1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics: Gradient of Adiabat in PV diagram

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data

    The Equation of State and the expression for the entropy for a sample of salt water is given by:

    [tex] V = V_{0}(1 + \beta(T - T_{0}) - \gamma(P - P_{0})) [/tex] [tex] S = S_{0} + C_{v}ln(T - T_{0}) + \frac{\beta}{\gamma}(V - V_{0})[/tex]

    where the subscript 0 denotes a reference state, the coefficients [tex] \beta [/tex] and [tex] \gamma [/tex] are constants and [tex] C_{v} [/tex] is the heat capacity of the salt water at constant volume.

    Derive an expression for the gradient of an adiabat in a PV diagram.

    2. Relevant equations
    Listed above.

    3. The attempt at a solution

    I struggle to write the attempts I've made trying to answer this question. I understand that in this case we have V(T,P) and S(T,V), the gradient will be (dP/dV) and using the fact that an adiabat occurs when there is no change in heat energy. Also aware of the fact that Cv can be written as a differential in terms of (dU/dT), which is possibly relevant.

    The real issue is I have no idea of the best way to put all of this information together and find a logical pathway to answer. Do I want to get to: [tex] dP = \frac{\partial{P}}{\partial{T}}dT + \frac{\partial{P}}{\partial{V}}dV [/tex] and substitute a concoction of the above information to get to the gradient?

    Been banging my head against this problem for a couple of weeks, and would be very grateful for someone to point me in the right direction!

    EDIT:
    Probably should have included that in my explanation. I get that for an adiabatic change, dS is 0 assuming reversibility. Similarly, I got to the point where I have:

    [tex]dV = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP[/tex]

    and

    [tex]dS = \frac{\partial{S}}{\partial{T}}dT + \frac{\partial{S}}{\partial{V}}dV = 0[/tex]

    However, none of these approaches seems to lead to a place where I can rearrange to get to dP/dV. Thats the part I'm a bit stuck on.
     
    Last edited: Mar 19, 2017
  2. jcsd
  3. Mar 19, 2017 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Construct dV from the first given equation and dS from the second equation. "Adiabatic" tells you something about dS.
     
  4. Mar 19, 2017 #3
    Probably should have included that in my explanation. I get that for an adiabatic change, dS is 0 assuming reversibility. Similarly, I got to the point where I have:

    [tex]dV = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP[/tex]

    and

    [tex]dS = \frac{\partial{S}}{\partial{T}}dT + \frac{\partial{S}}{\partial{V}}dV = 0[/tex]

    However, none of these approaches seems to lead to a place where I can rearrange to get to dP/dV. Thats the part I'm a bit stuck on.


    Thanks for the reply though.
     
  5. Mar 19, 2017 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Can you solve your dS equation for dT and sub into the first equation?
     
  6. Mar 19, 2017 #5
    Solve the S equation for ##T-T_0## and substitute into the V equation. Then take the derivative of P with respect to V at constant S.
     
  7. Mar 20, 2017 #6
    So solving dS gives me:

    [tex] dS = \frac{dT}{T-T_{0}} + \frac{\beta}{\gamma} = 0 [/tex]

    and subbing into dV gives:

    [tex] -\frac{\gamma}{\beta}\frac{dT}{T-T_0} = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP [/tex]

    But from here, and I'm probably being thick, I find myself with a lot of differentials and not a lot of directions. Calculating the partials gives:

    [tex] -\frac{\gamma}{\beta}\frac{dT}{T-T_0} =\beta{V_{0}}dT - \gamma{P_{0}}dP [/tex]

    but then there's no Vs anywhere in sight...
     
  8. Mar 20, 2017 #7
    Who said anything about differentiating the S equation? Not me.
     
  9. Mar 20, 2017 #8
    It all dawned on me at once, I was leading myself down the garden path. Thanks!
     
  10. Mar 20, 2017 #9
    And just for future reference for anyone who stumbles across a similar problem, I ended up with:

    [tex] P = P_{0} + \frac{1}{\gamma{V}} + \frac{\beta}{\gamma{V}}e^{-C_{v}}e^{-\frac{\beta}{\gamma}(V-V_{0})} [/tex]

    dP/dV of that should give the slope of the adiabat!
     
  11. Mar 20, 2017 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You have been given two different approaches. My approach is to use the equation for S and the fact that dS = 0 to get an expression for dT in terms of dV. Then when you take the differential of the V equation, you can express it in terms of just the differentials dV and dP. From that you can get an expression for dP/dV in terms of V, P, and T. But you can eliminate T by using the V equation.

    The equation that you were given for S doesn't look correct to me. You have a logarithm of the dimensional quantity T- To. The argument of a logarithm is generally dimensionless. Also, the equation for S is undefined at the reference state where T = To. It looks to me that the argument of the logarithm should maybe be T/To. The the equation for S will yield S = So at the reference state.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Thermodynamics: Gradient of Adiabat in PV diagram
  1. Pv diagrams (Replies: 1)

  2. Thermal (PV diagram) (Replies: 21)

  3. Thermodynamic Adiabats (Replies: 3)

Loading...