## Potential to do work in Irreversible vs. Reversible Expansion

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Amanda Wu 2C
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

### Potential to do work in Irreversible vs. Reversible Expansion

From reading the textbook and from the calculus (integral) point of view, I understand that work done by a system in irreversible expansion is less than that of the system in reversible expansion, which the book attributes to a less than maximal possible opposing pressure at each stage of expansion (in irreversible expansion). I was wondering if anyone can explain why or what causes the opposing pressure to be less than maximal in irreversible expansion as opposed to in reversible expansion.

Tim Nguyen 2J
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

### Re: Potential to do work in Irreversible vs. Reversible Expansion

So in irreversible processes, the tendency is for the reaction to go toward completion as quickly as possible. In the image from an older response in the link below that I found, you can see that it abruptly changes to a lower than optimal constant pressure being used in the irreversible reaction (it is much more direct, but the trade off is the lost energy in the form of heat because the expansion is working against less external pressure). By having an external pressure that matches with the internal pressure doing expansion work, the most work is done here in the reversible process at every step.

Original Post (not my post)
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=5062

Image (not my work, all credit to the authors contributing to the post)