Express dH(V,T) as a state function

[grandpa2390]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

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

 

[Chestermiller]

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.

 

[grandpa2390]

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.

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.

 

[Chestermiller]

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.

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

Chet

 

[grandpa2390]

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.

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

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

Chet

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

 

[grandpa2390]

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.

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

 

[Chestermiller]

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

No. But you can also start with your equation for dU interns of dT and dV, and add PdV+VdP to get dH.

 

[grandpa2390]

No. But you can also start with your equation for dU interns of dT and dV, and add PdV+VdP to get dH.

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

 

[grandpa2390]

No. But you can also start with your equation for dU interns of dT and dV, and add PdV+VdP to get dH.

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 ...

 

[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...

 

[grandpa2390]

No. But you can also start with your equation for dU interns of dT and dV, and add PdV+VdP to get dH.

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.

 

[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...

 

[Chestermiller]

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.