Question

What is the vapor pressure of ethanol at 16 ∘C?

Ethanol has a heat of vaporization of 38.56kJ/mol and a normal boiling point of 78.4 ∘C.

Answer in torr please

Thanks!

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)

DATA:

P1 = 760 tor ( 1 atm) T = 78.4 °C = 78.4 +273 = 351.4 K

P2 = P2 ...T2 = 16°C = 16+273 = 289 K

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

ln(P2/760) = 38560/8.314*(1/351.4 -1/289 )

solve for P2

ln(P2/760) = -2.8497

P2 = 760 exp(-2.8497)

P2 = 43.974 torr