krit.club logo

The Amazing World of Solutes, Solvents, and Solutions - Solubility of Gases

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Gas solubility is the maximum amount of a gaseous solute that can dissolve in a specific volume of liquid solvent at a specific temperature and pressure.

•

Effect of Temperature: The solubility of most gases in liquids decreases as the temperature increases (T↑  ⟹  S↓T \uparrow \implies S \downarrow). This is because increasing temperature increases the kinetic energy of the gas molecules, allowing them to break the intermolecular bonds with the solvent and escape.

•

Effect of Pressure: The solubility of gases is directly proportional to the pressure of the gas above the liquid (P↑  ⟹  S↑P \uparrow \implies S \uparrow). This is why carbonated drinks are sealed under high pressure.

•

Henry's Law: This law quantitatively relates pressure and gas solubility, stating that the amount of dissolved gas is proportional to its partial pressure in the gas phase.

•

Biological Significance: Aquatic organisms rely on dissolved oxygen (O2O_{2}). Since O2O_{2} is more soluble in cold water, aquatic life is generally more active and healthy in cooler environments compared to warm, depleted waters.

📐Formulae

S∝PS \propto P

C=kH⋅PC = k_{H} \cdot P

Solubility=Mass of dissolved gasVolume of solvent\text{Solubility} = \frac{\text{Mass of dissolved gas}}{\text{Volume of solvent}}

💡Examples

Problem 1:

A bottle of sparkling water is pressurized with CO2CO_{2} at 3 atm3\text{ atm}. When the cap is removed, the pressure drops to 1 atm1\text{ atm}. If the initial solubility was 4.5 g/L4.5\text{ g/L}, calculate the new solubility at the lower pressure.

Solution:

Using the direct proportion from Henry's Law: S1P1=S2P2\frac{S_{1}}{P_{1}} = \frac{S_{2}}{P_{2}} Substituting the values: 4.53=S21\frac{4.5}{3} = \frac{S_{2}}{1} S2=1.5 g/LS_{2} = 1.5\text{ g/L}

Explanation:

Since pressure and solubility are directly proportional, reducing the pressure to one-third (from 3 atm3\text{ atm} to 1 atm1\text{ atm}) reduces the solubility to one-third of its original value.

Problem 2:

Explain why 'fizzing' occurs when a soda can is opened using the concepts of gas solubility.

Solution:

Inside a sealed can, CO2CO_{2} is under high pressure (PhighP_{high}), leading to high solubility. Upon opening, the pressure drops to atmospheric pressure (Plow≈1 atmP_{low} \approx 1\text{ atm}).

Explanation:

Because solubility is directly proportional to pressure (S∝PS \propto P), the sudden drop in pressure causes the excess dissolved CO2CO_{2} to rapidly leave the solution in the form of bubbles, creating the 'fizz' effect.

Problem 3:

Compare the amount of dissolved Oxygen (O2O_{2}) in a lake during Winter (5∘C5^{\circ}C) and Summer (30∘C30^{\circ}C).

Solution:

SO2 at 5∘C>SO2 at 30∘CS_{O_{2}} \text{ at } 5^{\circ}C > S_{O_{2}} \text{ at } 30^{\circ}C

Explanation:

The solubility of gases in liquids is inversely related to temperature. As the temperature of the lake water increases in the summer, the gas molecules move faster and escape the water, leading to lower dissolved oxygen levels compared to the winter.