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Work and isothermal compressibility

  1. Mar 21, 2015 #1
    1. The problem statement, all variables and given/known data
    1 kg of water is at room temperature and the pressure is isothermally increased on the system from 1 atmosphere to 1000 atmospheres. What is the work done? What is the change in heat? What would be the temperature change if this was done adiabatically? The volumetric thermal expansion is 1.5 10-5/K and the isothermal compressibility is 4.9 10-12/Pa.

    2. Relevant equation
    PV=nRT
    dW=-PdV
    α=1/V(dV/dT)p = 1.5*10^-5/K (that's partial v partial t at constant p)
    and
    κ=-1/V(dV/dP)t (partial v partial p at constant t)

    3. The attempt at a solution

    So for part 1 I figure I have to get dW=-PdV into a form that has dP instead of dV, since we don't know the change in volume. I'm pretty much lost at that point. I'm not too clear on how to use Maxwell relations and partial derivatives are kind of confusing to me.
     
  2. jcsd
  3. Mar 22, 2015 #2

    stevendaryl

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    What is the definition of "isothermal compressibility"?
     
  4. Mar 22, 2015 #3
    Can you integrate the equation ##\frac{1}{V}\frac{dV}{dP}=-κ## from the initial pressure to any arbitrary pressure P to get the volume at pressure P? Can you integrate PdV by parts?

    Chet
     
  5. Mar 22, 2015 #4
    I assume isothermal compressibility is just how easy or hard it is to compress something at constant temperature.

    Can I integrate the equation for the compressibility? I'm not sure if you're allowed to separate dV and dP when they're partials at constant T. Like I said, I'm not comfortable with the partials yet. If you could separate them you'd get ln|V|=-κ P (would you not evaluate the integral for P at the limits 1 and 1000?). Then you'd exponentiate both sides so V = e-κP ?

    PdV by parts is PV-∫VdP .

    I'm still not seeing how that helps.
     
  6. Mar 22, 2015 #5

    stevendaryl

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    That's it, except when you integrate, you get an arbitrary constant. So you have:
    [itex]ln(V) = -\kappa P + C[/itex]
    [itex]V = e^C e^{-\kappa P}[/itex]

    You have to pick the constant [itex]C[/itex] so that when [itex]P = P_0[/itex], then [itex]V=V_0[/itex], where [itex]P_0[/itex] and [itex]V_0[/itex] are the initial pressure and volume.
     
  7. Mar 22, 2015 #6
    In addition to what what stevendaryl just said, if you know V as a function of P, you can integrate VdP.

    Chet
     
  8. Mar 22, 2015 #7
    Okay, I think this gives me enough to work on. Thanks.
     
  9. Mar 22, 2015 #8
    Actually I have one more question. When I do the integration by parts for -PdV I get -(PV-∫VdP). The first term in that equation should be evaluated at the limits 1 and 1000 Pa right? So I put V in terms of P, but then the P's cancel out (V=nRT/P) and there's nothing to evaluate. Is that right?
     
  10. Mar 22, 2015 #9
    No. Liquid water doesn't obey the ideal gas law.

    Chet
     
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