# Homework Help: Express dH(V,T) as a state function

1. Apr 7, 2016

### grandpa2390

1. The problem statement, all variables and given/known data
the entire problem is to show that the enthalpy as a function of volume and temperature can be expressed in the form:....
I am having trouble with the first step.

2. Relevant equations
dH = TdS + VdP
$dH = (\frac{∂H}{∂S})dS + (\frac{∂H}{∂P})dP$

3. The attempt at a solution

I think the second one is the state function but I can't find anything definitive.

2. Apr 7, 2016

### Staff: Mentor

I don't think this is what they want you to do. I think what they want is to start out with dH = TdS+VdP, and then use the the chain rule to express dS(V,T) and dP(V,T). The partial of S with respect to T is known in terms of Cv. The partial of S with respect to V is obtained from a Maxwell relationship.

3. Apr 7, 2016

### grandpa2390

I'm glad you're here. I'm not sure what I am supposed to do exactly. I'm trying to figure it out.

the problem is to show that enthalpy as a function of volume and temperature can be expressed in the form

$dH = (P + π_T - \frac{1}{k_T})dV + (C_v + V(\frac{∂P}{∂T})_v)dT$

by following the steps below.

the first step is to express dH(V,T) as a state function.

I thought the state functions were something to memorize.
I'm going to write out what you said, and think about it. Because that is the first step. thanks ;)

if you want the rest of the steps, I'll post them. I just don't want to seem like I am asking for the solution.

4. Apr 7, 2016

### Staff: Mentor

Yes. Seeing that solution, the steps that I recommended to follow are perfect.

Chet

5. Apr 7, 2016

### grandpa2390

Forgive me it's been so long I'm going to retake multivariable in the summer. But I don't know how to use the chain rule here.
I know if I was doing the chain rule of y = x*y it would d
ok. It's been so long since I took multivariable, I plan on retaking it during the summer. so I am going to have to look up what you mean by the chain rule in this case.
I'll be back with questions after I give your suggestion a good attempt :)

right now I only know the f'g + g'f, but I don't know how to use it here

6. Apr 7, 2016

### grandpa2390

are you saying to use the chain rule on dH to get the dS and dP?

7. Apr 7, 2016

8. Apr 7, 2016

### grandpa2390

ok so $dH = dU + PdV + VdP$

Ah!!! I got it! :)

I divide by dV and then use relations and identities to get the term in front of the dV

9. Apr 7, 2016

### grandpa2390

but I'm not getting why this goes in parenthesis before the dV.

what is the step in between ?

wait a minute... no nvm. I thought I had something for a moment...

well I have that mu_T = -dV + VdT ...

Last edited: Apr 7, 2016
10. Apr 7, 2016

### grandpa2390

I don't know. I am racking my brains to figure out the why you recommended works. I just can't see it. : (

and so I can't duplicate it for the other term

if I divide the entire dH = dU + PdV + VdP by dT then I get the the second term except for a PVa as well...

Last edited: Apr 7, 2016
11. Apr 7, 2016

### grandpa2390

edit: I think what threw me off this entire time was that my professor wrote this in the question: $dH = U + PV$, and I didn't know how to work that in. but I am questioning now if it isn't supposed to be $H = U + PV$ because that is what it was in the previous question...

well maybe:

$dH = U + PV$

$(\frac{∂H}{∂V}) = (\frac{∂U}{∂V})+P(\frac{∂V}{∂V})+V(\frac{∂P}{∂V})$ which of course simplifies to the left

for the second term
$dH = U + PV$

can I do this?

$(\frac{dH}{dT})_v = (\frac{∂U}{∂T}) + (\frac{∂(PV)}{∂T})$
v is a constant so
$(\frac{∂H}{∂T})_v = (\frac{∂U}{∂T}) + V(\frac{∂(P)}{∂T})$

$= C_v + V(\frac{∂(P)}{∂T})$

in both cases multiply both sides by the ∂V and ∂T at the end to get the dT and dV in each term?

This is the best I can come up with so far.

so with the revelation that my professor probably made a typo on the problem sheet, I am now thinking that this problem was real simple. If I want to express as a function of volume and temperature, take the function and take the derivative with respect to volume and temperature.

12. Apr 8, 2016

### grandpa2390

if that is correct
the second part is to show that for an ideal gas it simplifies to dH = C_p dT
I got the C_p dT part
but the first term simplifies to P-(1/P) dV. the only thing I can think is that since P is a constant, it is equal to zero due to the dV. but I'm not sure if that makes sense...

13. Apr 8, 2016

### Staff: Mentor

I know that you are familiar with the following 3 equations:
$$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV$$
$$dH=dU+PdV+VdP$$
and
$$dP=\left(\frac{\partial P}{\partial T}\right)_VdT+\left(\frac{\partial P}{\partial V}\right)_TdV$$

Just combine them, and you will have what you want.