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The solubility of a gas in water is 0.22 g/L at 20.0 kPa of pressure. What is the solubility when the pressure is increased to 115 kPa?

User Lutchman
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1 Answer

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Answer:

Solubility is 1.265g/L

Step-by-step explanation:

According to Henry's law which states that the solubility of a gas in liquid is directly proportional to the pressure of the gas above the solution.

Sg1 / P1 = Sg2/ P2,

where Sg is the gas solubility, P is the gas pressure.

Sg1 = 0.22

P1 = 20kPA

P2 = 115kPA

Sg2 =?

Sg2 = Sg1 P2 / P1

Sg2 = 0.22 * 115*10^3 / 20*10^3

Sg2 = 25.3*10^3 / 20*10^3

Sg2 = 1.265g/L

the solubility of the gas in water at the new pressure is 1.265g/L

User Ilyes Kooli
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