Boiling Points
Boiling Points
When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere.
The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). The figure below shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure.
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.)
Example
A Boiling Point at Reduced Pressure
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in the figure above to determine the boiling point of water at this elevation.
Solution
The graph of the vapor pressure of water versus temperature in the figure above 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.
The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation:
\(P=A{e}^{-\text{Δ}{H}_{\text{vap}}\text{/}RT}\)
where ΔHvap is the enthalpy of vaporization for the liquid, R is the gas constant, and A is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:
\(\text{ln}\phantom{\rule{0.2em}{0ex}}P=-\cfrac{\text{Δ}{H}_{\text{vap}}}{RT}\phantom{\rule{0.2em}{0ex}}+\text{ln}\phantom{\rule{0.2em}{0ex}}A\)
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 T1, the vapor pressure is P1, and at temperature T2, the vapor pressure is T2, the corresponding linear equations are:
\(\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{1}=-\cfrac{\text{Δ}{H}_{\text{vap}}}{R{T}_{1}}\phantom{\rule{0.2em}{0ex}}+\text{ln}\phantom{\rule{0.2em}{0ex}}A\phantom{\rule{5em}{0ex}}\text{and}\phantom{\rule{5em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{2}=-\cfrac{\text{Δ}{H}_{\text{vap}}}{R{T}_{2}}\phantom{\rule{0.2em}{0ex}}+\text{ln}\phantom{\rule{0.2em}{0ex}}A\)
Since the constant, A, is the same, these two equations may be rearranged to isolate ln A and then set them equal to one another:
\(\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{1}+\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}{H}_{\text{vap}}}{R{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\text{ln}\phantom{\rule{0.2em}{0ex}}{P}_{2}+\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}{H}_{\text{vap}}}{R{T}_{2}}\)
which can be combined into:
\(\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{{P}_{2}}{{P}_{1}}\right)=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}{H}_{\text{vap}}}{R}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{T}_{2}}\right)\)
Example
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, ΔHvap, can be determined by using the Clausius-Clapeyron equation:\(\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{{P}_{2}}{{P}_{1}}\right)=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}{H}_{\text{vap}}}{R}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{T}_{2}}\right)\)
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 ΔHvap. Rearranging the Clausius-Clapeyron equation and solving for ΔHvap yields:
\(\text{Δ}{H}_{\text{vap}}=\phantom{\rule{0.2em}{0ex}}\cfrac{R\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{{P}_{2}}{{P}_{1}}\right)}{\left(\cfrac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{T}_{2}}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(8.3145\phantom{\rule{0.2em}{0ex}}\text{J/mol}\text{⋅}\text{K}\right)\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{\text{100 kPa}}{\text{10.0 kPa}}\right)}{\left(\cfrac{1}{307.2\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{372.0\phantom{\rule{0.2em}{0ex}}\text{K}}\right)}\phantom{\rule{0.2em}{0ex}}=\text{33,800 J/mol}=\text{33.8 kJ/mol}\)
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.
Example
Estimating Temperature (or Vapor Pressure)
For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?Solution
If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, ΔHvap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:\(\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{{P}_{2}}{{P}_{1}}\right)=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}{H}_{\text{vap}}}{R}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{1}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{T}_{2}}\right)\)
Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (T1 = 80.1 °C = 353.3 K, P1 = 101.3 kPa, ΔHvap = 30.8 kJ/mol) and want to find the temperature (T2) that corresponds to vapor pressure P2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for T2. Rearranging the Clausius-Clapeyron equation and solving for T2 yields:
\({T}_{2}={\left(\cfrac{-R\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{{P}_{2}}{{P}_{1}}\right)}{\text{Δ}{H}_{\text{vap}}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{T}_{1}}\right)}^{-1}={\left(\cfrac{-\left(8.3145\phantom{\rule{0.2em}{0ex}}\text{J/mol}\text{⋅}\text{K}\right)\text{⋅}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{83.4\phantom{\rule{0.2em}{0ex}}\text{kPa}}{101.3\phantom{\rule{0.2em}{0ex}}\text{kPa}}\right)}{\text{30,800 J/mol}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{353.3\phantom{\rule{0.2em}{0ex}}\text{K}}\right)}^{-1}=\text{346.9 K or}\phantom{\rule{0.2em}{0ex}}{73.8}^{\circ }\text{C}\)
This lesson is part of:
Liquids and Solids