Oil reservoirs below ground frequently are in contact with underground water and, in connection with an oil drilling operation, you are asked to compute the solubility of water in a heavy oil at the underground conditions. These conditions are estimated to be 140°C and 410 bar. Experiments at 140°C and 1 bar indicate that the solubility of steam in the oil is x = 35 x 10-4 (x, is the mole fraction of steam). Assume Henry's law in the form f = H(T)x, where H(T) is a constant, dependent only on the temperature, and f, is the fu- gacity of H,O. Also assume that the vapor pressure of the oil is negligible at 140°C. Data for H,0 are given in the steam tables. 6.
Oil reservoirs below ground frequently are in contact with underground water and, in connection with an oil drilling operation, you are asked to compute the solubility of water in a heavy oil at the underground conditions. These conditions are estimated to be 140°C and 410 bar. Experiments at 140°C and 1 bar indicate that the solubility of steam in the oil is x = 35 x 10-4 (x, is the mole fraction of steam). Assume Henry's law in the form f = H(T)x, where H(T) is a constant, dependent only on the temperature, and f, is the fu- gacity of H,O. Also assume that the vapor pressure of the oil is negligible at 140°C. Data for H,0 are given in the steam tables. 6.
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter17: Spontaneity, Entropy, And Free Energy
Section: Chapter Questions
Problem 128IP: Some nonelectrolyte solute (molar mass = 142 g/mol) was dissolved in 150. mL of a solvent (density =...
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Transcribed Image Text:6.
Oil reservoirs below ground frequently are in contact with underground water and, in
connection with an oil drilling operation, you are asked to compute the solubility of water
in a heavy oil at the underground conditions. These conditions are estimated to be 140°C
and 410 bar. Experiments at 140°C and 1 bar indicate that the solubility of steam in the
oil is x = 35 x 10-4 (x, is the mole fraction of steam). Assume Henry's law in the form
f = H(T)x1, where H(T) is a constant, dependent only on the temperature, and f is the fu-
gacity of H,O. Also assume that the vapor pressure of the oil is negligible at 140°C. Data
for H,0 are given in the steam tables.
Expert Solution
Step 1
Solution
The solubility of water in oil is described by
f1 = H(T)4
Henry's constant can be evaluated at 1 bar where,
* f1 = 1 bar then,
H(T) = f1 / x1 (x1=35×10-4)
= 1/35×10-4 (t=140° C)
= 286 bar
* To obtain f1 at 410 bar and 140° C use the steam table alternatively get f at saturation (3.6 th bar) and use the pointing factor to correct to 410 bar.
at 140° C
RT In f1 = Δg1 → 410 bar = Δh1 → 410 bar => TΔS1 → 410 bar
From steam tables book
RT In f1 = 282 J/g
R=8.314 J/mole.k
=> 8.314 J/18 grams.k and
T = 413k since 1 mole of steam = 18 grams
In f1 = 1.48
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