Physics II / Work and the Joule Experiment
Practice question · Numerical answer

A gas is compressed at a constant external pressure of 200 kPa from 0.050 m3^3 to 0.020 m3^3. Compute the work done on the gas, in kJ.

Hints
  1. Work on the gas is minus the external pressure times the volume change.
  2. Compute -200 x (0.020 - 0.050), in kPa and cubic metres.
Show the answer

6 (answers within ±0.2 count)

Why

W=PΔV=200×(0.030)=+6.0W = -P\Delta V = -200 \times (-0.030) = +6.0 kJ. Compression means the surroundings do work on the gas, so the sign is positive from the system’s point of view.

Read the lesson: Work and the Joule Experiment →

Practise Work and the Joule Experiment

The app has 5 more questions on this lesson, and keeps your place in the course. Physics II is free to start.

More questions on Work and the Joule Experiment