Expression for volume as a function of pressure

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Homework Statement


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.


Homework Equations





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)].
 

Answers and Replies

  • #2
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Homework Statement


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.


Homework Equations





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)].
Start out by solving your equation for dV/dP.
 
  • #3
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δV/δP=-κV

I'm not sure what to do now.
 
  • #4
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δV/δP=-κV

I'm not sure what to do now.
Have you learned how to solve a differential equation like this by separation of variables or by integrating factor?
 
  • #5
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No, I haven't taken differential equations.
 
  • #11
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Yes, I'm following.
Does that mean dV/V=dln(V)?
 
  • #12
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-κV=δV/δP
-κδP=δV/V=δln(V)

I can get this far, but I don't know how to solve it now.
 
  • #13
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-κV=δV/δP
-κδP=δV/V=δln(V)

I can get this far, but I don't know how to solve it now.

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?
 
  • #14
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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?
Excellent job!!! Just get rid of that extra equal sign in the last equation.
 
  • #15
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Thanks for the help (and for catching my typo).
 

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