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A saturated solution of carbon dioxide in water contains 3.00 g of CO₂ when the CO₂ partial pressure is 8.00 atm. What mass of CO₂ will escape if the partial pressure is reduced to 4.00 atm?

Answer :

Using Henry's Law 1.50 g of [tex]CO_2[/tex] would escape from the solution when the partial pressure is reduced from 8.00 atm to 4.00 atm, assuming the solubility of [tex]CO_2[/tex] would be halved.

The problem presented is primarily a matter of applying Henry's Law, which relates the solubility of a gas in a liquid to the partial pressure of that gas above the liquid. The law is usually expressed in the form S = kH * P, where S is the solubility (concentration) of the gas, kH is Henry's Law constant, and P is the partial pressure of the gas.

In this case, a solution initially saturated with [tex]CO_2[/tex] at 8.00 atm has 3 grams of [tex]CO_2[/tex]. When the pressure is halved to 4.00 atm, Henry's Law suggests that the solubility of [tex]CO_2[/tex] would also be halved (assuming that the Henry's Law constant remains unchanged at this lower pressure).

Since the initial amount is 3.00 g at 8.00 atm, at 4.00 atm the solubility would be 1.50 g. Therefore, the mass of [tex]CO_2[/tex] that would escape is the difference: 3.00 g - 1.50 g = 1.50 g.

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