Cyclic Process

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A gas is taken around a complete thermodynamic cycle and returns to its original state after several processes, so what is the net change in its internal energy over the full cycle?

RankGuru · Accepted Answer

As emphasised in NCERT Class 11, Chapter 12 (Thermodynamics), internal energy is a state function, meaning its value depends only on the current state of the system, not on the path taken to reach it. In a complete cycle the gas returns to its initial state, so the initial and final internal energies are identical and $\Delta U_{	ext{cycle}} = 0$. Applying the First Law over the cycle, $\Delta Q = \Delta \cup + \Delta W = 0 + \Delta W$, so the net heat absorbed equals the net work done by the gas. The option that $\Delta U$ equals the net heat absorbed is wrong because that heat instead equals the net work, not the internal energy change. The option that $\Delta U$ equals the net work is wrong for the same reason; over a cycle $\Delta U$ is strictly zero. The option that it is always negative for a clockwise cycle confuses the net work (positive for clockwise on a P-V diagram) with internal energy. A consistency check: since $U$ depends only on state variables like $T$ for an ideal gas, returning to the same $T$ guarantees the same $U$, so the cyclic change must vanish.

Cyclic Process

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