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.