Chemical Thermodynamics: Diagram-Based Questions (7)

Question

A gas is taken through the cyclic process: A→B isobaric expansion, B→C isochoric pressure drop, C→A isothermal compression. Given $T_A = 300$ K, $V_A = 1$ L, $P_A = 2$ atm, and $V_B = 2$ L. Find $\Delta U$, $W$, and $Q$ for the full cycle.

Solution — Step by Step

Final answer: $\Delta U_{cycle} = 0$, $W_{cycle} \approx 62.2$ J, $Q_{cycle} \approx 62.2$ J.

Why This Works

Internal energy depends only on state, so any closed cycle has $\Delta U = 0$. The first law $\Delta U = Q - W$ then gives $Q = W$ for the cycle. The net work equals the area enclosed by the cycle on a $P$-$V$ diagram.

This is true for any cycle, regardless of the working substance (ideal gas, real gas, refrigerant) — a powerful conceptual shortcut.

Alternative Method

Compute $Q$ for each segment using $C_p$ and $C_v$, then sum. Slower but useful when individual segment heats are asked.