JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Heat & Thermodynamics

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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,

C_{v} = 2 R

. Here,

R

is the gas constant. Initially, each side has a volume

V_{0}

and temperature

T_{0}

. The left side has an electric heater, which is turned on at very low power to transfer heat

Q

to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to

V_{0} /2

. Consequently, the gas temperatures on the left and the right sides become

T_{L}

and

T_{R}

, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

The value of

\frac{Q}{R T _{0}}

(A) $4 (22 + 1)$
(B) $4 (22 - 1)$ ✓ Correct
(C) $(52 + 1)$
(D) $(52 - 1)$

Answer: (B)

Solution

From the previous question:

T_{R} = 2 T_{0}

Final volumes:

V_{L} = \frac{3 V _{0}}{2}

V_{R} = \frac{V _{0}}{2}

For the right side (adiabatic):

P_{R} (\frac{V _{0}}{2})^{3/2} = P_{0} V_{0}^{3/2}

P_{R} = P_{0} \cdot 2^{3/2} = 22 P_{0}

For the left side, using ideal gas law:

\frac{P _{R} V _{L}}{T _{L}} = \frac{P _{0} V _{0}}{T _{0}}

\frac{2 2 P _{0} \cdot \frac{3 V _{0}}{2}}{T _{L}} = \frac{P _{0} V _{0}}{T _{0}}

T_{L} = 32 T_{0}

Heat supplied to the system:

Q = n C_{v} Δ T_{L} + n C_{v} Δ T_{R}

Q = 1 \times 2 R \times (32 - 1) T_{0} + 1 \times 2 R \times (2 - 1) T_{0}

Q = 2 R [(32 - 1) + (2 - 1)] T_{0}

Q = 2 R [42 - 2] T_{0} = 4 R (22 - 1) T_{0}

Therefore:

\frac{Q}{R T _{0}} = 4 (22 - 1)

Correct answer: (B)

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Single correct option

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Heat & Thermodynamics

Solution

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