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### Topic: In Thermodynamics, dw = Pdv?  (Read 34007 times)

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#### Dev

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##### Re: In Thermodynamics, dw = Pdv?
« Reply #15 on: January 28, 2013, 10:11:18 AM »
Hi.
Well, I started off believing that W = I PdV.
It was someone who suggested that W is actually I d(PV)

The thing that would make me happiest is finding out W = I PdV,
but no one here who has posted before this said that W = PdV. They always added that this is true only when pressure is constant.
What I need to hear is that W = I PdV for everything.

#### curiouscat

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##### Re: In Thermodynamics, dw = Pdv?
« Reply #16 on: January 28, 2013, 10:15:38 AM »
The thing that would make me happiest is finding out W = I PdV,
but no one here who has posted before this said that W = PdV.

Quote
They always added that this is true only when pressure is constant.
What I need to hear is that W = I PdV for everything.

I'll say it. But I may be wrong.

#### Dev

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• Posts: 23
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##### Re: In Thermodynamics, dw = Pdv?
« Reply #17 on: January 28, 2013, 10:48:14 AM »
While I wish you wouldn't have said that last part, I'm going to take it.

W is indeed equal to I PdV for everything.
Thanks.

#### Ali Durrani

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##### Re: In Thermodynamics, dw = Pdv?
« Reply #18 on: October 29, 2015, 02:48:18 PM »
You are confusing 2 or three processes taking place on or by the system,
the correct equation is for non isenthalpic process
combining 1st and second law of thermodynamics
dH=dU+pdV+Vdp
this can be easily transformed from equation of enthalpy
H=U+pV
taking differential operator on both sides
dH=dU+d(PV)
dH=dU+pdV+Vdp
work can also be written in terms of enthalpy (In case of Steam Cycles)
dW=H-H=dU+pdV+Vdp
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