## Reversible expansion

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

Gabrielle Schwab 1G
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am

### Reversible expansion

Can someone explain how the work done during a reversible expansion of a gas is the maximum expansion work possible? I know it explains this in the book but it didn't make much sense to me.

Alli Foreman 2H
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am

### Re: Reversible expansion

You get more work out of a process when it is done slowly because less heat is lost to the surroundings. Since a reversible process is done infinitely slowly, you lose the least possible amount of work. Therefore a reversible process does the maximum work.

Yuanming Mao 3J
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

### Re: Reversible expansion

The work done by the gas in expansion equals Pex*delta V. During reversible reaction, at each stage of expansion the external pressure must be equal to internal pressure. Since the internal pressure is higher than the external pressure(that's why the gas expands), the Pin will decrease and Pex will increase gradually at each stage. That's why at each stage the Pexis the maximum. On the contrary, during irreversible reaction, the opposing pressure is less than the maximum possible(always the same and never increases) at each stage of the expansion, so that the potential of a system to do work is lost.

Chem_Mod
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### Re: Reversible expansion

Both of the explanation works. Well done!