## Reversible Expansion

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

Ray Guo 4C
Posts: 90
Joined: Fri Sep 28, 2018 12:15 am

### Reversible Expansion

I understand mathematically and graphically why a reversible expansion does more work, but does gas undergoing irreversible expansion ends up with more internal energy because it does less work?

Te Jung Yang 4K
Posts: 63
Joined: Fri Sep 28, 2018 12:18 am

### Re: Reversible Expansion

It depends on the system at hand. If the system is isolated, heat cannot come into the reversible expansion and replace the additional energy loss, so the first law of thermodynamics tells us that in the case that both systems (reversible and irreversible) are isolated, the irreversible one would indeed have less internal energy (based off of U=q+w).

Ray Guo 4C
Posts: 90
Joined: Fri Sep 28, 2018 12:15 am

### Re: Reversible Expansion

Te Jung Yang 4K wrote:It depends on the system at hand. If the system is isolated, heat cannot come into the reversible expansion and replace the additional energy loss, so the first law of thermodynamics tells us that in the case that both systems (reversible and irreversible) are isolated, the irreversible one would indeed have less internal energy (based off of U=q+w).

Where does the energy goes if the two systems end up with a difference in their energy levels? For the one with lower internal energy at the end, what is the additional energy expenditure used for?

505211599
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### Re: Reversible Expansion

If the system is isolated, then the reversible expansion wil result in a lower internal energy.