Question

The solubility of a gas is 0.890 8/1 at a pressure of 121 kPa. What
is the solubility of the gas if the pressure is increased to 150 kPa,
given that the temperature is held constant?

Answer 1

Answer:

1.10 g/L

Explanation:

Step 1: Calculate Henry's constant (k)

The solubility of a gas (C) is 0.890 g/L at a pressure (P) of 121 kPa. Solubility and pressure are related through Henry's law.

C = k × P

k = C / P

k = (0.890 g/L) / 121 kPa = 7.36 × 10⁻³ g/L.kPa

Step 2: Calculate the solubility of the gas if the pressure is increased to 150 kPa

We will use Henry's law.

C = k × P

C = (7.36 × 10⁻³ g/L.kPa) × 150 kPa = 1.10 g/L

Answer 2

Solubility of the gas, if the temperature is held constant and pressure is increased to 150 kPa from 121 kPa, is 1.10 g/L.

What is Henry's law?

Henry's law of gas states at solubility (C) of the dissolved gas is directly proportional to the partial pressure (P) of the gas.

C ∝ P

C = kP, where

k = Henry's constant

Let first we calculate the Henry's constant, when the solubility of a gas is 0.890 g/L at a pressure of 121 kPa is:

k = (0.890 g/L) / (121 kPa)

k = 7.36 × 10⁻³ g/L.kPa

Now we calculate the solubility of the gas, if the pressure is increased to 150 kPa as:

C = (7.36 × 10⁻³ g/L.kPa) (150 kPa)

C = 1.10 g/L

Hence, required solubility is 1.10 g/L.

To know more about Henry's constant, visit the below link:

https://brainly.com/question/7007748