Why is Work Zero at Constant Volume?

[Xyius]

I am having a little trouble understanding this and I was wondering if I could get some help.

I understand that work done at constant volume is zero because since the volume isn't changing, there is no movement on the boundaries and therefore no work. But it isn't coming out mathematically. This is what I mean.

dU=\delta Q - \delta W
dU=\delta Q - d(PV)
dU=\delta Q - VdP-PdV
dU=\delta Q - VdP

Yet, my professor is saying that at constant volume, work is just...
dU=\delta Q

This makes sense conceptually, but why isn't it making sense mathematically? How could I make VdP=0?

 

[Dadface]

By definition work done involves movement and in this example movement would result in a change of volume.If the volume is constant(dV=0) no work is done.The work done is given by the integral of PdV not VdP.

 

[Xyius]

To me that doesn't make sense, it seems like they are ignoring the other integral. Do you not get pdv + vdp from differentiating vp? And even if you didn't want to do the product rule, and say from the get go that v= constant, it would be..

d(vp)=vdp
By pulling the constant out of the derivative.

So I do not understand why this integral is being left out. Is it because most experiments are done at atmospheric pressure which is constant?

 

[Dadface]

The work done is zero not if the pressure is constant but if the volume is constant.

Work done = force times distance moved in direction of force.Imagine a gas at pressure P in a cylinder trapped by a moveable piston of area A so that force on the piston=PA.If the piston is pushed out a distance dx then the work done =PAdx=PdV.If the piston is firmly fixed in position so that the gas is unable to expand then no work is done even if the pressure changes.

 

[Xyius]

I understand that completely. My question is asking, what is the meaning of the other integral? Mathematically it HAS to be there, because when you differentiate the first law that is what you get.

I found something similar on wiki, but explaining enthalpy. It seems to confirm my guess, but I just want to make sure.
http://en.wikipedia.org/wiki/Enthalpy#Relationship_to_heat

 

[Redbelly98]

By definition, work is ∫F·dx. This is equivalent to

∫(F/A)·Adx
= ∫P·dV​

So work is not equal to ∫d(P·V), that is an erroneous statement. The "other integral", ∫V·dP, has no physical meaning in terms of calculating work.

I found something similar on wiki, but explaining enthalpy. It seems to confirm my guess, but I just want to make sure.
http://en.wikipedia.org/wiki/Enthalpy#Relationship_to_heat

According to that link,

...the energy added to the system through expansion work is δW = − pdV

Ignoring the minus sign, which can occur depending whether one means the work done on or done by the system, this would contradict your earlier statement that δW = d(pV)

 

[nasu]

This is what I mean.

dU=\delta Q - \delta W
dU=\delta Q - d(PV)
dU=\delta Q - VdP-PdV
dU=\delta Q - VdP

How do you justify the second step (second line)?
Do you assume W=PV and then change the \delta W into differential?
The (infinitesimal) amount of work done is pdV and not PV.

 

[Xyius]

Okay I believe I understand now, I think was confusing the definition of enthalpy with the first law. Since enthalpy is H=U+PV and the first law is U=Q+W where W=pdv. (Which I fully understand now.) I would see the PV in the definition for enthalpy and write H=U+W, which I now know is wrong. Thanks for clearing things up.