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A while ago at university we did this experiment:

http://dev.physicslab.org/Document.aspx?doctype=2&filename=Thermodynamics_HeatEngineLab.xml

enter image description here

In the experiment we had a piston with air inside, an air tank connected to the piston through a hose, and an electric sensor measuring both the volume and pressure inside the piston. A software then drew a PV diagram as we inserted a weight on top of the piston, immersed to air tank to hot water, removed the weight after the expanding air raised it to maximum height, removed the air tank from hot water, and waited it to lower again to the low position.

Here is a picture from the same website showing the resulting diagram:

enter image description here

Our diagram after the experiment was very close, neglecting some expected noise.

But we were left wondering: Why is the pressure constant during expansion (between B-C and D-A)?

We knew that obviously the pressure increases as the air heats from the hot water, but the line was pretty much perfectly straight. I would have assumed that many factors affect how the pressure rises (such as heat conduction properties of the tank, the exact temperature of the water). But somehow the pressure was (as far as we could tell with our naked eyes) constant. Somehow as the temperature of the air inside the piston rises, the piston moves up in perfect proportion to increase volume and lower the pressure just the right amount to account for the rise in pressure due to the temperature increase. Why is it so? Many others also repeated this experiment, showing the same results even though exact pressures and temperatures varied.

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  • $\begingroup$ Can you make up something from the fact that the process is slow (Quasistatic)? $\endgroup$
    – Das
    Oct 8, 2018 at 16:08
  • $\begingroup$ @SwapnilDas Well, kind of; just that as the temperature increases very slightly, the pressure increases slightly, moves the piston up very slightly, and the pressure stays the same. But it doesn't really help me with intuition here.. $\endgroup$
    – S. Rotos
    Oct 8, 2018 at 16:56

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The comment from @SwapnilDas was a good hint for you.

For a slow process (quasi-static), the force pushing down on the piston is constant, equal to the weight of the piston and block.

The opposing force is the air pressure times the piston area. The piston area and weights are constant, so the air pressure must also be constant.

$$F=P\cdot{}A$$

In other words, the pressure must be constant because the piston force is constant.

If instead the piston were held in place as the water is heated, then both the pressure and the piston force would increase by the same ratio.

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