A real gas within a closed chamber at $27^{\circ} \mathrm{C}$ undergoes the cyclic process as shown in figure. The gas obeys $P V^{3}=\mathrm{RT}$ equation for the path $A$ to $B$. The net work done in the complete cycle is (assuming $R=8 \mathrm{~J} / \mathrm{molK}$ ):

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Heat & Thermodynamics

A real gas within a closed chamber at

2 7^{\circ} C

undergoes the cyclic process as shown in figure. The gas obeys

P V^{3} = RT

equation for the path

A

B

. The net work done in the complete cycle is (assuming

R = 8 J / molK

(A)225 J
(B)205 J✓ Correct
(C)20 J
(D) $- 20 J$

Answer: (B)

Solution

From the P–V diagram:

A (V = 2, P = 20) \to B (V = 4)

along

P V^{3} = R T

;

B \to C (V = 2, P = 10)

at constant

P = 10

;

C \to A

at constant

V = 2

.

Work from A to B, using

P = \frac{R T}{V ^{3}}

with

T = 2 7^{\circ} C = 300

R = 8

J/molK, so

R T = 2400

W_{A B} = \int_{2}^{4} \frac{R T}{V ^{3}} d V = R T [- \frac{1}{2 V ^{2}}]_{2}^{4} = R T (\frac{1}{8} - \frac{1}{32}) = 2400 \times \frac{3}{32} = 225 J

Work from B to C (constant pressure,

V

decreasing

4 \to 2

W_{B C} = P Δ V = 10 (2 - 4) = - 20 J

Work from C to A (constant volume) is zero.

W_{cy c l e} = 225 - 20 = 205 J

Note: the value of

R T

used here (from

R = 8

J/molK,

T = 300

K) is not the same as

P V^{3}

evaluated at point A (

20 \times 2^{3} = 160

), so the given data for the A→B leg is internally inconsistent — this was flagged by the official source as open to challenge. Using the official approach (as above) gives net work

= 205

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Heat & Thermodynamics

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