# Arguing work is an inexact differential

1. Feb 5, 2012

### Sekonda

Hey,

Given the fundamental thermodynamic relation (dN=0) ; TdS=pdV+dU, I have to argue that pdV is inexact. I know that dS and dU are exact differentials and that the integral of an exact differential around a closed loop will give a zero result, not too sure where to start, the question is only a couple of marks so the answer need only be brief but my understanding is limited as to why pdV, the work, has to be an inexact differential.

Any help appreciated!

Thanks.

2. Feb 5, 2012

### tiny-tim

Hey Sekonda!

What would you add to pdV to make it an exact differential?

3. Feb 5, 2012

### Sekonda

Vdp?

I'm still not sure I'm going to have to think about this a bit more!

4. Feb 5, 2012

### Sekonda

Is this the correct term to add to pdV to make it an exact differential; if so, how does this imply that pdV is inexact?

5. Feb 5, 2012

### tiny-tim

Yes, pdV + Vdp is an exact differential … it's exactly d(pV).

So that means that (unless Vdp = 0, ie p is constant) pdV is not exact.

6. Feb 5, 2012

### Sekonda

Ahh yes that does make sense! I'm always bad at 'arguing', 'showing' or 'proofing' questions; thanks man.

Cheers!