The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)
A boiling point at reduced pressure
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in
[link] to determine the boiling point of water at this elevation.
Solution
The graph of the vapor pressure of water versus temperature in
[link] indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.
Check your learning
The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use
[link] to determine the approximate atmospheric pressure at the camp.
The quantitative relation between a substance’s vapor pressure and its temperature is described by the
Clausius-Clapeyron equation :
where Δ
Hvap is the enthalpy of vaporization for the liquid,
R is the gas constant, and ln
A is a constant whose value depends on the chemical identity of the substance. This equation is often rearranged into logarithmic form to yield the linear equation:
This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T
1 , the vapor pressure is P
1 , and at temperature T
2 , the vapor pressure is T
2 , the corresponding linear equations are:
Since the constant, ln
A , is the same, these two equations may be rearranged to isolate ln
A and then set them equal to one another:
which can be combined into:
Estimating enthalpy of vaporization
Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.
Solution
The enthalpy of vaporization, Δ
Hvap , can be determined by using the Clausius-Clapeyron equation:
Since we have two vapor pressure-temperature values (
T1 = 34.0 °C = 307.2 K,
P1 = 10.0 kPa and
T2 = 98.8 °C = 372.0 K,
P2 = 100 kPa), we can substitute them into this equation and solve for Δ
Hvap . Rearranging the Clausius-Clapeyron equation and solving for Δ
Hvap yields:
Note that the pressure can be in any units, so long as they agree for both
P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.
Check your learning
At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.
