# 5.5: Saturation, Dew Point and Bubble Point of Liquid Mixtures

• • R.L. Cerro, B. G. Higgins, S Whitaker
• Professors (Chemical Engineering) at University of Alabama at Huntsville & University of California at Davis

When a pure liquid phase (species $$A$$) is in equilibrium with the pure gas phase, the vapor pressure of the component in the gas phase is equal to the vapor pressure of the pure component.

$p_{A} =p_{A, vap} \label{35}$

This result is consistent with setting $$x_{A} = 1$$ in Equation $$(5.4.5)$$ so that the liquid is pure component $$A$$. In general, when air is the gas phase we will assume that the vapor pressure in the gas phase is saturated with the liquid component and that the concentration of the air in the liquid is negligible. When the liquid phase is a mixture, Raoult’s law (Equation $$(5.4.5)$$) must be used to compute the composition of the gas phase. If a liquid mixture is in equilibrium with its own vapors, then the overall pressure is equal to the sum of the vapor pressures of the individual components.

$p=\sum_{A = 1}^{A = N}p_{A, vap} x_{A} \label{36}$

When a liquid mixture is heated, the vapor pressure of the components in the mixture increases and the sum of the partial pressures, given by Equation \ref{36}, increases accordingly. If the liquid mixture is in contact with air at atmospheric pressure, the partial pressure of the components of the mixture is also given by Equation \ref{36}. When the sum of the partial pressures of the components of the mixture is equal to the atmospheric pressure, the liquid mixture boils. The difference between a liquid mixture and a pure liquid is that the boiling temperature of a mixture is not constant. For a mixture in equilibrium with its own vapors, the bubble point of a mixture is the pressure at which the liquid starts to vaporize. Similarly, for a vapor mixture the dew point of the mixture is the pressure at which the vapors start to condense. These terms are also used when the liquid is in contact with air, i.e. it is customary to refer to the bubble point as the temperature at which the liquid mixture starts to boil and dew point as the temperature at which the first condensed liquid appears.

Example $$\PageIndex{1}$$: Bubble point of a water-alcohol mixture

A mixture of ethanol ($$\ce{C2H5OH}$$) and water ($$\ce{H2O}$$) with the mole fractions given by $$x_{\text{Et}} = x_{\ce{H2O}} =0.5$$ is slowly heated under well-stirred conditions in an open beaker. Using Antoine’s equation to determine the vapor pressure of the components and Raoult’s Law to estimate the partial pressures, we can estimate the bubble point of this mixture, i.e. the temperature at which the first bubbles will start forming at the bottom of the beaker as well as the composition of the first bubbles.

The vapor pressure of pure ethanol and water can be computed using Antoine’s equation. The partial pressure of the components in the gas phase in equilibrium with the liquid mixture are computed using Raoult’s law given by Equation $$(5.4.10)$$. The bubble point will be determined as the temperature at which the sum of the partial pressures of ethanol and water is equal to atmospheric pressure, i.e.,

$p_{\ce{H2O}} +p_{ \text{Et }} =p_{atm} \quad \text{ or } \quad x_{\ce{H2O}} p_{\ce{H2O},vap} + x_{ \text{Et }} p_{ \text{Et},vap} =760 \text{ mmHg} \label{1}\tag{1}$

This problem, in principle, can be solved by substitution of Antoine’s equation for the vapor pressures of the pure components in Equation \ref{1}, and then solving for the temperature. A much simpler route consists of guessing values of the temperature until we satisfy Equation \ref{1}. This procedure is easily done using a spreadsheet as illustrated in Table $$\PageIndex{1}$$. The values of Antoine’s coefficients for water and ethanol are available in Table A3 of Appendix A and are given by

$\text{Water: } \qquad A=7.94915, \quad B=1657.46, \quad \theta =227.03 \label{2}\tag{2}$

$\text{Ethanol: } \qquad A=8.1629, \quad B=1623.22, \quad \theta =227.03 \label{3}\tag{3}$

The computed values of the vapor pressure are listed in Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Computational determination of the boiling point of a water-ethanol mixture
Computation of dew point of Ethanol and water mixture
Temp $$p_{\ce{H2O}}$$ $$p_{ \text{Et}}$$ Residue
Degrees C mmHg mmHg mmHg
60 74.7483588 175.7127 -509.539
70 116.9527 270.8179 -372.229
80 177.72766 405.8736 -176.399
86 225.542371 511.0501 -23.4076
86.8 232.662087 526.6464 -0.6915
86.9 233.565109 528.6235 2.188576
87 234.471058 530.6067 5.077746
90 263.047594 593.0441 96.09172

We could continue the computation by inserting additional rows between the temperatures $$T=86.8$$ C and $$T=86.9$$ C. However, for the purpose of this example we will accept the boiling point of the mixture as $$T=86.85\pm 0.05$$ C.

## Humidity

In air-water mixtures the humidity is often used as a measure of concentration that is vaguely described according to

$\text{humidity } = \frac{ \text{mass of water}}{\text{mass of dry air}} \label{37}$

In Sec. 4.5.1 we have been more precise in terms of measures of concentration and there we have identified point concentrations, area-average concentrations, and volume-average concentrations. An analogous set of definitions exists for the humidity. For example, the point version of the humidity is given explicitly by

$\text{ point humidity } = \frac{\rho_{\ce{H2O}} }{\rho_{\text{dry air}} } = \frac{\rho_{\ce{H2O}} }{\rho_{\ce{O2}} + \rho_{\ce{N2}} } \label{38}$

and it will be left as an exercise for the student to explore the other forms discussed in Sec. 4.5.1. In addition, it will be left as an exercise for the student to show that

$\text{ point humidity } = \frac{MW_{\ce{H2O}} p_{\ce{H2O}} }{MW_{\text{dry air}} \left(p-p_{\ce{H2O}} \right)} \label{39}$

in which $$p$$ is the total pressure and $$p_{\ce{H2O}}$$ is the partial pressure of the water vapor. This result can be derived from the definition given by Equation \ref{38} only if the air-water mixture is treated as an ideal gas. The percent relative humidity is often used as a measure of concentration since our personal comfort may be closely connected to this quantity. It is defined by

$\% \text{ relative humidity } = \frac{p_{\ce{H2O}} }{p_{\ce{H2O} , vap} } \times 100 \label{40}$

where $$p_{\ce{H2O} , vap}$$ is the vapor pressure of water at the temperature of the system. When the percent relative humidity is 100% the air is completely saturated and the addition of further water will result in condensation. Values of the vapor pressure of water are listed in Table $$\PageIndex{2}$$ as a function of the temperature.

Table $$\PageIndex{2}$$: Vapor Pressure of Water as a Function of Temperature

$$T$$, C Vapor Pressure, mm Hg $$T$$, F Vapor Pressure, in. Hg
0 4.579 32 0.180
5 6.543 40 0.248
10 9.209 50 0.363
15 12.788 60 0.522
20 17.535 70 0.739
25 23.756 80 1.032
30 31.824 90 1.422
35 42.175 100 1.932
40 55.324 110 2.596
45 71.88 120 3.446
50 92.51 130 4.525
55 118.04 140 5.881
60 149.38 150 7.569
65 187.54 160 9.652
70 233.7 170 12.199
75 289.1 180 15.291
80 355.1 190 19.014
85 433.6 200 23.467
90 525.76 212 29.922
95 633.90 220 34.992
100 760.00 230 42.308
105 906.07 240 50.837
110 1074.56 250 60.725
115 1267.98 260 72.134
120 1489.14 270 85.225
125 1740.93 280 100.18
130 2026.16 290 117.19
135 2347.26 300 136.44

This table provides crucial information for the solution of any air-water system and several examples are given in the following paragraphs.

Example $$\PageIndex{2}$$: Humid Air Flow

Humid air exits a dryer at atmospheric pressure, 75 C, 25% relative humidity, and at a volumetric flow rate of 100 $$m^3$$/min. In this example we wish to determine:

1. Absolute humidity of the air in kg water/kg air.
2. Molar flow rates of water and dry air.

The vapor pressure of water at 75 C is found in Table $$\PageIndex{2}$$ to be $$p_{\ce{H2O} , vap} =289.1 \text{ mm Hg}$$. The density of mercury is found in Table A2 of Appendix A. We convert all parameters into SI units according to

$p_{\ce{H2O}} =\left(0.25\right) \frac{289.1 \text{ mmHg}}{1000 \ { mm/m}} \left(9.81 \ { m/s}^{ 2} \right) \left(13,546 \ { kg/m}^{ 3} \right)=9605 \ { Pa} \tag{1} \label{a}$

and we use Equation $$(5.4.9)$$ to compute the molar fraction of water in the air as

$y_{\ce{H2O}} =\frac{p_{\ce{H2O}} }{p} =\frac{9605 \ { Pa}}{101,300 \ { Pa}} =0.095 \tag{2} \label{b}$

In order to determine the absolute humidity, we use Equation \ref{37} in the more precise form given by Equation \ref{38}

$\text{ humidity } = \frac{ \text{mass of water}}{ \text{mass of dry air}} =\frac{ \text{mass of water/volume}}{ \text{mass of dry air/volume}}\tag{3}\label{c}$

and this leads to an expression for the humidity given by

$\text{ humidity } = \frac{\rho_{\ce{H2O}} }{\rho_{\text{dry air}} } = \frac{MW_{\ce{H2O}} y_{\ce{H2O}} }{MW_{\text{dry air}} y_{air} } = \frac{MW_{\ce{H2O}} y_{\ce{H2O}} }{MW_{\text{dry air}} (1 - y_{\ce{H2O}} ) } \\ =\frac{\left( 18.05 \ g \ water/mol \right)\left( 0.095 \right)}{\left( 28.85 \text{ g dry air/mol}\right) (1 - 0.095) } =0.066 \text{ g water/g dry air} \tag{4}\label{d}$

Assuming that water vapor and air behave as ideal gases at atmospheric pressure, we use the ideal gas law given by Equation $$(5.2.2)$$ to compute the total concentration of the mixture. The concentration of the gas mixture is the total number of moles of air and water per unit volume of the mixture. This can be expressed as

$c = \frac{n}{V} =\frac{p }{R T} =\frac{101,300 \ { Pa} }{8.314 \frac{ m^{ 3} Pa}{ mol \ K} \times 348.16 \ { K}} =35 \ { mol/m}^{ 3}\label{e}\tag{5}$

This result gives the total number of moles of gas per unit volume of mixture. If one considers a material volume of a flowing gas mixture, and the gases satisfy the ideal gas law, the concentration of the flowing mixture in the material volume should be equal to the concentration computed using Equation \ref{e}. In order to determine the molar flow rates of water and dry air, we note that

$\dot{M}_{\ce{H2O}} =c_{\ce{H2O}} Q=y_{\ce{H2O}} c Q=0.095 \times 35 \ mol/m^{ 3} \times 100 \ m^{ 3}/min \\ = 332.5 \text{ mol water/min} \tag{6a} \label{f}$

$\dot{M}_{\text{dry air}} =c_{\text{dry air}} Q=y_{\text{dry air}} c Q=(1 - y_{\ce{H2O}} ) c Q=3167.5 \text{ mol dry air/min} \label{g} \tag{6b}$

## Modified mole fraction

In general, the most useful measures of concentration are the molar concentration $$c_{A}$$ and the species density $$\rho_{A}$$. Associated with these concentrations are the mole fraction defined by Equation $$(5.1.6)$$ and the mass fraction defined by Equation $$(5.1.3)$$. Sometimes it is convenient to use a modified mole fraction or mole ratio which is based on all the species except one. If we identify that one species as species $$N$$, we express the modified mole fraction as

$X_{A} ={c_{A} / \sum_{B = 1}^{B = N-1}c_{B} } \label{41}$

When it is convenient to work in terms of this modified mole fraction, one usually needs to be able to convert from $$x_{A}$$ to $$X_{A}$$ and it will be left as an exercise for the student to show that this relation is given by

$X_{A} ={x_{A} / \left(1-x_{N} \right)} , \quad A=1, 2,...N \label{42}$

In some types of analysis it is convenient to choose species $$N$$ to be species $$A$$. Under those circumstances we need to express Equation \ref{41} as

$X_{D} ={c_{D} / \sum_{\begin{array}{l}{B = 1}\\{B \neq A}\end{array}}^{B = N}c_{B} } \label{43}$

and Equation \ref{41} takes the form

$X_{D} ={x_{D} / \left(1-x_{A} \right)} , \quad D=1, 2,...N \label{44}$

Similar relations can be developed for the modified mass fraction $$\Omega_{A}$$ and they will be left as exercises for the student. It is important to note that the sum over all species of the modified mass or mole fraction is not one.