## Reversible vs irreversible

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

Jeremiah Hutauruk
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### Reversible vs irreversible

Does w=-PdeltaV only work when there is constant pressure and when the reaction is irreversible?

Chem_Mod
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### Re: Reversible vs irreversible

That is correct.

Cameron_Greenberg_3C
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### Re: Reversible vs irreversible

KimGiang2F
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### Re: Reversible vs irreversible

When there is constant pressure and the reaction is irreversible, the equation w=-Pex∆V is derived from P=F/A, in which to find the force of the equation you can use the relationship and get F=PexA. The work needed to drive the piston out through a distance d is, therefore, work= PexA •d (work=force•distance). From volume = area • height, it can be assumed that the product of area and distance moved is equal to the change in volume of the sample (A•d=∆V). So to get the final equation, you can replace A•d in the work equation with ∆V, resulting in w=-Pex∆V. The resulting equation is negative as when a system expands, it loses energy as work. Therefore, if ∆V is positive, w is negative.

Samantha Chang 2K
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### Re: Reversible vs irreversible

Yes, and remember that an irreversible reaction will always require less work.

Bianca Barcelo 4I
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### Re: Reversible vs irreversible

Will reversible expansion always do more work?

Ian Marquez 2K
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### Re: Reversible vs irreversible

Reversible expansion will always do more work. It is also clear to make this distinction by looking at some of the graphs and the area underneath the curve when comparing reversible and irreversible expansion.