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Thermodynamics

JEE Main level, single correct, medium

52 of 190

Question 52 of 190

A sample of an ideal gas is taken through the cyclic process $abca$ as shown in the figure. The change in the internal energy of the gas along the path $ca$ is $-180J.$ The gas absorbs $250J$ of heat along the path $ab$ and $60J$ along the path $bc$ . The work done by the gas along the path $abc$ is:

Show answer and solution

Answer: (A) $130J$

In a cyclic process,

\Delta U=0

From 1st law of thermodynamics

Q=\Delta U+W

Q=W

{Q}_{ab}+{Q}_{bc}+{Q}_{ca}={W}_{abc}+{W}_{ca}

{W}_{abc}={Q}_{ab}+{Q}_{bc}+{Q}_{ca}-{W}_{ca}

{W}_{abc}={Q}_{ab}+{Q}_{bc}+{U}_{ca}

{W}_{abc}=250+60-180=130J

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JEE Main 2025 Apr 2 Shift 1 question paper

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