Methanol, CH3OH, a colorless, volatile liquid, was formerly known as wood alcohol. It boils at 65.0°C and has a heat of vaporization of 37.4 kJ/mol. What is its vapor pressure at 22.0°C?
Use the Clausius-Clapeyron equation.
p1 = 760 mm
T1 = 65.0 + 273.15
P2 = unknown
T2 = 22.0 + 273.15
delta Hvap = 37400 J/mol
REMEMBER: The vapor pressure of a liquid at its boiling point is ALWAYS atmospheric pressure and standard pressure is 760 mm.
109 mmHg
To calculate the vapor pressure of methanol at 22.0°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔH_vap/R) * (1/T2 - 1/T1)
Where:
P1 = vapor pressure at boiling point = 1 atm
T1 = boiling point = 65.0°C = 338.15 K
T2 = temperature for which we need to find the vapor pressure = 22.0°C = 295.15 K
ΔH_vap = heat of vaporization = 37.4 kJ/mol
R = ideal gas constant = 8.314 J/(mol∙K)
Plugging in the values:
ln(P2/1 atm) = -(37.4 kJ/mol / 8.314 J/(mol∙K)) * (1/295.15 K - 1/338.15 K)
Simplifying the equation:
ln(P2/1 atm) = -0.00450
Taking the exponential of both sides:
e^(ln(P2/1 atm)) = e^(-0.00450)
P2/1 atm = 0.9955
Multiplying both sides by 1 atm:
P2 = 0.9955 atm
Therefore, the vapor pressure of methanol at 22.0°C is approximately 0.9955 atm.
To find the vapor pressure of methanol at 22.0°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R)(1/T2 - 1/T1)
Where:
P1 = vapor pressure at the given boiling point (65.0°C or 338.15K)
P2 = vapor pressure at the desired temperature (22.0°C or 295.15K)
ΔHvap = heat of vaporization
R = ideal gas constant (8.314 J/(mol·K))
T1 = boiling point temperature in Kelvin
T2 = desired temperature in Kelvin
First, we need to convert the boiling point and desired temperature to Kelvin:
T1 = 338.15K
T2 = 295.15K
Now, we can substitute the values into the equation:
ln(P2/P1) = (-37.4 kJ/mol / 8.314 J/(mol·K))(1/295.15K - 1/338.15K)
Simplifying further:
ln(P2/P1) = (-4.5071)(0.00340055 - 0.0029579)
Calculating the values inside the parentheses:
ln(P2/P1) = -0.0014231
To find P2/P1, we can take the antilogarithm (exponential) of both sides:
P2/P1 = e^(-0.0014231)
Calculating the right side of the equation:
P2/P1 ≈ 0.9985774
Now, we can solve for P2 (vapor pressure at 22.0°C) by multiplying both sides of the equation by P1:
P2 = (0.9985774)(P1)
Substituting the boiling point vapor pressure:
P2 = (0.9985774)(1 atm)
Calculating the vapor pressure of methanol at 22.0°C:
P2 ≈ 0.9985774 atm
Therefore, the vapor pressure of methanol is approximately 0.9986 atm at 22.0°C.