## Midterm 4a

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

JennyCKim1J
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### Midterm 4a

For 4A, how do I know that I'm using irreversible eq. instead of reversible eq?

Emma Li 2C
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### Re: Midterm 4a

I think it's because a constant external pressure is characteristic of an irreversible process, and the pressures of the system vs surroundings being at equilibrium is characteristic of a reversible process.

Joshua Hughes 1L
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### Re: Midterm 4a

usually when there is a large volume change due to a difference or change in pressure(ie the external pressure). Large and/or fast reactions would likely be irreversible

reversible reactions are smaller, continuous little fluctuations/changes and often (in this class) are isothermal.
When doing this on the exam though, the moment I saw a constant External Pressure that it was expanding against I immediately thought of the equation w = - P(external) x Delta V and so I just went with it

Pooja Nair 1C
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### Re: Midterm 4a

If it helps, draw the two diagrams of reversible and irreversible reaction. Since there is no change in pressure (they don't give you a second value), then you don't have to integrate to find work, and you can just use w = -P * delta V

Rachel Wang
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### Re: Midterm 4a

I assumed I use w=-PdeltaV when it's isothermal and isobaric.