Henry's Law - UNSOLVED PRACTICE SET

Q7. State Henry's law mathematically. What does each term represent?

Q8. Why does Henry's law constant (KH) increase with temperature? What does this imply about gas solubility?

Q9. The KH value for CO₂ in water at 298 K is 1.67 × 10⁸ Pa. Calculate the solubility of CO₂ in water at 298 K when the partial pressure of CO₂ is 2.5 atm.

[Given: 1 atm = 1.013 × 10⁵ Pa]

Q10. Why do deep-sea divers use a mixture of helium and oxygen instead of nitrogen and oxygen? Explain using Henry's law.

Q11. Your friend opens a cold bottle of soda and it fizzes vigorously, but a warm bottle fizzes even more. Explain both observations using Henry's law.

Q12. Why does Henry's law fail for gases like NH₃ and HCl in water?

Q13. Discuss Henry's law in detail:

(a) Statement and mathematical formulations (p = KH · x, C = k · p)

(b) Graphical representation — solubility vs. pressure

(c) Factors affecting Henry's law constant — temperature, nature of gas and solvent

(d) Limitations of Henry's law

(e) Relationship between KH and solubility — higher KH means lower solubility

Illustrate with numerical examples.

Q14. Explain the applications of Henry's law:

(a) Carbonated beverages — why CO₂ is dissolved under high pressure

(b) Respiration — exchange of O₂ and CO₂ in lungs and blood

(c) Deep-sea diving and decompression sickness ("the bends")

(d) High-altitude sickness and the role of oxygen partial pressure

(e) Opening of soft drink bottles and effervescence

For each application, write the relevant Henry's law equation.

Q15. Henry's law has critical applications in Indian contexts. Discuss:

(a) Why packaged drinking water in India must be stored away from heat to prevent loss of dissolved gases and taste

(b) The design of oxygen cylinders for patients at high altitudes in Ladakh and the Himalayas

(c) Why scuba diving is rare in India but commercial divers in Mumbai harbour face decompression risks

(d) The science behind traditional Indian drinks like nimbu pani and aam panna — how carbonation and flavour extraction relate to gas solubility

Q16. Henry's law constant for O₂ in water at 298 K is 4.34 × 10⁴ atm.

(a) Calculate the solubility of O₂ in water at 298 K and 1 atm partial pressure of O₂.

(b) Calculate the solubility at 0.2 atm partial pressure of O₂ (approximate in air).

(c) Air contains 21% O₂ by volume. Calculate the mass of O₂ dissolved in 1 L of water exposed to air at 1 atm and 298 K.

[Given: Molar mass of O₂ = 32 g/mol]

Q17. The Henry's law constant for CO₂ in water at 298 K is 1.67 × 10⁸ Pa. A soft drink is bottled with CO₂ at 4 atm pressure.

(a) Calculate the mole fraction of CO₂ in the soft drink.

(b) Calculate the molarity of CO₂ in the drink (density of water = 1000 kg/m³, molar mass of water = 18 g/mol).

(c) When the bottle is opened, the pressure drops to 1 atm. Calculate the volume of CO₂ released from 500 mL of soft drink at STP.

[Given: 1 atm = 1.013 × 10⁵ Pa]

Q18. A diver descends to 30 m depth in the Arabian Sea where the pressure is 4 atm. The diver breathes air containing 79% N₂.

(a) Calculate the partial pressure of N₂ at this depth.

(b) If Henry's law constant for N₂ at body temperature (310 K) is 8.42 × 10⁴ atm, calculate the solubility of N₂ in blood at this depth.

(c) If the diver ascends rapidly, calculate the volume of N₂ bubbles (at STP) that could form from 5 L of blood. Explain why this causes "the bends."

[Given: Molar volume at STP = 22.4 L/mol]