1. Limited time only! Sign up for a free 30min personal 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!

Heat Capacities Given Equation of State

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the equation of state ##V(P,T)=V_1\cdot exp(\frac{T}{T_1}-\frac{P}{P_1})## where ##V_1\;,T_1\;,P_1## are constants:
    a. derive an equivalent equation ##P(V,T)##;
    b. given ##C_V=DT^3## where D is a const, calculate the entropy of the system ##s(V,T)## up to a const;
    c. find heat capacity ##C_P##.

    2. Relevant equations
    Definitions of heat and heat capacities.
    First law? Enthalpy?

    3. The attempt at a solution
    Starting with a, we get via a simple calculation ##P(V,T)=P_1\cdot (\frac{T}{T_1}-ln(\frac{V}{V_1}))##.
    Then, using heat and heat cap. definitions we get
    $$ds=dQ/T=CdT/T$$$$\Rightarrow (\frac{\partial S}{\partial T})_V = \frac{C_V}{T} = DT^2$$$$\Rightarrow s = \frac{1}{3}DT^3 + S_0$$
    Of this step I'm not entirely sure. It looks okay, but s is not a function of the volume, and I think I need that dependence to work out ##C_P##. So this is where I'm stuck.

    Thanks!
     
  2. jcsd
  3. Jan 13, 2016 #2
    What is the partial derivative of entropy with respect to volume at constant temperature?
     
  4. Jan 13, 2016 #3
    Hmm... from Maxwell's relations $$(\frac{\partial S}{\partial V})_T = (\frac{\partial P}{\partial T})_V$$
    which equals ##\frac{P_1}{T_1}## from the given equation of state, and then $$s = \frac{P_1}{T_1}V + s_0'$$
    The heat cap. is given by ##C_P = T(\frac{\partial s}{\partial T})_P##.
    I could try rewriting $$(\frac{\partial s}{\partial T})_P = -\frac{(\frac{\partial P}{\partial T})_s}{(\frac{\partial P}{\partial s})_T} = -\frac{(\frac{\partial S}{\partial V})_P}{(\frac{\partial P}{\partial s})_T}$$
    But I can't see how these changes in variables get me where I want to go.
    Maybe I can somehow relate ##s_0## to ##s_0'##, but it doesn't seem relevant...
     
  5. Jan 13, 2016 #4
    What happened to the DT3/3?
     
  6. Jan 13, 2016 #5
    Ok, I think I got it while writing this reply. What I started typing is:
    It's part of the constant of integration ##s_0'##, or if we neglect a constant that has nothing to do with the temp. and volume, it is exactly ##s_0'##.
    This I understand, but even writing it explicitly, I'm not sure on how to differentiate the entropy with respect to the temp. while holding P const, as V is free to change to keep the given pressure constant.

    But..! We are given V in terms of both T and P. So plugging it to s, we finally get $$s = \frac{1}{3}DT^3 + \frac{P_1}{T_1}(
    V_1⋅exp(\frac{T}{T_1}−\frac{P}{P_1}))+s_0$$ where ##s_0## is now a true constant that we can entirely ignore and differentiate happily to get $$C_P = T(\frac{\partial s}{\partial T})_P = DT^3 + \frac{P_1V_1T}{T_1^2}exp(\frac{T}{T_1}-\frac{P}{P_1})$$
    Is that correct?!
    And thank you very much!
     
  7. Jan 13, 2016 #6
    Correct. Nice job.
     
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: Heat Capacities Given Equation of State
  1. Heat Capacity (Replies: 1)

  2. Heat capacity (Replies: 7)

  3. Heat capacity (Replies: 3)

  4. Heat capacity (Replies: 16)

Loading...