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JEE Advanced 2021 · Paper 2, Physics question 16

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A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, CV=2RC_V = 2R. Here, RR is the gas constant. Initially, each side has a volume V0V_0 and temperature T0T_0. The left side has an electric heater, which is turned on at very low power to transfer heat QQ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V0/2V_0/2. Consequently, the gas temperatures on the left and the right sides become TLT_L and TRT_R, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

An insulated cylinder divided by an insulating movable partition into two equal compartments, each marked T0, V0. A heater coil is inside the left compartment.

The value of QRT0\dfrac{Q}{RT_0} is

Choose one option
Answer and worked solution

Official answer: B

Final pressure. The right side is compressed adiabatically:
p=p0 2γ=22 p0.p=p_0\,2^\gamma=2\sqrt2\,p_0.
The frictionless partition makes both sides equal in pressure.

Left temperature. The left side has volume 32V0\frac32V_0, so
TL=pVLR=22⋅32⋅T0=32 T0.T_L=\frac{pV_L}R=2\sqrt2\cdot\frac32\cdot T_0=3\sqrt2\,T_0.

Heat supplied. The whole system does no external work, so all the heat goes into internal energy:
Q=2R(32−1)T0+2R(2−1)T0=4RT0(22−1).Q=2R(3\sqrt2-1)T_0+2R(\sqrt2-1)T_0=4RT_0(2\sqrt2-1).

So QRT0=4(22−1)\dfrac Q{RT_0}=4(2\sqrt2-1). (B)

Topics:First law and internal energyThermodynamic processesIdeal gas equation
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