Thermodynamics - partial differential

[Beer-monster]

Homework Statement

Edit: Some more details.

The question concerns Joule-Thompson Expansion and I need to derive the relationship:

\left(\frac{\partial H}{\partial P}\right)_T =V+T\left(\frac{\partial S}{\partial P}\right)_TV is not constant as it is expansion but the enthalpy is specied as H(P,T) for the problem. Therefore I would assume that Volume is a function of P and T and not independent. However, I'm not 100% sure on this.

The Attempt at a Solution

dH=TdS+VdP

Therefore:

\left(\frac{\partial H}{\partial P}\right)_T = T\left(\frac{\partial S}{\partial P}\right)_T + V\left(\frac{\partial P}{\partial P}\right)_T = V+T\left(\frac{\partial S}{\partial P}\right)_T

Is this right? I know it would not be if I'm right and V is a function of P and T. Could someone please clear this up for me.Thanks

BM

 

[Mapes]

You got it, as long as you've implicitly used the chain rule and ignored the differentials as compared to non-infinitesimal values. That is,

\left(\frac{\partial}{\partial P}\right)_T\left(V\,dP\right)=\left(\frac{\partial V}{\partial P}\right)_T dP+V\left(\frac{\partial P}{\partial P}\right)_T=V.

The differentials can be ignored in any mix of differential and non-infinitesimal values. Does this make sense?

 

[Beer-monster]

I think I get it. As V is not constant and probably depends on P we can't ignore it in the differential and need to apply the product rule.

But doesn't the differential of V only equal 0 if V is constant or independent of P. In which case, why would we need to explicitly use the product rule?

 

[Mapes]

I'm not sure if this answers your question, but we must use the product rule every time we differentiate a product. It's safest to assume every variable depends on every other variable unless it's been specifically shown not to. In this case, (\partial V/\partial P)_T\,dP goes away not because V has any certain value, but because dP is an infinitesimal value that vanishes when compared with the finite value V.

In summary, it's safer to use the product rule and cancel the appropriate terms than to work "fast and loose."

 

[Beer-monster]

So because dp is so small (approximately zero) compared to the differential <br /> (\partial V/\partial P)_T<br /> we can ignore it provided the differential is of something non-infintesimal?

 

[Mapes]

Exactly. We can't automatically do the same thing for an equation like dH=T\,dS+V\,dP because everything is infinitesimal.

 

[Beer-monster]

Yeah, that occurred to me and gave me a bit of a pause. But though differentials are made of infintesimal elements they are not themselves infintesimal so would be reduced to nothing if multiplied by dP.

However, if this was worked out in a problem the product rule would have to be expressed explicitly otherwise one would think that you were treating V as constant.