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Differential of X [ dX = U/V dV + U/p dp ] Internal Energy?

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data

    dX = U/V dV + U/p dp

    1. Write the differential of X in terms of the independent variables.

    2. Prove that this is an exact differential.

    3. Use the ideal gas equation of state to verify that X is actually the internal energy and that it satisfies the above equation. Would you expect the internal energy to be a state variable for the van der Waals gas?
    2. Relevant equations
    Internal Energy
    dU = TdS - PdV


    3. The attempt at a solution
    dX = U/V dV + U/p dp ... I'm a bit too embarrassed to type anything that I've tried. I'm not really sure where to start.

    dX = U / -V2 + U / -p2
    = - U / V2 - U / p2

    Partial Derivative Attempt

    dX = U/V dV + U/p dp

    dX = [ U / V dV ]p + [ U / p dp ]V

    Any help to put me in the right direction would be much appreciated. Thanks
     
  2. jcsd
  3. May 13, 2015 #2
    If dX is going to be an exact differential, then a good place to start on this is to write down the chain rule expression for dX in terms of dP and dV. Then compare that equation with your starting differential equation.

    Chet
     
  4. May 13, 2015 #3
    dU / dX = dU/dV dU/dp - Chain Rule expression for dX in terms of dP and dV?

    M =(V,U) dV N= (U,p) dp

    ( ∂U / ∂V = dV ) + ( ∂U / ∂p dp ) = dX

    X (V,p)
    ∴ X = U (Internal Energy)
     
  5. May 13, 2015 #4
    What I had in mind was:
    $$dX=\left(\frac{\partial X}{\partial v}\right)_pdv+\left(\frac{\partial X}{\partial p}\right)_vdp$$
    and setting U = U(p,v), so, if the equation represents an exact differential, then
    $$\left(\frac{\partial X}{\partial v}\right)_p=\frac{U(p,v)}{v}$$
    $$\left(\frac{\partial X}{\partial p}\right)_v=\frac{U(p,v)}{p}$$

    Chet
     
  6. May 13, 2015 #5
    Ok. Exact and partial differentials are still pretty new to me so I think I'll do a bit more reading and come back to this. Thank you for your reply!
     
  7. May 13, 2015 #6
    Yes. You need to be able to work with them to do many thermo analyses.

    Chet
     
  8. May 13, 2015 #7
    Doesn't this mean that internal energy is not a state variable for the van der Waals gas?

    van der Waals equation of state:

    waals.gif

    Ideal Gas Equation of State:

    pV = nRT
     
  9. May 14, 2015 #8
    Internal energy is a state variable for all materials.

    Chet
     
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