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Express dH(V,T) as a state function

  1. Apr 7, 2016 #1

    grandpa2390

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    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. jcsd
  3. Apr 7, 2016 #2
    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.
     
  4. Apr 7, 2016 #3

    grandpa2390

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    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.
     
  5. Apr 7, 2016 #4
    Yes. Seeing that solution, the steps that I recommended to follow are perfect.

    Chet
     
  6. Apr 7, 2016 #5

    grandpa2390

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    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
     
  7. Apr 7, 2016 #6

    grandpa2390

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    are you saying to use the chain rule on dH to get the dS and dP?
     
  8. Apr 7, 2016 #7
    No. But you can also start with your equation for dU interns of dT and dV, and add PdV+VdP to get dH.
     
  9. Apr 7, 2016 #8

    grandpa2390

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    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
     
  10. Apr 7, 2016 #9

    grandpa2390

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    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
  11. Apr 7, 2016 #10

    grandpa2390

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    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
  12. Apr 7, 2016 #11

    grandpa2390

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    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.
     
  13. Apr 8, 2016 #12

    grandpa2390

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    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...
     
  14. Apr 8, 2016 #13
    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.
     
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