Question

The boiling point using the Clausius-Clapeyron equation

Answer 1

I've collected experimental data in order to determine methanol's thermodynamic parameters. Now I've come to the wish I want to determine the boiling point, and surely wiki had an equation, but I don't dare to use it, as I can't see how it is derived.
Attached is found my data.

Answer 2

The enthalpy of vaporization has been found using the Clausius-Clapeyron:
\[\large -R \frac{ d[ \ln(p)] }{ d(T^{-1} )}= \Delta _{vap}H\]
Ones I've found the boiling point I can determine the entropy of vaporization from
\[\Large \Delta_{vap}S(T _{b})=\frac{ \Delta _{vap}H(T _{b}) }{ T _{b} }\]

Answer 3

The equation of interest on the wiki is from here:
http://en.wikipedia.org/wiki/Boiling_point#Saturation_temperature_and_pressure
I just always want a second pressure term in the equation.

Answer 4

"Now I've come to the wish I want to determine the boiling point"

What do you mean by this?
You already have a curve that gives you Tb with respect to pressure.
If you choose any pressure, you can read Tb at that pressure on the graph.

Answer 5

I wanted to try avoiding doing specific readings for the data, but you suggest I simply use my regression data and do a prediction?
Eg. lets take the atmospheric boiling point:
We assume data follow the Clausius-Clapeyron relation and thereby write the regression line as.
\[\Large \ln(p)=-\frac{ \Delta _{vap}H }{ R } T ^{-1}+C\]
Here is C an integration constant, then solve for the temperature for \(\large p=1 atm\)?

Answer 6

Sure it makes sense! I would so no matter what.

Answer 7

Now if you want Tb as a function of pressure, sure, Clapeyron's relation is your saviour.

Answer 8

What you posted 3 lines above is correct. Use a known value to find C.
Be careful with the units though. In your equation, \(\Delta_{vap}H\) must be a molar quantity.

Answer 9

Yes I thought about it, but luck for my the differential coefficient is in kelvin. So a kelvin unit multiplied with the gas constant give joule/mol

Answer 10

but lucky*

Answer 11

Your equation is correct on a small temperature interval where \(\Delta_{vap}H\) is constant.
On longer intervals, we usually choose to write : \(\Delta_{vap}H = A-B.T\)

Answer 12

Oh are A and B empirical constants?
So far I've written my assumption as:
\[\Large \frac{ d(\Delta _{vap}H) }{ dT }=0\]

Answer 13

Yes, they are. This is because the real curve looks like that:


On any interval, the linear approximation is great, and you can still integrate to find ln(P) as a function of T.

Answer 14

Ah I see. Of curiosity how does one determine the empirical constant for the molar heat capacity?

Answer 15

All my book write is that a good approximation is to fit so:
\[\Large C _{p.m}=a+bT+\frac{ c }{ T^2 }\]

Answer 16

Do a lot of enthalpy vs temperature plots and fit?