An excerpt from my textbook:

It is assumed that there is no macroscopic motion of the system. Therefore, $\Delta E_{mech} = 0$ and the first law of thermodynamics can be written as $$\Delta E_{sys} = \Delta E_{mech} + \Delta E_{th} = \Delta E_{th} = W + Q$$

But $\Delta E_{mech} = \Delta K + \Delta U$, the mechanical kinetic and potential energy. And so if, for example, a piston compresses an ideal gas and it is not an isobaric process, would the increase in pressure not increase the system's mechanical potential energy, since pressure is a restoring force?

More specifically I found this question in my textbook:

A gas cylinder and piston are covered with heavy insulation. The piston is pushed into the cylinder, compressing the gas. In this process the gas temperature: a) Increases. b) Decreases. c) Doesn't change. d) There's not sufficient information to tell.

The solution given is a), the temperature increases. But how would we know that the work done on the gas is not transferred to mechanical potential energy in the form of an increase in pressure of the gas?


1 Answer 1


First of all,

Internal Energy

The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes "potential + kinetic energy".

So, it doesn't matter whether it is potential energy or kinetic energy, both contribute to the internal energy of the system.

Now let's come to your textbook question

It is quite simple to understand.

As we know, the internal energy of an ideal gas is only a function of its temperature.

Here, Work is done on the gas to compress it and it increases the internal energy which further helps in increasing the temperature of the gas.

I hope, you understood the answer.

Thank you


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