10.4 Vapor Pressure

The vapor pressure of a liquid is defined as the pressure exerted by the molecules that escapes from the liquid to form a separate vapor phase above the liquid surface. The vapor pressure of water at room temperature (25° C) is 0.0313 atm, or 23.8 mm of mercury (760 mm Hg = 1 atm).

Vapor pressure is constant when there is an equilibrium of water molecules moving between the liquid phase and the gaseous phase, in a closed container.

The vapor pressure of a liquid is the point at which equilibrium pressure is reached, in a closed container, between molecules leaving the liquid and going into the gaseous phase and molecules leaving the gaseous phase and entering the liquid phase.

Note the mention of a “closed container”. In an open container the molecules in the gaseous phase will just fly off and an equilibrium would not be reached, as many fewer gaseous molecules would be re-entering the liquid phase. Also note that at equilibrium the movement of molecules between liquid and gas does not stop, but the number of molecules in the gaseous phase stays the same—there is always movement between phases. So, at equilibrium there is a certain concentration of molecules in the gaseous phase; the pressure the gas is exerting is the vapor pressure. As for vapor pressure being higher at higher temperatures, when the temperature of a liquid is raised, the added energy in the liquid gives the molecules more energy and they have greater ability to escape the liquid phase and go into the gaseous phase.

 Fig.10.53

Source: commons.wikimedia.org/

Liquids and solids molecules of condensed phases as permanently confined within them, these molecules still possess some thermal energy, and there is always a chance that one that is near the surface will occasionally fly loose and escape into the space outside the solid or liquid. As more molecules escape the liquid, the pressure they exert increases. The liquid and vapor reach a state of dynamic equilibrium:  liquid molecules evaporate and vapor molecules condense at the same rate.

Fig.10.54

Source: commons.wikimedia.org/

Fig.10.55

Source: https://www.chem1.com/acad/webtext/virtualtextbook.html

Eventually they stabilize at a fixed value P_vap that depends on the substance and on the temperature and is known as the equilibrium vapor pressure, or simply as the “vapor pressure” of the liquid or solid. The vapor pressure is a direct measure of the escaping tendency of molecules from a condensed state of matter.

Note carefully that if the container is left open to the air, it is unlikely that many of the molecules in the vapor phase will return to the liquid phase. They will simply escape from the entire system and the partial pressure of water vapor P_w will never reach P_vap; the liquid will simply evaporate without any kind of equilibrium ever being achieved.

Liquids – Vapor Pressure

Approximate vapor pressure for temperatures in the range 20 ^oC – 25 ^oC (68 ^oF – 77 ^oF).

Table 10.5

• Ref: Engineering ToolBox, (2006). Vapor Pressures for some common Liquids. [online] Available at: https://www.engineeringtoolbox.com/vapor-pressure-d_312.html [Accessed Day Mo. Year].

National weather service allows to determine vapor pressure using interactive data. Here is the link: https://www.weather.gov/epz/wxcalc_vaporpressure

Table 10.6

Vapor Pressure of Water from 0 °C to 100 °C

Reference: Lange’s Handbook, pps. 5-28 to 5-29.

Vapor pressure plots and boiling points

The escaping tendency of molecules from a phase always increases with the temperature; therefore the vapor pressure of a liquid or solid will be greater at higher temperatures. The variation of the vapor pressure with the temperature is not linear.

It’s important that you be able to interpret vapor pressure plots such as the three shown here. Take special note of how boiling points can be found from these plots. You will recall that the normal boiling point is the temperature at which the liquid is in equilibrium with its vapor at a partial pressure of 1 atm (760 torr). Thus the intercepts of each curve with the blue dashed 760-torr line indicate the normal boiling points of each liquid. Similarly, you can easily estimate the boiling points these liquids would have in Denver, Colorado where the atmospheric pressure is 630 torr by simply constructing a horizontal line corresponding to this pressure.

Fig.10.56

Source: www.Openstax.org/

Fig.10.57

Source: commons.wikimedia.org

Clausius-Clapeyron Equation

The equilibrium between a liquid and its vapor depends on the temperature of the system; a rise in temperature causes a corresponding rise in the vapor pressure of its liquid. The Clausius-Clapeyron equation gives the quantitative relation between a substance’s vapor pressure (P) and its temperature (T); it predicts the rate at which vapor pressure increases per unit increase in temperature.

Fig.10.58

where ΔH_vap 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. However, since the relationship between vapor pressure and temperature is not linear, the equation is often rearranged into logarithmic form to yield the linear equation:

For any liquid, if the enthalpy of vaporization and vapor pressure at a particular temperature is known, the Clausius-Clapeyron equation allows to determine the liquid’s vapor pressure at a different temperature. To do this, the linear equation may be expressed in a two-point format. If at temperature T₁, the vapor pressure is P₁, and at temperature T₂, the vapor pressure is P₂, the corresponding linear equations are:

Fig.10.55

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:

Fig.10.56

which can be combined into:

Fig.10.57

Fig.10.58

Fig.10.59

Source: Commons.wikimedia.org/

Practice Problem: If the vapor pressure of water at 293 K is 17.5 mmHg, what is the vapor pressure of water at 300 K?

Solution

Step 1: Use the Clausius Clapeyron equation (Equation 11). Assume 293 K to be T₁ and 17.5 mmHg to be P₁ and 300 K to be T₂. We know the enthalpy of vaporization of water is 44000 J mol^-1. Therefore we plug in everything we are given into the equation.

Step 2: Calculate everything possible, which is everything on the right side of the equation.

Step 3: To isolate the variable, we need to get rid of the natural log function on the left side. To do so, we must exponentiate both sides of the equation after calculating the numerical value of the right side of the equation:

Step 4: To solve for P₂, we multiply both sides of the equation by 17.5mmHg.

=26.7 mm of Hg

The vapor pressure of water at 298K is 23.8 mmHg. At what temperature is the vapor pressure of water 1075 mmHg?

Solution

Step 1: Once again this can be solved using the Clausius-Clapeyron equation. If we set P1 = 23.8 mmHg, T1 = 298K, and P2 = 1075 mmHg, all we need to do is to solve for T₂. So plug everything we know into the Clausius-Clapeyron Equation and we get:

Step 2: We can solve the right side of the equation for a numerical answer and we can simplify the right side of the equation to:

Step 3: Further simplifying the equation by distributing the 5291.96 K, we get:

Step 4: Solving for T₂ we get:

T2 =383K

*Note: There may be some slight variations in this answer if you try working this problem out. The rounding in this problem will make a relatively large difference in mmHg (10-20 mmHg).

Here is how enthalpy of vaporization is determined experimentally: