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

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Discussion Overview

The discussion centers around the appropriate units for pressure when using the Clausius-Clapeyron equation to determine the boiling point of a substance. Participants explore the implications of different units on the equation and its application, including the relationship between pressure, temperature, and enthalpy of vaporization.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether pressure should be in torr or atm when applying the Clausius-Clapeyron equation, noting that using R=8.314 leads to unit cancellation.
  • Another participant clarifies that using P=1 atm results in a solvable equation for temperature, challenging the initial claim of cancellation.
  • Concerns are raised about the units of temperature and the intercept 'b' in the equation, with one participant suggesting that the equation may be limited to finding the slope only.
  • A participant introduces the concept of activity and its relationship to pressure, suggesting that pressure is an approximation of activity, which is unitless.
  • One participant recommends using SI units, stating that pressure should be in Pascals when using R in J/K, and emphasizes that ln(P) remains unitless regardless of the pressure units used.
  • There is a discussion about the dimensionality of Hvap/RT, with participants agreeing that it is dimensionless.

Areas of Agreement / Disagreement

Participants express differing views on the implications of using various units for pressure in the Clausius-Clapeyron equation. While some agree on the dimensionality of certain terms, there is no consensus on the best approach to take regarding units or the necessity of using different forms of the equation.

Contextual Notes

Participants mention potential limitations regarding the assumptions made about pressure and temperature units, as well as the interpretation of the intercept 'b'. There is also uncertainty about the applicability of the equation in different contexts.

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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