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Adiabatic equation in dependence on volume and pressure

  1. Jul 9, 2015 #1
    1. The problem statement, all variables and given/known data

    Find: (a) Equation of state $$f (p, V, T)$$ and (b) Adiabatic equation in dependence on volume and pressure. Internal energy $$U(V, S)=\frac{1}{aV} ln(\frac{S}{\gamma})$$ where a and ##\gamma## are positive constants.

    2. Relevant equations

    (a) ##dU=TdS-pdV \rightarrow##
    ##T=(\frac{dU}{dS})_V##
    ##p=-(\frac{dU}{dV})_{S}##

    ##T=(\frac{d}{dS})_V \cdot \frac{1}{aV} ln(\frac{S}{\gamma})=\frac{1}{aV} \cdot \frac{\gamma}{S} \cdot \frac{1}{\gamma}=\frac{1}{a \gamma V} ##
    ##p=-(\frac{d}{dV})_S \frac{1}{aV} ln(\frac{S}{\gamma})=- \frac{1}{a} \cdot (- \frac{1}{V^2}) \cdot ln(\frac{S}{\gamma})=\frac{1}{aV^2} ln(\frac{S}{\gamma}) ##

    (b) ##pV^{\gamma}=NkT##

    ##\gamma=?##
    ##\gamma=\frac{C_p}{C_V}##
    ##C_p=(\frac{dU}{dT})_p##
    ##C_V=(\frac{dU}{dT})_V##

    3. The attempt at a solution

    (b) ##C_p=(\frac{d}{dT})_p \frac{1}{aV} ln(\frac{S}{\gamma})##

    ##T=\frac{1}{a \gamma V} \rightarrow T \gamma=\frac{1}{a V}##
    ##C_p=(\frac{d}{dT})_p T \gamma ln(\frac{S}{\gamma})##
    ##p=\frac{1}{aV^2} ln(\frac{S}{\gamma}) \rightarrow paV^2=ln(\frac{S}{\gamma})##
    ##C_p=(\frac{d}{dT})_p T \gamma paV^2=\gamma paV^2##

    ##C_V=(\frac{d}{dT})_V \frac{1}{aV} ln(\frac{S}{\gamma})##
    ##C_V=(\frac{d}{dT})_V T \gamma ln(\frac{S}{\gamma})##
    ##C_V=(\frac{d}{dT})_V T \gamma paV^2=\gamma paV^2##

    ##\gamma=\frac{C_p}{C_V}= \frac{\gamma paV^2}{\gamma paV^2}=1##

    I need help with (b) part of the problem. I know this is not correct and I assume I did something wrong while substituting, but I have no idea what. If you know something please post.
    Thanks.
     
  2. jcsd
  3. Jul 9, 2015 #2
    You need help on part a also. Your equation for T is incorrect. After you get that corrected, you will need to elimate S between the equations for S and T.

    In part b, your equation for Cp is incorrect.

    Chet
     
  4. Jul 9, 2015 #3
    With regard to part b: how does S vary along an adiabatic reversible path?
     
  5. Jul 10, 2015 #4
    Regarding a mistake with (a) ##T=\frac{1}{aV} \cdot \frac{\gamma}{S} \cdot \frac{1}{\gamma}=\frac{1}{aSV}##
    Thank you for noticing.
    This leads to: ##C_V=(\frac{d}{dT})_V TSpaV^2=SpaV^2##

    (b) The right formula: ##C_V=(\frac{dH}{dT})_p##

    ##H=U+pV=U+U=2U##
    ##C_p=(\frac{d}{dT})_p (2U)=2 \cdot C_V=2SpaV^2##

    ##\gamma=\frac{C_p}{C_V}=\frac{2SpaV^2}{SpaV^2}=2##

    So, adiabatic equation is: ##pV^2=NkT##

    (Answer to your question: Entropy is constant.)

    Thank you very much for your answer. It's been really helpful.
     
  6. Jul 10, 2015 #5
    For part a,

    $$S=γe^{aV^2p}=\frac{1}{aVT}$$

    This is the p-V-T relationship they were looking for.

    For part b,
    From your final equation for p, you could have immediately seen that pV2 is constant if S is constant along a reversible adiabat. I'm not so sure about the NkT part, however. I don't think that that is correct.

    Chet
     
  7. Jul 10, 2015 #6
    I see. I could have stated that: ##pV^2=const##
    and then from the pressure equation: ## const=\frac{1}{a}ln(\frac{S}{\gamma})##
    Then I substitute variables in constant in order to get a dependence on T:
    ##pV^2=\frac{1}{a}ln(\frac{S}{\gamma}) \rightarrow apV^2=ln(\frac{S}{\gamma}) \rightarrow e^{apV^{\gamma}}=\frac{S}{\gamma} ##
    ## \rightarrow \gamma e^{ap V^{\gamma}}=S ##

    and: ##T=\frac{1}{aSV} \rightarrow S=\frac{1}{aTV} ##

    which leads to: ##\gamma e^{apV^{\gamma}}= \frac{1}{aTV} \rightarrow e^{apV^{\gamma}}= \frac{1}{a \gamma TV} \rightarrow apV^\gamma ##
    ## = ln(\frac{1}{a \gamma TV}) \rightarrow pV^{\gamma}= \frac{1}{a}ln(\frac{1}{a \gamma TV}) ##

    Solution: ## pV^2= \frac{1}{a}ln(\frac{1}{2aTV}) ##

    Thank you one more time.
     
    Last edited: Jul 10, 2015
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