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Find the general expression for the adiabatic relationship

  1. Apr 9, 2016 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    Find the general expression for the adiabatic relationship between P and V

    2. Relevant equations
    start with ##(\frac{∂P}{∂V})_s##
    expression for adiabatic relationship between V and T: ##(\frac{∂V}{∂T})_s = \frac{-C_v}{\frac{RT}{V}}##
    Relationship between S and T ##(\frac{∂S}{∂T})_p = \frac{C_p}{T}##
    Final Relationship ##(\frac{∂P}{∂V})_s = \frac{C_p}{C_v}(\frac{∂P}{∂V})_T##

    3. The attempt at a solution

    it is hard to input every step along the way. so I am going to put what I get at step 5 and if you can tell me if I need to rework the first few steps. It would be helpful. because 6 is where I get lost. I don't end up wit two partials to use a permuter on (in reverse)

    1 apply permuted
    2 find two maxwell relations to replace the resulting partial derivative expression
    insert the expression we derived in class for the adiabatic relationship
    3 use permuter on other partial derivative
    4 plug in the relationship between S and T
    5 Use maxwell relation on remaining partial derivative term
    ##\frac{R}{V}\frac{C_p}{C_v}(\frac{∂T}{∂V})_p##
    6 apply the permuter (in reverse) to the two partial derivatives to finally arrive at the general expression.
    my attempt is to say that I can say ##(\frac{∂P}{∂V})_T = (\frac{∂P}{∂T})_V(\frac{∂T}{∂V})_p##

    then I divide by R/V and the other partial derivative to get the answer that belongs on the right... but the left side. I don't know what I am dividing by these two in order to get the partial of P with respect to V.
     
  2. jcsd
  3. Apr 10, 2016 #2

    stevendaryl

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    It's not clear whether you are assuming an ideal gas, or not. I think that your expression

    [itex]\frac{\partial V}{\partial T}|_S = -frac{C_V}{RT}[/itex]

    is only valid for an ideal gas.

    The equation for internal energy [itex]U[/itex] is:

    [itex]dU = -P dV + T dS[/itex]

    From this, you get: [itex]\frac{\partial U}{\partial T}|_S = - P \frac{\partial V}{\partial T}|_S[/itex]

    So I get:

    [itex]\frac{\partial V}{\partial T}|_S = -\frac{\frac{\partial U}{\partial T}|_S }{P}[/itex]

    which reduces to your expression when [itex]\frac{\partial U}{\partial T}|_S = C_v[/itex] and [itex]P = \frac{RT}{V}[/itex]

    (Are you sure there isn't a mistake? For an ideal gas, it should be that [itex]P = \frac{NRT}{V}[/itex], where [itex]N[/itex] is the number of particles.
     
  4. Apr 10, 2016 #3

    grandpa2390

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    I assume it is an ideal gas. the problem doesn't say.
    mistake where?
     
  5. Apr 10, 2016 #4

    stevendaryl

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    I think your expression for [itex]\frac{\partial V}{\partial T}|_S[/itex] is mistaken. I think it should be [itex]\frac{\partial V}{\partial T}|_S = - \frac{C_v}{\frac{NRT}{V}}[/itex], with an [itex]N[/itex] in the denominator. Or you could just leave it as [itex]\frac{\partial V}{\partial T}|_S = - \frac{C_v}{P}[/itex]
     
  6. Apr 10, 2016 #5
    What do the words "general expression" mean to you?
     
  7. Apr 10, 2016 #6

    grandpa2390

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    I don't know. I copied it right from the professor's slides...
     
  8. Apr 10, 2016 #7

    grandpa2390

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    to it me it means that it is an expression that should generally work. In my mind that would mean that this is an ideal gas and the expression I am trying to prove is an ideal gas expression. But when people who know more than I do come in and start questioning... I become unsure. I don't want to argue with anyone... when there is a higher than average chance that I will be wrong.

    but if I am to give a definite answer. then yes, this is for an ideal gas.
     
  9. Apr 10, 2016 #8
    No. It should not be constrained to an ideal gas. So now what?
     
  10. Apr 10, 2016 #9

    grandpa2390

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    I'm sorry :(
    the general expression mean it works for any gas. not any specific gas like an ideal gas.

    as for the previous part. the question says we have already shown for an ideal gas... now derive a general expression.. so not an ideal gas for this problem?
     
    Last edited: Apr 10, 2016
  11. Apr 10, 2016 #10

    grandpa2390

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    Thank you. You have given me exactly what I needed. I'm having trouble reading your work here. but the concept of inserting the solution of the partial is one that I really need to grasp. Every question I have asked in the last couple of days has usually ended with someone telling me to substitute in.

    But yeah I am not sure if ##(\frac{\partial P}{\partial V})_S = \frac{C_P}{C_V}(\frac{\partial P}{\partial V})_T## is correct. I'm not going to dare question him... that already got me in trouble...

    but after substituting in for the other partial I do wind up with ##\frac{1}{n}\frac{C_P}{C_V}(\frac{\partial P}{\partial V})_T##
    and whether or not that n is extraneous, I'll get most of the credit.

    the ##\frac{C_p}{C_v}## comes from that being the value of gamma (heat capacity ration) which is the ideal gas adiabats PV(gamma) = PV( gamma) initial = final.

    Maybe @Chestermiller can help with the n? I'm going to move on to the next problem for now :)
     
  12. Apr 10, 2016 #11

    grandpa2390

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    oh! the value for that I got by doing a reverse permuter.

    I found that I could say ##(\frac{\partial P}{\partial V})_T = (\frac{\partial P}{\partial T})_V(\frac{\partial T}{\partial V})_P##
    then said that ##(\frac{\frac{\partial P}{\partial V})_T}{(\frac{\partial P}{\partial T})_V} = (\frac{\partial T}{\partial V})_P##

    ##(\frac{\partial P}{\partial T})_V = \frac{nR}{V}##

    the R and the V cancel leaving with 1/n

    except for the n... it gets me what he wants... I don't know if it is correct, but as I said before, I am not about question him. He is vastly superior to me in this, and I have a way of sounding skeptical when asking questions. It makes people not like me, and I don't want to anger the person who will be grading my test...
    I have three classes. I want to graduate :(
     
  13. Apr 10, 2016 #12
    Yes.

    The presence of Cp and Cv in the answer should be an indication to you the the analysis should start with equations for both dH and dU.

    $$dH=TdS+VdP=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$
    $$dU=TdS-PdV=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$
     
    Last edited: Apr 10, 2016
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