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Simple integral of dlnP/dt problem

  • Thread starter UCLphysics
  • Start date
  • #1

Homework Statement



Integrate the Clausius Clapeyron equation to get the saturated partial pressure equation.



Homework Equations



dln(P)/dT = L/RT2

Es= Es0 exp(-L/R (1/T - 1/T0))

The Attempt at a Solution



P=Es0

∫L/RT (limits -> T &T0) = -L/R * (1/T - 1/T0)

dln(P)/dT = 1/P*dP/dT?



I've managed to confused myself to the point where these are the only coherent workings I have. I feel that I'm close and missing a trick? My problem is how I get to the exp and the Es0 constant. I know this is to do with dln(P)/dT, but the maths text book and the notes for the course I have don't help at all.
 

Answers and Replies

  • #2
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On the left hand side you have a derivative of a function with respect to T. On the right hand side you have a function of T. You can integrate both sides with respect to T. What do you get?
 
  • #3
if you use the dln(P)/dT = 1/P*dP/dT relation

P= -P (L/R * (1/T - 1/T0))

I understand you are trying to get me to work through the problem. The problem is my maths are wrong and need to be shown how to do it.
 
Last edited:
  • #4
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390
You have [tex]\frac {d} {dT} \ln P = \frac L {RT^2}[/tex] You integrate that from [itex]T_0[/itex] to [itex]T[/itex]: [tex]\int_{T_0}^T\frac {d} {dT} \ln P dT = \int_{T_0}^T \frac L {RT^2} dT[/tex] The integral on the left is an integral of a derivative, so integration cancels differentiation: [tex]\int_{T_0}^T\frac {d} {dT} \ln P dT = \left[\ln P\right]_{T_0}^T = \ln P - \ln P_0 = \ln \frac {P}{P_0}[/tex] Can you figure out the rest?
 
Last edited:
  • #5
Yes, thank you for the help. My brain just seemed to shut down down on this problem.
 
  • #6
HallsofIvy
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Homework Helper
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One thing that is confusing is the use of "t" on the left and "T" on the right. I assumed that "t" was "time" and "T" was temperature, which makes the problem very difficult!
 
  • #7
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390
The t vs T thing was my fault, I just wrote the differentiation symbol in the autopilot mode. Sorry about that! I have corrected that.
 

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