Methanol, CH3OH a colorless voltile liquid was formerly known as wood alcohol. It boils at 65 degrees C and has a heat of vaporization of 37.4 kJ/mol. What is its vapor pressure at 22.0 degrees C?
Try the Clausius-Clapeyron equation.
To determine the vapor pressure of methanol at 22.0 degrees C, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at a certain temperature to its boiling point, heat of vaporization, and the temperature of interest. The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where:
- P2 and P1 are the vapor pressures at temperatures T2 and T1, respectively
- ΔHvap is the heat of vaporization
- R is the ideal gas constant (8.314 J/mol*K)
- T2 and T1 are the temperatures at which the vapor pressures are measured, in Kelvin
First, we need to convert the temperatures to Kelvin:
- T1 = 65 C + 273.15 = 338.15 K (boiling point of methanol)
- T2 = 22 C + 273.15 = 295.15 K (temperature of interest)
Now, let's calculate the vapor pressure at 22.0 degrees C.
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
ln(P2/1 atm) = (-(37.4 kJ/mol) * (1/8.314 J/mol*K)) * (1/295.15 K - 1/338.15 K)
Simplifying:
ln(P2/1 atm) = -0.1805
Now, we can solve for P2 by taking the exponential of both sides of the equation:
P2/1 atm = e^(-0.1805)
Finally, solving for P2:
P2 = 1 atm * e^(-0.1805)
Using a scientific calculator or software, we find:
P2 ≈ 0.835 atm
Therefore, the vapor pressure of methanol at 22.0 degrees C is approximately 0.835 atm.