A liquid has a delta(Hvap) of 35.5 kJ/mol and a boiling point of 122 C at 1.00atm. What is its vapor pressure at 113 C?
To find the vapor pressure of a liquid at a given temperature, you can use the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = enthalpy of vaporization
R = ideal gas constant
T1 = initial temperature
T2 = final temperature
In this case, we are given:
ΔHvap = 35.5 kJ/mol
T1 = 122 °C = 395 K
T2 = 113 °C = 386 K
We need to convert the given enthalpy of vaporization from kJ/mol to J/mol:
ΔHvap = 35.5 kJ/mol = 35,500 J/mol
Now, let's plug the values into the Clausius-Clapeyron equation:
ln(P2/1.00atm) = -35,500 J/mol / (8.314 J/(mol·K)) * (1/386 K - 1/395 K)
ln(P2/1.00atm) = -4.2762 * (-0.002311)
Simplifying this equation:
ln(P2/1.00atm) = 0.009891
To solve for P2, we need to exponentiate both sides of the equation:
P2/1.00atm = e^(0.009891)
Now, we can solve for P2 by multiplying both sides of the equation by 1.00atm:
P2 = 1.00atm * e^(0.009891)
Using a scientific calculator or computer software to evaluate the exponential function:
P2 ≈ 1.01atm
Therefore, the vapor pressure of the liquid at 113 °C is approximately 1.01 atm.