Bonus VLE Question:

A mixture of 70 mole % n-butane, 4 mole % water, and 26 mole % nitrogen is at 75°C and 50 psia. Antoine constants are given below:

Form: log₁₀Psat = A - B / (T + C)

where T is the temperature in °C

Psat is the vapor pressure in mmHg

Compound	A	B	C
n-butane	6.85353	955.43	241.0
water	8.10765	1750.286	235.0

1. Liquid mixtures of butane and water can be characterized as:

A. ideal

B. non-ideal

C. highly non-ideal

D. cannot be characterized

2. At 50 psia, as the temperature is reduced, which of the following observations best describes the essential nature of the problem?

A. high boilers condense first

B. Low boilers condense first

C. there exists a dew point

D. there exist multiple dew points.

3. At what temperature will the first drop of liquid form?

A. 32°C

B. 35°C

C. 49°C

4. 52°C

  4. The composition of the first drop of liquid will be:

A. water

B. butane

C. nitrogen

E. 70% butane, 4% water, 26% nitrogen

 5. At 50 psia, essentially no butane will condense until the temperature is reduced below:

A. 0°C

B. 26°C

C. 32°C

D. 52°C

Answers:

1.      C. The butane / water system is extremely non-ideal. In fact, they are essentially immiscible and form two liquid phases.

2.      D. Because nitrogen is above its critical temperature, it is essentially non-condensible although a very small amount (often but not always negligible) will be dissolved in the liquid. Also, butane and water can be considered to be immiscible liquids. Therefore, there will be two dew points. The first will be water condensing and the second will be butane.

3.      D. As shown above, liquid water will be in equilibrium with water in the vapor when P_w = P_w^sat. The partial pressure of water is (0.04)(50 psia) = 2 pis or 103 mmHg. Trial and error using the Antoine equation shows that the vapor pressure of water will be 103 mmHg at 52 degrees C. This water dew point is higher than the butane dew point calculated below.

4.      A. see 3.

5.      B. Since water is condensing, the partial pressure of butane will change, but a first guess is that 1 mole% water will be left in the vapor, thus leaving 73% butane. The butane partial pressure would then be (0.73)(50 psia) = 36.5 psi or 1888 mmHg. The Antoine equation shows that the vapor pressure of butane will be 1888 mmHg at 26 degrees C. At this temperature, the vapor pressure of water is 25 mmHg, so that the vapor mole fraction of water is about 0.014 which his close to our assumption.