The constituent partial pressures of a gas in equilibrium with a liquid solution

Question

The constituent partial pressures of a gas in equilibrium with a liquid solution at 30oC and 1 atm
containing 2 lbm SO2/100 lbm H2O are pH2O = 31.6 mm Hg and pSO2 = 176 mm Hg. The
balance of the gas is air.

(a) Calculate the partial pressure of air. If you make any assumptions, state what they are.

(b) Suppose the only data available on this system gave pSO2 = 176 mm Hg, but there was no
information given on the equilibrium partial pressure of water. Use Raoult's law to estimate
a value for this quantity. Assuming that the value given in the problem statement is correct,
what percentage error results from using Raoult's law?

(c) The same system was examined in Example 6.4-1. What percentage errors in the two calcu-
lated quantities would result from using Raoult's law for the partial pressure of water?

The text-book as PDF

http://197.14.51.10:81/pmb/CHIMIE/Elementary%20Principles%20of%20Chemical%20Processes%203.pdf

This question from Chapter 6 page 255 & 256

Explanation / Answer

(1)

Pressure exerted by SO2 = 176 mm Hg

Pressure exerted by H2O = 31.6 mm Hg

Pressure exerted by air = 760-(176+31.6) = 552.4 mm Hg

(2)

Mass of SO2 = 2 lbm = 907.18 g

Moles of SO2 = Mass / MW = 907.18/64 = 14

Mass of H2O = 100 lbm = 45359.2 g

Moles of H2O = Mass / MW = 45359.2/18 = 2520

So,

XSO2 = 14/(14+2520) = 0.0055

XH2O = 2520/(14+2520) = 0.9945

Now,

P0H2O = 31.8 mm Hg ( at 300C . I have taken this standard value form literature )

So, from Raoult's law

PH2O = P0H2O * XH2O = 31.62 mm Hg

% error = [ (31.62-31.6) / 31.6 ] * 100 = 0.063 %

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.