Question

At its normal boiling point of 126 degree C, octane, C H has a vapor pressure of 760 mm Hg. What is its vapor pressure at 25 degree C? The enthalpy of vaporization of 39.07 kJ/mol point (in degree C) of a solution prepared by dissolving 7.40 g Assume ideal behavior. K_fp for H_2 O is -1.86 degree C/m.

Answer 1

Question 10.

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(P2/760) = 39070/8.314*(1/(126+273)-1/(25+273))

P2/760 = exp(-3.99177)

P2 = 760*exp(-3.99177)

P2 = 14.03 mm Hg