Question

The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 1 atm at 64.7 oC. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 55.5 oC? Give your answer in atmospheres, to the third decimal point.

Answer 1

Answer: 55.5 oC is 0.014 atm (3rd decimal point)

Step-by-step explanation:

The Clausius-Clapeyron equation is given as:

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

where:

P1 = vapor pressure at temperature T1

P2 = vapor pressure at temperature T2

ΔH_vap = enthalpy of vaporization

R = gas constant = 8.314 J/(mol*K)

Converting the enthalpy of vaporization to J/mol:

ΔH_vap = 35.2 kJ/mol = 35,200 J/mol

Converting temperatures to Kelvin:

T1 = 64.7 + 273.15 = 337.85 K

T2 = 55.5 + 273.15 = 328.65 K

Substituting the values into the equation and solving for P2:

ln(P2/1 atm) = -(35,200 J/mol / 8.314 J/(mol*K)) * (1/328.65 K - 1/337.85 K)

ln(P2/1 atm) = -4.231

P2/1 atm = e^(-4.231)

P2 = 0.014 atm

Therefore, the vapor pressure for methanol at 55.5 oC is 0.014 atm, to the third decimal point.