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Thermodynamics and Hooke's Law

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data
    *1* The Gibbs function G(T; P) of a certain gas is:
    G = nRT ln P + A + BP + (1/2)*CP^2 + (1/3)DP^3

    where A, B, C and D are constants. Find the equation of state of the gas.


    2. Relevant equations

    G = nRT ln P + A + BP + (1/2)*CP^2 + (1/3)*DP^3

    3. The attempt at a solution

    I think I have a solution for this, but it seemed to be too easy.

    The definition for Gibbs free energy is defined to be:
    1) G = U - TS + PV and 2) dG = -SdT + vdP

    solving 2) for dG/dP yields: dG/dP = -SdT/dP + V 3)

    I thought to take 3) at constant temperature yielding: dG/dP = V
    Now, taking dG/dP at constant T from the given formula yields:

    dG/dP =nRT/P + B + CP + DP^2 4)

    Setting 3) = 4) and solving for PV yields:

    PV = nRT + PB + CP^2 + DP^3
    That just seemed too easy for me >.>



    The second part of this homework is:

    1. The problem statement, all variables and given/known data
    Consider a spring which follows Hookes law; namely the displacement x from equilibrium
    position is proportional to the tension X when it is pulled at a constant temperature. The spring constant is temperature dependent, k = k(T). Determine the free energy F, the internal energy U, and the entropy S, as a function of T and x. Neglect the thermal expansion of spring. Use F0(T) ´ F(T; x = 0); U0(T) ´ U(T; x = 0); S0(T) ´ S(T; x = 0) where necessary.

    2. Relevant equations

    k = k(T)
    F[0](T) = F(T; x = 0)
    U[0](T) = U(T; x = 0)
    S[0](T) =S(T; x = 0)

    where the brackets denote subscripts


    3. The attempt at a solution

    The force of the spring can be obtained by: F = -k(T) * x
    And the work: W = int(-k(T) * x) dx = -(1/2)*k(T)*x^2

    and dW = k(T) *xdx

    Using the first law of thermodynamics: dU = dQ - dW, so

    dU = dQ + k(T)*xdx
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 11, 2008 #2

    Mapes

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    The elongation of the spring would add another term to your energy equation:

    [tex]U=TS-PV+Fx=TS-PV+k(T)x^2[/tex]

    [tex]F=U-TS=-PV+k(T)x^2[/tex]

    [tex]S=-\left(\frac{\partial F}{\partial T}\right)_V=-\frac{\partial k(T)}{\partial T} x^2[/tex]

    Does this help?
     
  4. Mar 11, 2008 #3
    That makes sense to me, but what about those 3 initial conditions? This just seems too easy to me >.>

    F[0](T) = F(T; x = 0)
    U[0](T) = U(T; x = 0)
    S[0](T) =S(T; x = 0)

    What do I do with those?
     
  5. Mar 11, 2008 #4

    Mapes

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    I cheated a bit by ignoring the temperature dependence of PV and by setting my energy baseline at [itex]\mu\,N[/itex]. The initial conditions make these constants go away anyway. Wouldn't you end up with, for example,

    [tex]S=S_0-\frac{\partial k(T)}{\partial T} x^2[/tex]

    I'll leave the rest for you.
     
  6. Mar 13, 2008 #5
    Thanks a bunch! This helps alot! :P
     
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