JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Heat & Thermodynamics

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One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are :
[R is the gas constant]

(A)Work done in this thermodynamic cycle $(1 \to 2 \to 3 \to 4 \to 1)$ is $∣ W ∣ = \frac{1}{2} RT_{0}$ ✓ Correct
(B)The ratio of heat transfer during processes $1 \to 2$ and $2 \to 3$ is $\frac{Q _{1 \to 2}}{Q _{2 \to 3}} = \frac{5}{3}$ ✓ Correct
(C)The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
(D)The ratio of heat transfer during processes $1 \to 2$ and $3 \to 4$ is $\frac{Q _{1 \to 2}}{Q _{3 \to 4}} = \frac{1}{2}$

Answer: (A), (B)

Solution

From the V-T graph:

Process $1 \to 2$ : isobaric with $P = \frac{R T _{0}}{V _{0}}$
Process $2 \to 3$ : isochoric with $V = 2 V_{0}$
Process $3 \to 4$ : isobaric with $P = \frac{R T _{0}}{2 V _{0}}$
Process $4 \to 1$ : isochoric with $V = V_{0}$

Work in cycle:

W = \frac{R T _{0}}{V _{0}} \cdot V_{0} - \frac{R T _{0}}{2 V _{0}} \cdot V_{0} = \frac{R T _{0}}{2}

Heat transfers:

Q_{1 - 2} = n C_{p} Δ T = 1 \cdot \frac{5 R}{2} \cdot T_{0}

Q_{2 - 3} = n C_{v} Δ T = 1 \cdot \frac{3 R}{2} \cdot T_{0}

\frac{Q _{1 - 2}}{Q _{2 - 3}} = \frac{5}{3}

Q_{3 - 4} = n C_{p} Δ T = 1 \cdot \frac{5 R}{2} \cdot \frac{T _{0}}{2}

\frac{Q _{1 - 2}}{Q _{3 - 4}} = 2

The cycle has isobaric and isochoric processes, not adiabatic.

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Heat & Thermodynamics

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