## Work with changeP and constant V

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

Emilie Hoffman 1E
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

### Work with changeP and constant V

Why is there no work done when a systems pressure changes but its volume does not?

Riya Pathare 2E
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### Re: Work with changeP and constant V

work is equal to -P*deltaV so when there is no volume change delta V will be 0 making work also equal to 0

SantanaRodriguezDis1G
Posts: 56
Joined: Sat Jul 22, 2017 3:00 am

### Re: Work with changeP and constant V

if delta v is equal to zero then work is equal to zero by the equation w=-pdeltav

SPandya1F
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### Re: Work with changeP and constant V

A gas enclosed by a piston in a cylinder can do work on the piston, the work being the pressure multiplied by the change in volume. If the volume doesn't change, no work is done. If there is no change in volume ($\Delta$V=0), work will equal 0.

Nora Sharp 1C
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

### Re: Work with changeP and constant V

expansion work in a irreversible change = -P*deltaV, and when there is no change in V, deltaV = 0 so work = 0. This only holds true if we only consider expansion work.

expansion work in a reversible change = -nRT*(ln (Vfinal/Vinitial))
if vfinal = vinitial, then any expansion work also is zero.