Question

The enthalpy of vaporization for water is 40.7 kJ/mol. Water has a vapor pressure of 101.3 kPa at 100.0 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 73.5°C? Give your answer in KPa, to the first decimal point.

Answer 1

The Clausius-Clapeyron equation is given by:

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

where P1 and T1 are the vapor pressure and temperature of the substance at one point, P2 and T2 are the vapor pressure and temperature at another point, ΔHvap is the enthalpy of vaporization, and R is the gas constant.

We can use this equation to find the vapor pressure of methanol at 73.5°C, given the vapor pressure of water at 100.0°C.

First, we convert the temperatures to Kelvin:

T1 = 100.0°C = 373.2 K

T2 = 73.5°C = 346.7 K

Next, we substitute the values into the equation, along with the enthalpy of vaporization for methanol and the gas constant:

ln(P2/101.3 kPa) = -(35.2 kJ/mol / 8.314 J/(mol*K)) * (1/346.7 K - 1/373.2 K)

Simplifying, we get:

ln(P2/101.3 kPa) = -5.631

Taking the exponential of both sides, we get:

P2/101.3 kPa = e^(-5.631)

P2 = 101.3 kPa * e^(-5.631)

P2 = 2.784 kPa

Therefore, the vapor pressure of methanol at 73.5°C is approximately 2.784 kPa, to the first decimal point.