Question

The comforting smell of roasted coffee can be attributed to a combination of compounds; one

of which is 2-furfurylthiol, C5H6OS. The heat of vaporization of 2-furfurylthiol is 44.7 kJ/mol

and at 20.0 °C it has a vapor pressure of only 3.98 torr. What is its vapor pressure of this

compound in a hot cup of coffee at 71.1 °C?

Answer 1

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(P2/3.98) = 44700/8.314*(1/(20+273) - 1/(71.1+273))

P2 = 3.98*exp( 44700/8.314*(1/(20+273) - 1/(71.1+273)))

P2 = 60.720 torr