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Dereive an expression for an isobaric/isothermal change in entropy

  1. Nov 2, 2009 #1
    I already dereived the following expressions for the Van der Waals gas:
    molar energy:
    [tex] U=3RT - \frac{aP}{RT} [/tex]
    and the expression for the PV, where V is the molar volume:
    [tex] PV= RT +(b - \frac {a}{RT})P [/tex]
    Using these and the central equation,
    [tex] TdS = dU + PdV [/tex]

    i am to dereive an expression for dS in two cases:
    1) as the substance undergoes an isobaric change from temperature [tex] T_{i} [/tex] to [tex] T_{f} [/tex] at pressure [tex] P [/tex].
    2) as the substance undergoes an isothermal change from pressure [tex] P_{i} [/tex] to [tex] P_{f} [/tex] at temperature [tex] T [/tex].
     
  2. jcsd
  3. Nov 2, 2009 #2

    gabbagabbahey

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    Okay; and your attempt at a solution is.....?
     
  4. Nov 2, 2009 #3
    On the example of the isobaric:
    [tex] dS = \frac {dU}{T}+\frac{P}{T}dV [/tex]
    using the expression for the molar internal energy:
    [tex] dU = 3RdT+ \frac {aP}{RT^{2}}dT - \frac {a}{RT}dP [/tex]
    and using the expression for the PV product:
    [tex]V= \frac {RT}{P} +b - \frac{a}{RT} [/tex]
    then [tex] dV=\frac {R}{P}dT+ \frac{a}{RT^{2}}dT- \frac{RT}{P^{2}}dP [/tex]
    Substituting into dS (taking dP=0):
    [tex] dS= 4R \frac{dT}{T}+\frac {2aPdT}{RT^{3}} [/tex]
    and integrating:
    [tex]dS=4Rln( \frac{T_{f}}{T_{i}})+ \frac {aP}{R(T_{i}-T_{f})} [/tex]
    I have no idea whether this is correct though.....
     
  5. Nov 2, 2009 #4

    gabbagabbahey

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    Looks good so far!:approve:

    When you integrate [itex]dS[/itex] from the initial state to the final state, you get [itex]\Delta S\equiv S_f-S_i[/itex], not [itex]dS[/itex].
     
    Last edited: Nov 2, 2009
  6. Nov 2, 2009 #5
    I found slightly a different equation for the van-der-Waals gas. Also in many places you seem to introduce the ideal gas law for a moment. That's where little differences come from, but maybe the difference isn't big though.

    The van-der-Waals equation as I found it is
    [tex]
    \left(p+\frac{a}{V^2}\right)(V-b)=RT
    [/tex]
    In any case for a van-der-Waals gas the energy is
    [tex]
    U=f(T)-\frac{a}{V}
    [/tex]
    You have assumed that [itex]f(T)=3RT[/itex], which is strictly speaking an addition to the van-der-Waals equation, but probably justified.

    Now
    [tex]
    T\mathrm{d}S=\mathrm{d}U+p\mathdm{d}V
    =f'\mathrm{d}T+\frac{a}{V^2}\mathrm{d}V+p\mathrm{d}V
    =f'\mathrm{d}T+\frac{RT}{V-b}\mathrm{d}V
    [/tex]
    so
    [tex]
    \mathrm{d}S=f'\frac{\mathrm{d}T}{T}+\frac{R\mathrm{d}V}{V-b}
    [/tex]
    [tex]
    \Delta S=\int \frac{\mathrm{d}f}{T}+R\ln\left|\frac{V_1-b}{V_0-b}\right|
    [/tex]
    With your assumption [itex]f(T)=3RT[/itex] this gives
    [tex]
    \Delta S=3R\ln\frac{T_1}{T_0}+R\ln\left(\frac{V_1-b}{V_0-b}\right)
    [/tex]
    Anyone agrees if that is OK? This derivation has not made the ideal gas law substitutions [itex]pV=RT[/itex] that seem to be in your calculation.
     
    Last edited: Nov 2, 2009
  7. Mar 1, 2010 #6
    http://theory.phy.umist.ac.uk/~judith/stat_therm/node51.html [Broken]
     
    Last edited by a moderator: May 4, 2017
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