## Question 8.11 Reversible v Irreversible actions

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

Sophia Bozone 2G
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

### Question 8.11 Reversible v Irreversible actions

In Question 8.11, What is the difference between the (a) and the (b) in which it expands reversible and isothermally?
Thanks

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: Question 8.11 Reversible v Irreversible actions

In part a, the process is irreversible meaning there is an external, constant pressure and the amount of work is usually less than in part b, in which it's a reversible process, where more work is done. Equation for part a to use: w = -P*deltaV and for part b it's: w = -nRT ln (Vf/Vi).

In a reversible process, the system's usually in equilibrium so for a infinitesimal pressure change, there will be an infinitesimal change in volume to counteract change in pressure to try and keep system and surroundings at equilibrium.

In an irreversible process, there will always be an entropy change.

I would suggest looking at Lavelle's graphs for irreversible vs. reversible processes (which is posted near the outline for chapter 8) to understand why reversible processes always end up doing more work than irreversible processes (good rule of thumb to know).

Hope that helps!

-Yashwi

Yang Chen 2E
Posts: 36
Joined: Thu Jul 13, 2017 3:00 am

### Re: Question 8.11 Reversible v Irreversible actions

Basically a reversible process means that if an infinitely small change happens to a variable, the process can be reversed. Page 265 of the textbook goes into depth about this.