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Specified equation of state from heat capacity

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    The constant-volume heat capacity of a particular simple system is [tex]
    c_v = AT^3 [/tex]
    where A is a constant. In addition the equation of state is known to be of the form
    [tex]
    (v-v_0)p = B(T)
    [/tex]
    where [itex]B(T)[/itex] is an unspecified function of T. Evaluate the permissible functional form of [itex]B(T)[/itex].

    2. Relevant equations

    3. The attempt at a solution


    So we have
    [tex]
    {\frac{\partial S}{\partial v}}
    _U = \frac{B(T)}{T(v-v_0)}[/tex]
    and
    [tex]
    {\frac{\partial S}{\partial T}}_v = AT^2
    [/tex]
    I apply the first derivative to the second equation and vice versa. I equate and get
    [tex]
    \frac{\partial}{\partial T} \left ( \frac{B(T)}{T} \right) \frac{1}{v-v_0} = 2A T {\frac{\partial T}{\partial v}}_u
    [/tex]
    The rightmost term can be rewritten as
    [tex]
    {\frac{\partial T}{\partial v}}_u = -\frac{ {\frac{\partial u}{\partial v}}_T}{{\frac{\partial u}{\partial T}}_v} = -\frac{T \frac{\partial s}{\partial v}_T - p}{c_v}= -\frac{T \frac{\partial p}{\partial T}_v - p}{c_v}[/tex] so that
    [tex]
    \frac{2}{T^2} \left [ -\frac{T B'(T)}{v-v_0} + \frac{B(T)}{v-v_0} \right] = \frac{\partial}{\partial T} \left ( \frac{B(T)}{T} \right) \frac{1}{v-v_0}
    [/tex]
    which I solve for [itex]B(T)[/itex] and get [itex]B(T) = ET [/itex] with [itex]E[/itex] a constant. Now
    [tex]
    c_p = c_v + \frac{Tv\alpha^2}{\kappa_T}
    [/tex]
    where [itex]\alpha[/itex] is the isobaric compressability with temperature and [itex]\kappa_T[/itex] is the isothermal compressability with pressure. Thus with
    [tex]
    v = \frac{B(T)}{p} + v_0
    [/tex]
    this should equal (I think) to
    [tex]
    c_v + T \left(\frac{\partial V }{\partial T}_p\right)^2\left/\right. \left(\frac{\partial V}{\partial p}\right)_T =c_v+ T\frac{B'(T)^2}{B(T)}
    [/tex]
    which in my case would be simply [itex]c_v + E[/itex]. The answers give
    [tex]
    c_v + (T^3/DT + E)
    [/tex].
    Now when solving for their [itex]B(T)[/itex] I get a very complicated expression. I do not see where my mistake lies, except I am not 100% sure if the two partial derivatives I had at the beginning commute.
     
  2. jcsd
  3. Aug 8, 2017 #2
    I think that it might be easier to start with the Maxwell relationship: $$\left(\frac{\partial S}{\partial v}\right)_T=\left(\frac{\partial p}{\partial T}\right)_v$$
     
  4. Aug 8, 2017 #3
    That's definitely a good idea! Integrating [itex]c_v/T[/itex] we find
    [tex]
    s = \frac{AT^3}{3} +f(v) [/tex]
    for some f. Then with the Maxwell relation
    [tex]
    f'(v) = \frac{B'(T)}{v-v_0}
    [/tex]
    which means that [itex]B'(T) = E[/itex] for some E. So this confirms what I found before (much more easily).
     
  5. Aug 8, 2017 #4
    I would have done it a little differently: $$dS=AT^2dT+\frac{B'(T)}{v-v_0}dv$$So,
    $$\frac{\partial^2 S}{\partial T \partial v}=\frac{\partial^2 S}{\partial v \partial T}=\frac{\partial}{\partial v}(AT^2)=0=\frac{\partial }{\partial T}\left(\frac{B'(T)}{v-v_0}\right)=\frac{B''(T)}{v-v_0}$$So, $$B''(T) = 0$$So, B(T) is a linear function of T.
     
  6. Aug 9, 2017 #5
    Maybe it was too late for me to be posting before but my first method misses the extra constant you'd get.
     
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