# Clausius Clapeyron Equation:

by LogicX
Tags: clapeyron, clausius, equation
 P: 181 1. The problem statement, all variables and given/known data Trying to figure out the boiling point for a substance using this equation. I have 3 temperatures along with the pressure at those three temperatures. 2. Relevant equations lnP=−ΔHvap/RT+b 3. The attempt at a solution Ok, so obviously a plot of ln(P) vs. (1/T) gives you a slope of −ΔHvap/R. But what units should pressure be in for this equation? I don't get it, if you use R=8.314, all the units cancel out and ln(P) is unitless. I don't know whether to plot it with P in torr or atm. If I use P= 1 atm, ln(P) will equal zero and everything will cancel out. So can I not use atm here? EDIT: Ok I realize that it doesn't matter what units you plot it in, the slope is the same. But then actually using that value, what if I want to find what the temperature will be at 1 atm? EDIT 2: Do I have to use the equation in the other form of the Clausius Clapeyon equation, i.e. ln(p2/p1)=-H/R (1/T1-1/T2)?
P: 22,656
 Quote by LogicX If I use P= 1 atm, ln(P) will equal zero and everything will cancel out.
Huh? What do you mean by "everything will cancel out"? You will get

$$\frac {\Delta H_{vap}}{RT} + b = 0$$

which you can easily solve for T.
P: 181
 Quote by Borek Huh? What do you mean by "everything will cancel out"? You will get $$\frac {\Delta H_{vap}}{RT} + b = 0$$ which you can easily solve for T.
Ok fine, but I still think there is something odd about this equation that you have to use the other form that I posted above in order to cancel out pressure units.

What if P=/= 1 atm? What are your units for temperature? atm/K minus a unitless constant? I don't even know what units b would be in. I feel like this equation is limited to finding the slope and that is it, and you have to use the other version to solve it for a different temperature.

P: 22,656

## Clausius Clapeyron Equation:

What units is H/RT?

Pressure thing is slightly convoluted, and you will see several explanations. The one I stick to is that in general, pressure as used is only an approximation of activity, which is unitless (thanks to the fact it is multiplied by activity coefficient, which has in this case units of pressure-1; as a first approximation activity coefficient is just 1, but it can take different values as well). As we assume standard state to be 1 atm, activity of 1 refers to 1 atm.
 P: 69 Whenever you are in doubt about units, stick to the SI system. The SI unit of pressure is Pascal (1 atm ~ 10^5 Pa). So if you are using R = 8.314 J/K, use the pressure in Pa, temp in K and Hvap in J. b is the intercept that the line makes with the y-axis (lnP). It is dimensionless. Hvap/RT is also dimensionless (see the units of R, H and T above). Also, while P is in Pa, lnP is unitless. For example, if P = 1 atm (10^5 Pa), lnP ~ 7.515, which is just a unitless number. If you can draw a straight line you can extrapolate it to find the temperature at different pressures, or the pressure at different temperatures. I hope this solves your confusion.