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Homework Help: Calculate the molar entropy

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    They gave me a system with [itex]u=\frac{3}{2} pv[/itex] and [itex]u^{1/2}=bTv^{1/3}[/itex]
    So i know that [itex]p=\frac{2b^{2}T^{2}}{3v^{1/3}}[/itex]

    And it says i have to use the third law of thermodynamics to obtain the integrating factor


    2. Relevant equations
    [itex]\lim_{T\to0}S=0[/itex]

    [itex]\left(\frac{\partial u}{\partial v}\right)_{T}=T\left(\frac{\partial s}{\partial v}\right)_{T}-p[/itex]





    3. The attempt at a solution

    So what i did is

    [itex]\left(\frac{\partial u}{\partial v}\right)_{T}=T\left(\frac{\partial s}{\partial v}\right)_{T}-p\Rightarrow\left(\frac{\partial s}{\partial v}\right)_{T}=\frac{1}{T}\left(\frac{\partial u}{\partial v}\right)_{T}+\frac{p}{T}[/itex]

    So we end with
    [itex]ds=\frac{4b^{2}T}{3v^{1/3}}dv[/itex]
    so ..
    [itex]s=2b^{2}Tv^{2/3}+f(T)[/itex]

    I know the solution is just [itex]s=2b^{2}Tv^{2/3}[/itex]

    But i don't know how to use the third law ot obtain the integrating factor and obtain the molar entropy without the uknown function of T


    I also tried using the differential function of [itex]S(T,v)[/itex] like this
    [itex]ds=\underbrace{\left(\frac{\partial s}{\partial T}\right)_{v}}_{c_{v}/T}dT+\left(\frac{\partial s}{\partial v}\right)_{T}dv[/itex]

    But i don't know if [itex]c_{v}[/itex] its constant, also even if so, it still wrong.

    I don't know how to really solve this, please help, and THX ! :D
     
  2. jcsd
  3. Feb 4, 2012 #2
    All right i just solved it, so what i did is:

    [itex]dS=\frac{dU}{T}+\frac{p}{T}dV[/itex]

    knowing that

    [itex]\frac{p}{T}=\frac{2}{3}\frac{b^{2}T}{v^{1/3}}[/itex]

    and

    [itex]du=\frac{2 b^2 T^2}{3 v^{1/3}}dv[/itex]

    so

    [itex]ds=\frac{2b^{2}T}{3v^{1/3}}dv[/itex]

    to get finally

    [itex]s=2b^2 T v^{2/3}[/itex]

    Still didn't use the third law to obtain the integrating factor, i am puzzled why i should use it
     
    Last edited: Feb 4, 2012
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