A Closer Look at Heat and Work — Question 10

Question 10

Expansion into a vacuum has no resisting-pressure work

A gas occupies one side of a rigid insulated container, initially 2.00 L. The other 3.00 L is evacuated. An internal partition is removed, and the gas fills the entire 5.00 L container. Neglect energy stored in the partition and external motion.

1. Find boundary work using the whole-container boundary and using the vacuum resistance at the moving gas front.

2. Contrast with a separate quasistatic expansion from 2.00 to 5.00 L at constant external pressure 100 kPa. Does the same volume change guarantee the same work?

Question 10 - Solution

Expansion into a vacuum has no resisting-pressure work

(a) Method and calculation

For the fixed outer boundary,$$dV_{\mathrm{outer}}=0\ \Rightarrow\ W_{\mathrm{outer}}=0.$$At the ideal vacuum front,$$p_{\mathrm{ext}}=0\ \Rightarrow\ W_{\mathrm{front}}=\int_2^5 0\,dV=0.$$The initially nonzero gas pressure does not imply a resisting load.

(b) Interpretation and further calculation

$$W_{100\,\mathrm{kPa}}=100(5-2)=300\,\mathrm J.$$$$\Delta V_{\mathrm{gas}}=3.00\,\mathrm L\quad\text{in both runs},\qquad W=0\ \text{or}\ 300\,\mathrm J.$$No uniform equilibrium pressure path should be invented during free expansion; the work calculation uses the actual boundary and resistance.

Physical check

The expansion volume alone supplies insufficient information to calculate mechanical work.