Question

The solubility of a gas in water is 0.16g/L at 104 kPa. What is the solubility when the pressure of the gas is increased to 288 kPa? Assume the temperature remains constant.

Answer 1

The solubility of the gas in water when the pressure is increased to 288 kPa is 0.44 g/L, calculated using Henry's law which states the direct proportionality of solubility to pressure at constant temperature.

Step-by-step explanation:

The solubility of a gas in water at a given temperature is directly proportional to the pressure of that gas above the solution. This relationship is described by Henry's law, which states that the solubility of a gas in a liquid at constant temperature is directly proportional to the partial pressure of the gas above the solution. Using Henry's law, we can calculate the new solubility when the pressure is increased from 104 kPa to 288 kPa, keeping the temperature constant.

The solubility equation from Henry's law can be expressed as: S1/P1 = S2/P2, where S1 and S2 are the solubilities at pressures P1 and P2, respectively. Given that the initial solubility (S1) is 0.16 g/L at a pressure (P1) of 104 kPa, we can find the new solubility (S2) at a pressure (P2) of 288 kPa by rearranging the equation to solve for S2: S2 = S1 * (P2/P1).

By plugging in the values, we find that the new solubility (S2) is 0.16 g/L * (288 kPa / 104 kPa) = 0.44 g/L. Therefore, when the pressure of the gas is increased to 288 kPa, the solubility in water increases to 0.44 g/L.

Answer 2

Answer:
new solubility = 0.443 g/l

Step-by-step explanation:
To solve this question, we will use Henry's law.
Henry's law states that, at constant temperature:
S1 / P1 = S2 / P2

We are given that:
S1 = 0.16 g/l
P1 = 104 kPa
S2 is the solubility we want to calculate
P2 = 288 kPa

Substitute with the givens in the above relation to get the new solubility (S2) as follows:
0.16 / 104 = S2 / 288
S2 = (0.16/104) * 288
S2 = 0.443 g/l

Hope this helps :)