Bubble- and Dew-Point Pressures from the Margules Model: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2013, Q6

Question 6: Bubble- and Dew-Point Pressures from the Margules Model (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Water (1) / acetic acid (2) at $T=100\,°\text{C}=373.15$ K. Two-parameter Margules with $A_{12}=1.705$, $A_{21}=0.9439$. Antoine constants as tabulated (ln form, mm Hg).

Find. (a) the bubble-point pressure of a liquid with $x_1=0.63$; (b) the dew-point pressure of the vapour whose first drop of condensate at 100 °C has composition $x_1=0.63$.

Figure 6 — Isothermal $P$–$x_1$–$y_1$ diagram at 100 °C from the Margules model. The bubble (blue) and dew (red) curves meet at a pressure-maximum azeotrope near $x_1\approx0.67$. Marked: the liquid $x_1=0.63$ (bubble point, part a) and the vapour $y_1=0.648$ in equilibrium with it, whose first drop is that same liquid (dew point, part b) — both at 818 mm Hg.

Approach. Use modified Raoult's law $y_iP=x_i\gamma_ip_i^{sat}$ with Margules $\gamma_i$. Part (a) is a direct BUBL P (liquid known). In part (b) the first-drop liquid is specified, and at fixed $T$ that liquid composition fixes the equilibrium state — so the dew-point vapour and pressure follow from the same relation.

Pure-component vapour pressures at 100 °C. From Antoine, $$p_1^{sat}=\exp\!\Big(18.3036-\tfrac{3816.44}{373.15-46.13}\Big)=759.9\ \text{mm Hg},\quad p_2^{sat}=427.7\ \text{mm Hg}.$$ (Water's $\approx760$ mm Hg confirms the constants — it boils at 100 °C, 1 atm.)
Margules activity coefficients at $x_1=0.63$. With $\ln\gamma_1=x_2^2[A_{12}+2(A_{21}-A_{12})x_1]$ and $\ln\gamma_2=x_1^2[A_{21}+2(A_{12}-A_{21})x_2]$, $$\gamma_1=1.108,\qquad \gamma_2=1.819.$$ Both exceed 1 — positive deviations from Raoult's law.
(a) Bubble-point pressure. The total pressure is the sum of the partial pressures over the known liquid: $$P=\sum x_i\gamma_ip_i^{sat}=0.63(1.1075)(759.9)+0.37(1.8188)(427.7)=530.2+287.8.$$ $$\boxed{P_{bubble}=818\ \text{mm Hg}},\qquad y_1=\frac{530.2}{818.1}=0.648.$$
(b) Dew-point pressure (first drop $x_1=0.63$). At the dew point the first drop of liquid is in equilibrium with the (essentially unchanged) vapour. The printed first-drop composition $x_1=0.63$ at 100 °C is the same liquid as in (a), so the same $\gamma_i$ apply, and modified Raoult's law gives the vapour being condensed and its dew pressure: $$y_1=\frac{x_1\gamma_1p_1^{sat}}{P}=0.648,\qquad P_{dew}=\Big[\sum_i\frac{y_i}{\gamma_ip_i^{sat}}\Big]^{-1}=\sum_i x_i\gamma_ip_i^{sat}.$$ $$\boxed{P_{dew}=818\ \text{mm Hg}}\quad(\text{vapour } y_1=0.648).$$ Check by the DEW P route: starting from $y_1=0.648$ and iterating $x_i=y_iP/(\gamma_ip_i^{sat})$ converges back to $x_1=0.630$ and $P=818$ mm Hg. The dew pressure equals the bubble pressure of (a) because both parts describe the same two-phase equilibrium state — (a) approaches it from the liquid side, (b) from the vapour side. Since $x_1=0.63$ lies just below this model's pressure-maximum azeotrope ($x_1\approx0.67$, 818.8 mm Hg), the vapour is only slightly richer in water than the liquid.

Quantity	Value
$p_1^{sat},\ p_2^{sat}$ at 100 °C	759.9, 427.7 mm Hg
$\gamma_1,\ \gamma_2$ at $x_1=0.63$	1.108, 1.819
(a) Bubble-point pressure	818 mm Hg (vapour $y_1=0.648$)
(b) Dew-point pressure	818 mm Hg (vapour $y_1=0.648$, first drop $x_1=0.63$)

Check (interpretation of part b): answered as printed — the 0.63 is the composition of the first liquid drop, so (b) returns the same equilibrium pressure as (a). If the examiner intended the more common DEW P pairing (vapour $y_1=0.63$ specified), iteration gives $P_{dew}=816$ mm Hg with a first drop of $x_1=0.592$ — within 0.3% of the printed-reading result.

Concept & Theory

Topic. Low-pressure VLE via modified Raoult's law with a Margules (activity-coefficient) model; bubble- vs dew-point pressure.

Key relations. $y_iP=x_i\gamma_ip_i^{sat}$; BUBL P: $P=\sum x_i\gamma_ip_i^{sat}$ (direct); DEW P: $P=1/\sum\big[y_i/(\gamma_ip_i^{sat})\big]$ (iterate because $\gamma_i=\gamma_i(x)$); Margules two-parameter $\ln\gamma$ expressions above.

Why this works. At these low pressures the vapour is essentially ideal, so non-ideality lives entirely in the liquid activity coefficients. A bubble calculation is explicit because the liquid — and hence every $\gamma_i$ — is known up front. A dew calculation is implicit: the liquid composition is the unknown, so the $\gamma_i$ must be found by iteration, converging in a few passes.

Common pitfall. Using $\ln\gamma_1=A_{12}x_2^2$ (the one-parameter form) — wrong here; the two-parameter Margules has the bracketed correction with $2(A_{21}-A_{12})x_1$. Also, forgetting that the dew calculation needs iteration and instead plugging the vapour mole fractions straight into the bubble formula.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §10.1–10.4 (modified Raoult's law; BUBL P / DEW P algorithms; Margules model).

Question 6 of 7: Bubble- and Dew-Point Pressures from the Margules Model

Question 6: Bubble- and Dew-Point Pressures from the Margules Model (equal value)