## Irreversible expansion

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

Cristian Carrasco 1F
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### Irreversible expansion

For irreversible expansion, why is w=P x delta V used instead of w= -nRt lnV2/V1?

Chem_Mod
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### Re: Irreversible expansion

For $W = -P\Delta V$
P is constant

For $W = -nRTln\frac{V_{_{2}}}{V_{1}}$
T is constant.

So they're used for different conditions.

Mike Vinci 2B
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### Re: Irreversible expansion

For W=-nRTln v2/v1, we account for the infinitesimal and consistent change in expansion of the system. In class, we saw the curve Professor Lavelle drew whose shape was that of exponential decay. This comes from the use of the natural log. However, the important aspect to understand is that such a curve like that increases the space beneath the curve, and for that reason shows an increase in the change of enthalpy. For w=-p * delta V, we have only the change of space given by delta v, and with constant pressure, the space becomes fixed and therefore represents a fixed rectangle as Professor Lavelle also drew.

Amelia Georgius 1K
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### Re: Irreversible expansion

When pressure is constant, you can use the equation w= -PdeltaV, but when temperature is constant, you can use the other equation, w= -nRT lnV2/V1.

Shanmitha Arun 1L
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### Re: Irreversible expansion

Simply put, it matters when P and T are constant. Based on that, certain equations are used.