The vapor pressure of a liquid increases as the temperature increases. This is why puddles left after a rainstorm evaporate faster on a hot day than on a cold day.

The vapor pressure of a liquid increases faster as the temperature nears the boiling point of the liquid- the data is not a straight line. However, it turns out that the plot of the log of the vapor pressure vs. 1/T is a straight line with a slope equal to that of -DH_vap/R. We can use this fact to derive a simple equation that relates the vapor pressure at certain temperatures to the heat of vaporization, the Clausius-Clapeyron Equation.

The equation for a straight line is y=mx+b, where m is the slope and b is the y-intercept. For our case, the y values are ln(P_vap) and the x values 1/T and the slope m = -DH_vap/R. Thus, the equation for the line is

ln(P_vap) = -DH_vap/RT + b

If we discover the vapor pressure at two separate temperatures, we have two points on the same line.

1. ln(P₂) = -DH_vap/RT₂ + b
2. ln(P₁) = -DH_vap/RT₁ + b

Example 1: Water has a vapor pressure of 24 mmHg at 25^oC and a heat of vaporization of 40.7 kJ/mol. What is the vapor pressure of water at 67^oC?

Solution: Simply use the Clausius-Clapeyron Equation to figure out the vapor pressure. We have to be a bit careful about the units of R: the units we're using are kJ, so R = 8.31x10^-3 kJ/mol K.

ln(P₂/P₁) = -DH_vap/R * (1/T₂- 1/T₁)
ln(P₂/24) = 40.7 kJ/8.31x10^-3 kJ/mol K *(1/340- 1/298)
ln(P₂/24) = 2.03
P₂/24 = 7.62
P₂ = 182 mmHg

Solution: Again, use the Clausius-Clapeyron Equation. Here, the only thing we don't know is DH_vap

ln(P₂/P₁) = -DH_vap/R * (1/T₂- 1/T₁)
ln(88/39) = -DH_vap/8.31x10^-3*(1/318 - 1/298)
DH_vap = 32.0 kJ

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