A Closer Look at Heat and Work — Question 5

Question 5

Piston friction creates a work difference on reversal

A piston of area $.0100\,\mathrm{m^2}$ moves slowly against a constant baseline load pressure 100 kPa. Dry friction has constant magnitude 100 N, opposing motion. Consider expansion by .00300 m$^3$ followed by compression through the same volume interval. Neglect changes in piston kinetic energy.

1. Find the required gas pressure on expansion and on compression, and gas work on each leg.

2. Find net gas work over the out-and-back displacement and verify it equals the energy dissipated by friction.

Question 5 - Solution

Piston friction creates a work difference on reversal

(a) Method and calculation

$$p_{\mathrm{exp}}=100000+100/.0100=110000\,\mathrm{Pa}.$$$$p_{\mathrm{comp}}=100000-100/.0100=90000\,\mathrm{Pa}.$$$$W_{\mathrm{exp}}=330\,\mathrm J,\quad W_{\mathrm{comp}}=-270\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{\mathrm{net}}=60.0\,\mathrm J,\quad\Delta x=\Delta V/A=.300\,\mathrm m.$$$$E_{\mathrm{friction}}=2F_f\Delta x=2(100)(.300)=60.0\,\mathrm J.$$The baseline load's net displacement work cancels; friction does not. A complete thermodynamic cycle would require restoration of all other state variables too.

Physical check

The pressure difference on reversal is $2F_f/A$, because friction reverses direction.