## Work for reversible vs irreversible process

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

Nancy Le - 1F
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Joined: Fri Sep 29, 2017 7:07 am

### Work for reversible vs irreversible process

Why is it that more work is done for a reversible process than for a irreversible process?

kaushalrao2H
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### Re: Work for reversible vs irreversible process

The book quotes, "If the external pressure were to be increased even infinitesimally at any stage of the (ideal gas) expansion, the piston would move in instead of out. Therefore, the work done during a reversible expansion of a gas is the maximum expansion work possible."

Nhan Nguyen 2F
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### Re: Work for reversible vs irreversible process

Reversible processes are idealized as being fast and efficient. They have more external pressure pushed upon them. Therefore, they perform really slow in transforming all heat into work (constantly maximizing work). However, no real processes are reversible since it just takes way too much time. We turn to irreversible processes to help us do the work faster, but faster reaction means that not all heat will be converted to work.

Janine Chan 2K
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Joined: Fri Sep 29, 2017 7:04 am

### Re: Work for reversible vs irreversible process

Reversible processes are very slow. Because it is done slowly, less energy is lost to the surrounding as heat. In the gas and piston example, an extremely small increase in external pressure would cause the gas to be compressed and the piston to move in. If the process were to be done quickly (aka irreversibly), you can think of it as creating friction and causing more heat loss. Hopefully that makes some kind of sense.