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Expression for volume as a function of pressure

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    When pressure is applied to a liquid, its volume decreases. Assuming that the isothermal compressibility κ=-1/V(δV/δP) is independent of pressure, derive an expression for the volume as a function of pressure.


    2. Relevant equations



    3. The attempt at a solution
    I don't know where to start, but I know the answer should be V2=V1 exp[-κ(P2-P1)].
     
  2. jcsd
  3. Jan 21, 2014 #2
    Start out by solving your equation for dV/dP.
     
  4. Jan 22, 2014 #3
    δV/δP=-κV

    I'm not sure what to do now.
     
  5. Jan 22, 2014 #4
    Have you learned how to solve a differential equation like this by separation of variables or by integrating factor?
     
  6. Jan 22, 2014 #5
    No, I haven't taken differential equations.
     
  7. Jan 22, 2014 #6
    Have you learned how to integrate dV/V?
     
  8. Jan 22, 2014 #7
    No, I haven't.
     
  9. Jan 22, 2014 #8
    Do you know what the derivative of ln(x) with respect to x is equal to?
     
  10. Jan 22, 2014 #9
    It's 1/x
     
  11. Jan 22, 2014 #10
    OK. This is where we start.
    If [tex]\frac{dln(x)}{dx}=\frac{1}{x}[/tex]
    then
    [tex]dln(x)=\frac{dx}{x}[/tex]
    Is this OK with you so far?
     
  12. Jan 22, 2014 #11
    Yes, I'm following.
    Does that mean dV/V=dln(V)?
     
  13. Jan 22, 2014 #12
    -κV=δV/δP
    -κδP=δV/V=δln(V)

    I can get this far, but I don't know how to solve it now.
     
  14. Jan 22, 2014 #13
    Do I just integrate both sides from 1 to 2?
    ∫(from 1 to 2) -κδP=∫(from 1 to 2) δln(V)
    -κ(P2-P1)=ln(V2)-ln(V1)=ln(V2/V1)
    V2/V1=e^(-κ(P2-P1))
    V2=V1=e^(-κ(P2-P1))

    Is this correct?
     
  15. Jan 22, 2014 #14
    Excellent job!!! Just get rid of that extra equal sign in the last equation.
     
  16. Jan 22, 2014 #15
    Thanks for the help (and for catching my typo).
     
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