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Derive the energy equation for a van der Waal gas.

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Derive the energy equation for a van der Waal gas when T and v are the independent variables.


    2. Relevant equations

    I'm required to start from: [itex]du=\frac{\partial u}{\partial T} dT +\frac{\partial u}{\partial v} dv[/itex]

    3. The attempt at a solution

    We'll when I first started this problem. I just started doing random derivatives that made sense to me but after looking it over I had no clue what I was doing.

    Does anyone know what the answer I should be getting after I fully derive it?
     
  2. jcsd
  3. Mar 21, 2014 #2

    vanhees71

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    I don't know, what's given. So I can't help to find an answer to your question. One way to derive the caloric equation of state, i.e., the internal energy [itex]U[/itex] is to use the canonical partition sum
    [tex]Z=\frac{z^N}{N!}[/tex]
    with the single-particle partition function
    [tex]z=\frac{V^*}{\Lambda^3} \exp \left (-\frac{\phi}{2 T} \right),[/tex]
    where [itex]V^*=V-2 \pi N d^3/3[/itex] is the available volume (geometrical volume minus excluded volume due to the hard-sphere model for the molecules), and [itex]\phi[/itex] is the interaction energy of the particles
    [tex]\phi=\frac{N}{V} \int_d^\infty \mathrm{d} r \; U_{\text{rel}}(r) 4 \pi r^2,[/tex]
    with the effective two-body potential
    [tex]U_{\text{rel}}(r)=\begin{cases}
    \infty & \text{for} \quad r \leq d ,\\
    -\alpha (d/r)^6 & \text{for} \quad r>d.
    \end{cases}
    [/tex]
    The Helmholtz free energy is given by
    [tex]A(T,V,N)=-T \ln Z,[/tex]
    and from this you can derive all thermodynamical quantities from the usual thermodynamic relations like
    [tex]p=-\left (\frac{\partial A}{\partial V} \right)_{T,N}[/tex]
    etc. The internal energy is given by the usual Legendre transformation
    [tex]U=A+T S.[/tex]
     
  4. Mar 22, 2014 #3
    [tex]dU=TdS-PdV=T\left(\frac{∂S}{∂T}\right)dT+T\left(\frac{∂S}{∂V}\right)dV-PdV=
    C_VdT+(T\left(\frac{∂S}{∂V}\right)-P)dV[/tex]
    The next step is to determine ∂S/∂V at constant T. This can be obtained from a Maxwell relation, starting from the equation dA=-SdT-PdV.

    S=-∂A/∂T
    P=-∂A/∂V

    So [tex]\frac{∂S}{∂V}=\frac{∂P}{∂T}[/tex]

    Therefore,
    [tex]dU=C_VdT+(T\left(\frac{∂P}{∂T}\right)-P)dV[/tex]

    The second term in this equation is zero for an ideal gas, but not for a real gas. Just substitute the van der Waals equation into the second term of this equation.

    chet
     
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