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

In summary, the equation allows you to find the slope of the line that connects ln(P) to (1/T). You can use this slope to find the temperature at different pressures or temperatures.
  • #1
LogicX
181
1

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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  • #2
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

[tex]\frac {\Delta H_{vap}}{RT} + b = 0[/tex]

which you can easily solve for T.
 
  • #3
Borek said:
Huh? What do you mean by "everything will cancel out"? You will get

[tex]\frac {\Delta H_{vap}}{RT} + b = 0[/tex]

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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
dpsguy said:
Hvap/RT is also dimensionless

I was hoping to guide LogicX to find it out.
 
  • #7
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 :)
 

1. What is the Clausius Clapeyron Equation?

The Clausius Clapeyron Equation is a thermodynamic equation that describes the relationship between temperature and vapor pressure of a substance. It is commonly used to predict the boiling point of a liquid at different pressures.

2. Who developed the Clausius Clapeyron Equation?

The equation was developed by two scientists, Rudolf Clausius and Benoit Paul Emile Clapeyron, in the mid-19th century. However, it is often referred to as just the Clausius Clapeyron Equation in recognition of their contributions.

3. What is the significance of the Clausius Clapeyron Equation?

The Clausius Clapeyron Equation is important in understanding the behavior of gases and vapors at different pressures and temperatures. It is also used in the study of phase transitions, such as the change from liquid to gas.

4. How is the Clausius Clapeyron Equation derived?

The equation is derived from the first and second laws of thermodynamics, specifically the relationships between heat, work, and internal energy. It can also be derived from the ideal gas law and the definition of enthalpy.

5. What are some practical applications of the Clausius Clapeyron Equation?

The equation is used in various fields, including chemistry, physics, and meteorology. It is used to predict the boiling point and vapor pressure of substances, as well as to study the behavior of gases and vapors in different environments. It is also useful in understanding weather patterns and predicting changes in atmospheric conditions.

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