Question

 Please help

The enthalpy of vaporization for water is 40.7 kJ/mol. Water has a vapor pressure of 101.3 ka at 100.0 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 70.0
°C? Give your answer in kPa, to the first decimal point.

Answer 1

The Clausius-Clapeyron equation is:

ln(P₂/P₁) = -ΔH_vap/R * (1/T₂ - 1/T₁)

where P₁ and T₁ are the vapor pressure and temperature of the first substance (water), P₂ and T₂ are the vapor pressure and temperature of the second substance (methanol), ΔH_vap is the enthalpy of vaporization, R is the gas constant (8.314 J/mol*K).

Using the given values:

P₁ = 101.3 kPa
T₁ = 100.0 + 273.15 = 373.15 K
ΔH_vap = 40.7 kJ/mol
R = 8.314 J/mol*K

We need to solve for P₂ at T₂ = 70.0 + 273.15 = 343.15 K.

ln(P₂/101.3) = -40700 J/mol / (8.314 J/mol*K) * (1/343.15 K - 1/373.15 K)

ln(P₂/101.3) = -3.948

P₂/101.3 = e^(-3.948)

P₂ = 16.1 kPa

Therefore, the vapor pressure for methanol at 70.0 °C is 16.1 kPa (to the first decimal point)