Pressure Exerted by Piston on Gas

Physics Asked by Scott Scott on November 29, 2020

I’m currently doing a lab where we are measuring the work done by weights on a gas, where the weights are added on top of a piston. Now, what I don’t understand is that work is equal to net force times distance. Therefore, if I were to draw a free body diagram of the piston, it has its own weight, the force due to the weights, and the force due to the atmospheric pressure all going down. The force going up is gas pressure, which is equal to all of those other three forces combined. Now, if I were to add some more weights, the net force at that point would solely be the added weights going down, so work should then equal added weights times the distance the piston moved down. But what my lab answer booklet says is work is equal to the weight of the piston and weights and force due to atmospheric pressure. But why? That’s not the net force acting on the piston. I’m so confused as to how work is equal to the entire force acting down without subtracting the force acting up from the gas. Please help.

One Answer

Your lab booklet is calculating the work done on the gas, whereas you are trying to calculate the work done on the piston. The pressure of the gas is not part of the calculation of the work done on the gas because the gas cannot do work on itself.

Answered by gandalf61 on November 29, 2020

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