# Simple integral of dlnP/dt problem

## 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.

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?

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.

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You have $$\frac {d} {dT} \ln P = \frac L {RT^2}$$ You integrate that from $T_0$ to $T$: $$\int_{T_0}^T\frac {d} {dT} \ln P dT = \int_{T_0}^T \frac L {RT^2} dT$$ The integral on the left is an integral of a derivative, so integration cancels differentiation: $$\int_{T_0}^T\frac {d} {dT} \ln P dT = \left[\ln P\right]_{T_0}^T = \ln P - \ln P_0 = \ln \frac {P}{P_0}$$ Can you figure out the rest?

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Yes, thank you for the help. My brain just seemed to shut down down on this problem.

HallsofIvy