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Thermodynamics-internal energy of a surface of a liquid

  1. Sep 18, 2010 #1
    [/tex]1. The problem statement, all variables and given/known data
    The differential of the internal energy of a surface of a liquid with surface tension [itex]\gamma[/itex] and area A may be written as
    [itex] dU=TdS+\gamma dA[/itex]

    Write down the corresponding form of the Helmholtz free energy F= U -TS. Using the fact that these equations involve exact differentials,derive the Maxwell relation
    [itex] (\frac{\partial S}{\partial A})_{T}=-(\frac{\partial \gamma}{\partial T})_{A}[/itex]
    The internal energy and the entropy are proportional to the area A. Show that the internal energy per unit area is
    [itex] u(T)= \frac{U}{A}=\gamma-T-(\frac{\partial \gamma}{\partial T})_{A}[/itex]

    2. Relevant equations

    The first part I solved in this way:

    F= U-TS
    dF= dU-Tds-Sdt
    dF= TdS+dW-Tds-Sdt=dW-Sdt
    I found on the internet that the mechanical work needed to increase a surface is [tex] dW=\gamma dA[/tex]
    so my dF= -SdT+[tex]\gamma dA[/tex]

    How to show the last expresion?
    Anyone willing to help?
     
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 18, 2010 #2
    For the maxwell relation use the differential form of the Helmholtz free energy you derived, and read off the partial derivatives, now differentiate them again and use young's theorem.
     
  4. Sep 18, 2010 #3
    hm I am not quite following....could you give me an example?
     
  5. Sep 18, 2010 #4
    [tex]dF=-SdT + \gamma dA [/tex]

    this is a differential for the function F(T,A). Hence we can see that:

    [tex]\frac{\partial F}{\partial T}\bigg |_A = -S[/tex]

    Now differentiate wrt A keeping T constant:

    [tex]\frac{\partial^2 F}{\partial A\partial T} = -\frac{\partial S}{\partial A}\bigg |_T[/tex]

    Now do the same for the other part:

    [tex]\frac{\partial F}{\partial A}\bigg |_T = \gamma [/tex]

    differentiate wrt. T keeping A constant and then use Young's theorem, and you are done
     
  6. Sep 19, 2010 #5
    oh that makes really sense

    is this correct

    [tex] \frac{\partial^2 F}{\partial T\partial A} = (\frac{\partial \gamma}{\partial T})_A[/tex]

    so [itex] (\frac{\partial S}{\partial A})_{T}=-(\frac{\partial \gamma}{\partial T})_{A}[/itex] by Young's theorem.

    and the last part

    [tex] u(T)= \frac{U}{A}= \frac{TS+\gamma A}{A}=\frac{TS}{A}+\gamma= \gamma -T(\frac{\partial \gamma}{\partial T})_{A}[/itex]
    is this correct?
     
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