What Units Should Pressure Be in for the Clausius Clapeyron Equation?

AI Thread Summary
The discussion centers on the appropriate units for pressure in the Clausius-Clapeyron equation when determining boiling points. It clarifies that pressure can be expressed in various units, such as torr or atm, but the slope of the plot remains consistent regardless of the chosen units. Using the ideal gas constant R = 8.314, the equation's terms become dimensionless, allowing for flexibility in pressure units. The conversation emphasizes that ln(P) is unitless, and when using standard SI units, pressure should be in Pascals, temperature in Kelvin, and enthalpy of vaporization in Joules. Ultimately, the equation can be manipulated to find temperatures at different pressures without losing its validity.
LogicX
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Homework Statement



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.

Homework Equations



lnP=−ΔHvap/RT+b

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)?
 
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LogicX said:
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.
 
Borek said:
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.
 
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.
 
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.
 
dpsguy said:
Hvap/RT is also dimensionless

I was hoping to guide LogicX to find it out.
 
Borek said:
I was hoping to guide LogicX to find it out.

Sorry for disturbing your plans. You are vastly more experienced in these things than I am :)
 
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