Question

What is the `DeltaH_"vap"` for ethanol and propane?

The vapor pressure for pure ethanol and pure propane is measured as a function of temperature. In each case, a graph of the ln of the vapor pressure versus 1/T is found to be a straight line. The slope of the line for ethanol is − 4643 K, and the slope of the line for propane is − 1888 K

Answer 1

Ethanol: `∆H_(vap) = 38.6kJ/(mol)`
Propane: `∆H_(vap) = 15.7kJ/(mol)`

Explanation:

Ethanol: -4643 'K slope = `-(∆H_(vap))/8.314`.
`∆H_(vap) = 4643* 8.314 = 38602 J/(mol)`.
`∆H_(vap) = 38.6kJ/(mol)`
Actual values (varies with temperature) here:
http://www.ddbst.com/en/EED/PCP/HVP_C11.php

Propane: -1888'K slope = `-(∆H_(vap))/8.314`.
`∆H_(vap) = 1888* 8.314 = 15697 J/(mol)`.
`∆H_(vap) = 15.7kJ/(mol)`
Actual values (varies with temperature) here:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C74986&Mask=4

The Clausius–Clapeyron equation describes the relationship between the vapor pressure of a substance, p, and the temperature at which it boils, T:
`ln(p) = - (DeltaH)/(R*T) + L`

Here ΔH stands for the enthalpy (heat) of vaporization, L is a substance-specific constant, and R is the universal gas constant, 8.314 J/(mol K).

Note that the variables you will be measuring in this experiment are p and T. The equation should yield a straight line if ln p is plotted along the vertical (y) axis and 1/T is plotted along the horizontal (x) axis.

The slope of this line, `(d(ln p))/(d(1/T))` equals `-(∆H_(vap))/R`.

Thus, if a numerical value of the slope is obtained from the graph of your experimental data, you should be able to calculate a value for ∆Hvap. The heat of vaporization represents the amount of thermal energy necessary to convert one mole of the liquid in question into the gaseous state. It is usually reported in kJ/mol.
https://www.macalester.edu/~kuwata/Classes/2001-02/Chem%2011/Vapor%20Pressure%20Lab.pdf