Question 48 of 50advanced🔧 ApplyNumerical3 marks

The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30×3.30 \times 10^{7} mm and 6.51×6.51 \times 10^{7}$$ mm respectively, calculate the composition of these gases in water.

Correct Answer

Mole fraction of oxygen in water = 4.61×104.61 \times 10⁻⁵ Mole fraction of nitrogen in water = 9.22×109.22 \times 10⁻⁵

Exercise: EXERCISES | Q: 1.39 | (Page 29)
For More Understanding

Explanation

This problem requires application of Henry's law (p = K_H × x) to find the mole fraction of dissolved gases. The partial pressures are calculated from the volume composition and total pressure, then converted to mm Hg to match the units of Henry's law constants.

Solution Steps

  1. Step 1: Calculate partial pressures of gases in air. Since volume percentage equals mole percentage for gases, partial pressure of O₂ = 20% × 10 atm = 2 atm and partial pressure of N₂ = 79% × 10 atm = 7.9 atm.

  2. Step 2: Convert partial pressures to mm Hg. p(O₂) = 2×7602 \times 760 = 1520 mm Hg. p(N₂) = 7.9×7607.9 \times 760 = 6004 mm Hg.

  3. Step 3: Apply Henry's law (p = K_H × x, so x = p/K_H). For O₂: x(O₂) = 1520 / (3.30×103.30 \times 10⁷) = 4.61×104.61 \times 10⁻⁵. For N₂: x(N₂) = 6004 / (6.51×106.51 \times 10⁷) = 9.22×109.22 \times 10⁻⁵.

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