Work done by a gas and the p-V diagram · Physics Class 11–12 | StemPy Lab
Course 11.1·Lesson 3 of 3
Work done by a gas and the p-V diagram
A gas that expands does work W=∫pdV, the area under its path on a pressure-volume graph. At a steady pressure W=pΔV
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Try the code and the simulations beside the lesson.
Work by a pushing gas
A gas pushes on a piston of area A with force pA. If the piston moves out a small distance dx, the gas does work pAdx=pdV. Over a whole expansion:
W=∫pdV
which is the area under the path on a graph of p against V. Expanding (moving right) gives positive work; squeezing gives negative work.
Two graphs of pressure against volume. On the left, a gas expands at a constant pressure from V one to V two, and the rectangle under the line is shaded. It is the work done by the gas, W equal to p times V two minus V one, and it is positive. On the right, a gas is compressed from V two back to V one along a curve, and the area under the curve is shaded. The work done by the gas is negative, because the volume decreases.
Two simple paths
Steady pressure (isobaric): a rectangle, W=pΔV. For an ideal gas this is also nRΔT.
Steady temperature (isothermal): p=VnRT, a curve, and W=nRTlnV1V2.
You should find that the steady-pressure path gives more work, because its pressure does not fall as the gas expands.
Paths and cycles
Between the same two states, different paths enclose different areas, so the work depends on the path. Go round a closed loop, and the net work is the area inside the loop: positive if the loop runs clockwise, negative if anticlockwise.
Watch out. The area must be in Pa×m3=J. Litres must become cubic metres.
A worked example
A gas at 3.0×105Pa and 2.0L expands to 6.0L. Find the work at a steady pressure and at a steady temperature.
Steady pressure.3.0×105Pa×4.0×10−3m3=1200J.
Steady temperature.pV=600J, so W=600Jln3=659J.
So the same start and end volumes give quite different work.
Practise with new numbers
Try it in Python
The program works out both paths of the example.
Remember these
W=∫pdV: the area under the path.
Steady p: pΔV; steady T: nRTlnV1V2.
Work depends on the path; a cycle's net work is the area enclosed.
Check yourself
Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.
A gas at a steady 2.0×105Pa shrinks from 5.0L to 3.0L. What work does the gas do?
A gas is heated in a sealed rigid tank. How much work does it do?
A loop on a p-V graph runs anticlockwise. Is the net work done by the gas positive or negative?
Why does the work of an isothermal expansion depend on lnV1V2, not on V2−V1?
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