Final answer:
To determine the vapor pressure of methanol at 12.0°C, we use the Clausius-Clapeyron equation with the given values for normal boiling point, vapor pressure, and enthalpy of vaporization. Temperatures are converted to Kelvin and values are substituted to solve for the desired vapor pressure.
Step-by-step explanation:
To find the vapor pressure of methanol at 12.0°C, we can use the Clausius-Clapeyron equation which relates the temperature and vapor pressure of a substance to its enthalpy of vaporization (ΔHvap). The normal boiling point of methanol is 65.6°C with a vapor pressure of 101.3 kPa and an enthalpy of vaporization of 35.2 kJ/mol. We can set up the equation as follows: ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1) where P1 is the vapor pressure at the normal boiling point, P2 is the vapor pressure at the desired temperature, T1 is the normal boiling point in Kelvin, and T2 is the desired temperature in Kelvin. First, we convert the temperatures from Celsius to Kelvin by adding 273.15. That gives us T1 = 65.6 + 273.15 = 338.75 K and T2 = 12.0 + 273.15 = 285.15 K. Next, we plug the values into the equation and solve for P2 ln(P2/101.3) = -(35.2 kJ/mol)/(8.314 J/mol·K) × (1/285.15 K - 1/338.75 K) After calculating the right side of the equation, we take the exponential of both sides to solve for P2. This will give us the vapor pressure of methanol at 12.0°C.