The First Law of Thermodynamics — Question 3

Question 3

Simultaneous electrical input and piston output must both be counted

Use $\Delta U=Q-W$, where $Q$ is heat into the system and $W$ is work done by it. A closed gas-plus-heater assembly absorbs 500 J through its outer thermal boundary. Electrical leads supply 900 J, and the gas pushes a piston outward at constant external pressure 150 kPa through .00400 m$^3$. The assembly is otherwise stationary, with no other transfers.

1. Find net work done by the assembly and its internal-energy change.

2. If the same assembly were mechanically clamped at fixed volume while receiving the same thermal and electrical inputs, find its internal-energy change and the difference between runs.

Question 3 - Solution

Simultaneous electrical input and piston output must both be counted

(a) Method and calculation

$$W_{pV}=p_{\mathrm{ext}}\Delta V=150000(.00400)=600\,\mathrm J.$$$$W_{\mathrm{total}}=W_{pV}+W_{\mathrm{electrical}}=600-900=-300\,\mathrm J.$$$$\Delta U=500-(-300)=800\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{\mathrm{clamped}}=0-900=-900\,\mathrm J,\quad\Delta U_{\mathrm{clamped}}=500+900=1400\,\mathrm J.$$$$\Delta U_{\mathrm{clamped}}-\Delta U_{\mathrm{moving}}=600\,\mathrm J.$$Suppressing the piston output retains that extra energy; it does not make the electrical work disappear.

Physical check

Fixed volume eliminates displacement work, not all possible work mechanisms.