Irreversible change - thermodynamics

[Grand]

Homework Statement

I'm reading a book on Thermodynamic processes, and they argue that for an irreversible change

dW\geq-PdV

and I can't explain to myself why is this. Can anyone help?

 

[Andrew Mason]

Homework Statement

I'm reading a book on Thermodynamic processes, and they argue that for an irreversible change

dW\geq-PdV

and I can't explain to myself why is this. Can anyone help?

A reversible expansion or compression is one that is quasi-static: it requires an external pressure that is essentially equal to (ie. infinitessimally lower or higher than) the internal pressure of the gas. So the work done BY the gas in a reversible change is always dW = PdV.

It appears that your book uses the old convention that dW = work done by surroundings ON the gas = -PdV in a reversible expansion / compression. I suggest you use dW = work done on the surroundings BY the gas. (ie dW = PdV for a reversible change). Then revert back to your convention. It is just less confusing.

If an expansion is not quasi-static (the external pressure is lower than the gas pressure by more than an infinitessimal amount ) the work done on the surroundings by the gas (external P x dV) will be less than the internal P x dV. So dW < PdV where P = internal gas pressure and dW = work done ON the surroundings BY the gas.

If you use the convention in your book, where dW = work done ON the gas by the surroundings, you have to add a - sign and that changes the inequality: dW > -PdV.

For an irreversible compression it is a little different. The external pressure does work on the gas. But the work done ON the gas is always internal P x dV whether the external pressure is greater than the internal pressure (irreversible) or equal to it (reversible).

So you end up with dW \geq -PdV where dW = work done ON the gas by the surroundings. If dW = work done BY the gas dW \leq PdV

AM