## Irreversible Pathway Picture

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

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RyanTran2F
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

### Irreversible Pathway Picture

Hi can someone clarify the picture we saw with the piston on Week 3 Friday's lecture? When it says "No work done" pointing from the piston to the right, is it just saying we do not consider the work being done to push the piston? However there IS work being done right (since we have the equation w = -PΔV)? Also why does it say "T changes along pathway" for irreversible but T is constant for reversible? Thanks :)

nickjadidian 1A
Posts: 20
Joined: Fri Sep 29, 2017 7:04 am

### Re: Irreversible Pathway Picture

There's physical movement, yes. But this doesn't mean there is "work" in the traditional sense. The change in the piston's position is a result of entropy. This only works because the piston is in a vacuum. delta U =0, q=0, and w=0. The point being made in this diagram was that there is no change in the internal energy of the system, but the volume shift is a function of the gas' tendency to take up a greater volume. It has a tendency to occupy as many possible positions as possible (entropy).

Rachel Formaker 1E
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Joined: Fri Sep 29, 2017 7:04 am
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### Re: Irreversible Pathway Picture

The reason that no work is done in this picture is because the gas is expanding into a vacuum. There is no external pressure for in the vacuum, so since there is nothing for the gas to push against as it expands, it is not doing any work.
The textbook calls this expansion against zero pressure "free expansion."

RyanTran2F
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

### Re: Irreversible Pathway Picture

So if the external environment was not a vacuum then, to move the piston, there would be work done and this would not be simply due to entropy but because work was done to push the piston against the external pressure? Thanks :)

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