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Module 4: Liquid Characteristics - Gas Absorption - Practice Problems
- Instructions:
- Work these problems on a sheet of paper and check your answers against those provided below.
-
#1
- How does an increase in the liquid temperature affect the solubility
of gases in a liquid?
- Solubility increases with increasing temperature.
- Solubility decreases with increasing temperature.
- Increasing temperature does not influence solubility.
- Solubility increases with increasing temperature.
- Answer: b. Solubility decreases with increasing temperature.
- An increase in the liquid temperature causes a decrease in the solubility of gases in the liquid.
-
#2
- Using Table 4 list the following compounds from the least soluble
to the most soluble in water. Air, methane (CH4), carbon
monoxide (CO), carbon dioxide (CO2), nitrogen oxide (NO).
-
-
- Answer:
- Air
- Carbon Monoxide
- Methane
- Nitrogen oxide
- Carbon dioxide
- Higher values for Henry's law constants indicate lower solubility.
Compare the Henry's law constant values of the gases in equivalent units
at the same temperature.
- For example at 20°C, the Henry's law constants are as follows:
-
#3
- Wet
scrubbers are a type of air pollution control device that can be
used to reduce sulfur dioxide (SO2) emissions. A wet scrubber
(e.g. a venturi scrubber, see Figure 9) sprays liquid
into the gas stream to create turbulent mixing between the liquid and
gas. A wet scrubber removes sulfur dioxide from the gas stream by absorption.
- City
water is a very inefficient liquid for controlling sulfur dioxide
emissions. If city water is recirculated in the scrubber system, which
of the following adjustment(s) should increase the scrubber's collection
efficiency? Select all that apply.
- Adding an alkaline material such as limestone to the recirculating
liquid
- Monitoring the pH of the liquid entering and exiting the scrubber
- Lowering the pH of the liquid entering the scrubber
- Increasing the temperature of the recirculating liquid
- Adding an alkaline material such as limestone to the recirculating
liquid
-
- Answer:
- Adding an alkaline material such as limestone to the recirculating
liquid
- Monitoring the pH of the liquid entering and exiting the scrubber
- Adding an alkaline material such as limestone to the recirculating
liquid
- Adding alkaline material, such as limestone, to the scrubber liquid
will reduce the acidity of the water entering the scrubber. The alkaline
material converts dissolved sulfur dioxide (sulfurous acid) into sulfite
and sulfate ions and alkaline materials neutralize the H+
ions formed from HCl dissociation. This step is effective because sulfur
dioxide cannot diffuse out of the liquid and therefore the equilibrium
limit is removed.
- Monitoring the pH levels will indicate the efficiency of sulfur dioxide
removal, prevent damage to equipment, and determine the amount of limestone
to add.
- Options c and d are incorrect. Lowering the pH of the recirculating liquid (i.e. making it more acidic), would make the absorption process less effective. Increasing the temperature of the liquid will lower the solubility capability.
-
#4
- The solubility graph for two gases (Compound A and Compound B)
in water at 20°C is shown below. Determine Henry's law constant
for each compound. Which compound is most readily absorbed by water?
-
-
- Answer:
- H (Compound A) = 66.7
- H (Compound B) = 133.3
- Compound A with its lower Henry's law constant will be more readily
absorbed in water.
- Solution:
- Since the graphed solubility curves for both compounds are straight,
Henry's law applies. Choose two points on each line to determine the
slope (see Figure 11).
- Find Henry's law constant, H, for Compound A.
Henry's law constant equals the slope of the line for Compound A. Points (0.003, 0.1) and (0.006, 0.3) are on this line. When the mole fraction of Compound A in water is 0.003, the mole fraction of Compound A in air is 0.1. Likewise, when the mole fraction of Compound A in water is 0.006, the mole fraction of Compound A in air is 0.3.
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- Find Henry's law constant, H, for Compound B.
Henry's law constant equals the slope of the line for Compound B. Points (0.003, 0.4) and (0.006, 0.8) are on this line. When the mole fraction of Compound B in water is 0.003, the mole fraction of Compound B in air is 0.4. Likewise, when the mole fraction of Compound B in water is 0.006, the mole fraction of Compound B in air is 0.8.
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Since Compound B has a higher Henry's law constant than Compound A, Compound B is less soluble in water. Therefore Compound A will be more readily absorbed in water.