Two moles of an ideal monoatomic gas undergo a cyclic process as shown in figure. The temperatures in different states are given as 6 T₁ = 3 T₂ = 2 T₄ = T₃ = 2400 K. The work done by the gas during the complete cycle is ( R = Universal gas constant) (a) -1600 R (b) 1600 R (c) -1200 R (d) 800 R

Answer: (a). T₁ = 400 K, T₂ = 800 K, T₃ = 2400 K, T₄ = 1200 K. 1 to 2 and 3 to 4 are at constant P (vertical lines on the T-P graph); 2 to 3 and 4 to 1 lie on lines through the origin ( T ∝ P), so they are isochoric with zero work. W₁₂ = nRΔ T = 2R(800 - 400) = 800R; W₃₄ = 2R(1200 - 2400) = -2400R. Wcycle = 800R - 2400R = -1600R So the answer is (a) -1600R.

Two moles of an ideal monoatomic gas undergo a cyclic process as… | MHT CET 2025 (23 Apr, Shift 1)

Q26MHT CET 2025 · 23 APR · SHIFT 1Difficult

P-V Diagrams & Cyclic Processes

Two moles of an ideal monoatomic gas undergo a cyclic process as shown in figure. The temperatures in different states are given as $6\,\mathrm{T}_1 = 3\,\mathrm{T}_2 = 2\,\mathrm{T}_4 = \mathrm{T}_3 = 2400\text{ K}$ . The work done by the gas during the complete cycle is ( $\mathrm{R} =$ Universal gas constant)

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Answer: (a)

$T_1 = 400$ K, $T_2 = 800$ K, $T_3 = 2400$ K, $T_4 = 1200$ K.
1 to 2 and 3 to 4 are at constant $P$ (vertical lines on the T-P graph); 2 to 3 and 4 to 1 lie on lines through the origin ( $T \propto P$ ), so they are isochoric with zero work.
$W_{12} = nR\Delta T = 2R(800 - 400) = 800R$ ; $W_{34} = 2R(1200 - 2400) = -2400R$ .
$W_{cycle} = 800R - 2400R = -1600R$
So the answer is (a) $-1600R$ .

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