The First Law of Thermodynamics — Question 2

Question 2

Different paths share an internal-energy difference

Use $\Delta U=Q-W$, where $Q$ is heat into the system and $W$ is work done by it. Path A from state X to Y transfers 1200 J inward as heat while the system performs 450 J of work. A second path B connects the same states and requires 200 J of work input.

1. Find the common $\Delta U$ and the heat transfer on path B.

2. Return from Y to X along path C, for which 300 J of heat leaves. Find $W_C$ and the net work for the cycle using A followed by C.

Question 2 - Solution

Different paths share an internal-energy difference

(a) Method and calculation

$$U_Y-U_X=Q_A-W_A=1200-450=750\,\mathrm J.$$$$Q_B=\Delta U+W_B=750-200=550\,\mathrm J.$$

(b) Interpretation and further calculation

$$\Delta U_C=-750\,\mathrm J,\quad W_C=Q_C-\Delta U_C=-300+750=450\,\mathrm J.$$$$W_{A+C}=450+450=900\,\mathrm J=Q_{A+C}=1200-300.$$Path C does positive work while internal energy falls; rejecting heat does not require work input.

Physical check

Only the internal-energy differences reverse automatically when endpoints reverse; the work depends on the chosen return path.