Question

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 50.66°C? (ΔHvap for ethanol is 39.3 kJ/mol.)

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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)

solve for P2

ln(P2/(100)) = 39300/8.314 * (1/(34.90+273.15) - 1/(50.66+273.15))

ln(P2/100) = 0.746839

P2 = 100*exp(0.746839)

P2 = 211.0318 mm Hg