Some Special Cases of the First Law of Thermodynamics — Question 9

Question 9

Insulated and rigid constraints eliminate different terms

Use $\Delta U=Q-W$, with work by the gas positive. Two closed stationary apparatuses each contain a substance with $\Delta U=C\Delta T$ and $C=100\,\mathrm{J/K}$. Apparatus A is rigid but thermally conducting: it receives 500 J of heat and has no other work channel. Apparatus B is insulated but has a movable piston: 500 J of compression work is done on it.

1. Find $(Q,W,\Delta U,\Delta T)$ for each apparatus.

2. A third apparatus is both rigid and insulated but receives 500 J through electrical leads. Find its energy and temperature changes. Which additional restriction would make both rigidity and insulation imply $\Delta U=0$?

Question 9 - Solution

Insulated and rigid constraints eliminate different terms

(a) Method and calculation

$$\text{A:}\quad Q=500,\quad W=0,\quad\Delta U=500\,\mathrm J,\quad\Delta T=5.00\,\mathrm K.$$$$\text{B:}\quad Q=0,\quad W=-500,\quad\Delta U=500\,\mathrm J,\quad\Delta T=5.00\,\mathrm K.$$Equal temperature changes can arise from different boundary-transfer mechanisms.

(b) Interpretation and further calculation

$$\text{C:}\quad Q=0,\quad W_{pV}=0,\quad W_{\mathrm{el}}=-500\,\mathrm J.$$$$\Delta U=500\,\mathrm J,\qquad\Delta T=5.00\,\mathrm K.$$To infer zero energy change, exclude electrical, shaft and all other work channels as well; a closed stationary system then has neither heat nor work exchange.

Physical check

Geometric and thermal constraints are independent; neither alone removes every possible transfer.