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A graph is shown where the x-axis is labeled “Temperature ( degree sign, C )” and has values of 200 to 1000 in increments of 200 and the y-axis is labeled “Pressure ( k P a )” and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled “ethyl ether,” begins at the point “0 , 200” and extends in a slight curve to point “45, 1000” while the second line, labeled “ethanol”, extends from point “0, 20” to point “88, 1000” in a more extreme curve. The third line, labeled “water,” begins at the point “0, 0” and extends in a curve to point “108, 1000” while the fourth line, labeled “ethylene glycol,” extends from point “80, 0” to point “140, 100” in a very shallow curve.
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.

Answer:

Approximately 40 kPa (0.4 atm)

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The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation    :

P = A e Δ H vap / R T

where Δ H vap 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:

ln P = Δ H vap R T + ln 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 T 1 , the vapor pressure is P 1 , and at temperature T 2 , the vapor pressure is T 2 , the corresponding linear equations are:

ln P 1 = Δ H vap R T 1 + ln A and ln P 2 = Δ H vap R T 2 + ln A

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:

ln P 1 + Δ H vap R T 1 = ln P 2 + Δ H vap R T 2

which can be combined into:

ln ( P 2 P 1 ) = Δ H vap R ( 1 T 1 1 T 2 )

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 H vap , can be determined by using the Clausius-Clapeyron equation:

ln ( P 2 P 1 ) = Δ H vap R ( 1 T 1 1 T 2 )

Since we have two vapor pressure-temperature values ( T 1 = 34.0 °C = 307.2 K, P 1 = 10.0 kPa and T 2 = 98.8 °C = 372.0 K, P 2 = 100 kPa), we can substitute them into this equation and solve for Δ H vap . Rearranging the Clausius-Clapeyron equation and solving for Δ H vap yields:

Δ H vap = R ln ( P 2 P 1 ) ( 1 T 1 1 T 2 ) = ( 8.3145 J/mol K ) ln ( 100 kPa 10.0 kPa ) ( 1 307.2 K 1 372.0 K ) = 33,800 J/mol = 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.

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.

Answer:

47,782 J/mol = 47.8 kJ/mol

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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