Methanol (CH3OH) is a colorless volatile liquid, formerly known as wood alcohol. It boils at 65.0 °C and has a heat of vaporization ΔHvap = 37.4 kJ/mol. What is methanol’s vapor pressure at 22.0 °C?
My answer: 109.5 torr
I obtained 109.22 but I think the 37.4 kJ number allows only 3 sig figures.
To calculate the vapor pressure of methanol at 22.0 °C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its boiling point and heat of vaporization.
The equation is as follows:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures, respectively
ΔHvap is the heat of vaporization
R is the ideal gas constant (8.314 J/mol*K)
T1 and T2 are the initial and final temperatures, respectively
Now let's plug in the values we know:
P1 = vapor pressure at boiling point = 1 atm (since boiling point is given in the question)
T1 = boiling point = 65.0 °C = 338.15 K
ΔHvap = 37.4 kJ/mol = 37,400 J/mol
T2 = 22.0 °C = 295.15 K
Plugging these values into the Clausius-Clapeyron equation:
ln(P2/1 atm) = (-37,400 J/mol / (8.314 J/mol*K)) * (1/295.15 K - 1/338.15 K)
Now we need to solve for P2, the vapor pressure at 22.0 °C.
Step 1: Calculate the right-hand side of the equation:
RHS = (-37,400 / 8.314) * (0.003393 - 0.002959) ≈ -1806.98
Step 2: Solve for ln(P2/1 atm):
ln(P2/1 atm) = -1806.98
Step 3: Solve for P2:
P2/1 atm = e^(-1806.98)
P2 = 1 atm * e^(-1806.98)
Now, evaluating this expression will give us the answer:
P2 ≈ 109.5 torr
Therefore, the vapor pressure of methanol at 22.0 °C is approximately 109.5 torr.