Topics:Work done by a gas Specific heats of gases Thermodynamic processes

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JEE Advanced 2014 · Paper 2, Physics question 12