Question

Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3 °C and the heat of vaporization (ΔHvap) is 39.9 kJ/mol.

Answer 1

To answer this question, we will use the equation of Clausius-Clapeyron which can be written as follows:
ln(P1/P2) = ΔH/R x (1/T2 - 1/T1)

Now, let's check the parameters in this equation:
P1 is the normal pressure of the vapor which is 4.24 kPa
T1 is the boiling point at this pressure which is 293 K
ΔH is the heat of vaporization given as 39.9 kJ/mol
T2 is given as 355.3 K
and we need to calculate P2

Substitute in the equation to get P2 as follows:
ln(4.24/P2) = 39.9/0.008314 x (1/355.3 - 1/293)
P2 = 101.2 kPa = 0.998766 atm