Question

3) 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?

Answer 1

1.27g/L is the new solubility when the pressure is increased

Step-by-step explanation:

Based on Henry's law, the solubility of a gas is directly proportional to the partial pressure of the gas in equilibrium.

The equation is:

S1 / P1 = S2 / P2

Where S is solubility and P the pressure of 1, initial state and 2, final state.

The values of the problem are:

S1 = 0.22g/L

P1 = 20.0kPa

S2 = 115kPa

P2 = ?

0.22g/L / 20.0kPa = S2 / 115kPa

S2 = 0.22g/L* 115kPa / 20.0kPa

S2 = 1.27g/L is the new solubility when the pressure is increased