Q.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 $3.30 \times 10^7$ mm and $6.51 \times 10^7$ mm respectively, calculate the composition of these gases in water.

The key idea is Henry’s Law: at constant temperature, the concentration of a gas dissolved in a liquid is proportional to its partial pressure above the liquid. For a gas mixture, each gas obeys Henry’s Law independently.

Step 1: Find the partial pressures of O₂ and N₂ in air at 10 atm.

Air is 20% O₂ and 79% N₂ by volume (the remaining 1% is negligible).

Partial pressure of O₂: $P_{\text{O}_2} = 0.20 \times 10 = 2.0 \text{ atm}$

Partial pressure of N₂: $P_{\text{N}_2} = 0.79 \times 10 = 7.9 \text{ atm}$

Step 2: Convert Henry’s constants to atm for consistency.

$1 \text{ atm} = 760 \text{ mm Hg}$

$K_{\text{H, O}_2} = \frac{3.30 \times 10^7}{760} = 4.342 \times 10^4 \text{ atm}$

$K_{\text{H, N}_2} = \frac{6.51 \times 10^7}{760} = 8.566 \times 10^4 \text{ atm}$ …

Why Henry’s Law?

When a gas mixture (like air) is in contact with water, each gas dissolves independently according to its own solubility. The amount that dissolves depends on the partial pressure of that gas above the liquid and Henry's constant, $K_H$.

Henry’s Law states:

where $p_{\text{gas}}$ is the partial pressure of the gas above the liquid, $x_{\text{gas}}$ is its mole fraction in the liquid, and $K_H$ is Henry’s constant (in the same pressure units). Both constants here are given in mm of Hg, so we must work entirely in mm Hg.

Step-by-step solution

1. Convert total pressure to mm Hg

2. Find partial pressures of O2 and N2

3. Apply Henry’s Law for each gas

For oxygen:

For nitrogen: …

Method: Henry's Law for Gas Mixtures

Henry's Law states that at constant temperature, the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid.

Key Concept

For a gas mixture, each gas behaves independently. The partial pressure of each gas is its mole fraction in the gas phase multiplied by the total pressure.

Steps

Step 1: Convert total pressure to mm Hg

Given total pressure = 10 atm

1 atm = 760 mm Hg

$P_{\text{total}} = 10 \times 760 = 7600 \text{ mm Hg}$

Step 2: Calculate partial pressures of each gas

• Oxygen: 20% by volume → 20% by mole fraction

  $P_{O_2} = 0.20 \times 7600 = 1520 \text{ mm Hg}$

• Nitrogen: 79% by volume → 79% by mole fraction

  $P_{N_2} = 0.79 \times 7600 = 6004 \text{ mm Hg}$

Step 3: Apply Henry's Law for each gas

Henry's Law: $x_{\text{gas}} = \frac{P_{\text{gas}}}{K_H}$

Where:

• $x_{\text{gas}}$ = mole fraction of gas in water
• $P_{\text{gas}}$ = partial pressure of gas (mm Hg)
• $K_H$ = Henry's Law constant (mm Hg)

For Oxygen:

$x_{O_2} = \frac{1520}{3.30 \times 10^7} = \mathbf{4.61 \times 10^{-5}}$

For Nitrogen:

$x_{N_2} = \frac{6004}{6.51 \times 10^7} = \mathbf{9.22 \times 10^{-5}}$

Step 4: Find composition in water

Total mole fraction of dissolved gases:

$x_{\text{total}} = 4.61 \times 10^{-5} + 9.22 \times 10^{-5} = 1.383 \times 10^{-4}$

Percentage composition in water: …

Here are the most common mistakes students make when solving this Henry’s Law problem, along with how to avoid each.

1. Forgetting to convert pressure units

The Mistake:

Students plug the total pressure (10 atm) directly into Henry’s Law without converting to mm Hg (since the given $K_H$ values are in mm).

Why it’s wrong:

Henry’s Law requires consistent units.

$K_H$ is in mm Hg, so pressure must also be in mm Hg.

How to avoid:

Always check units before substituting.

Convert:

So:

2. Using total pressure instead of partial pressure

The Mistake:

Using $P_{\text{total}} = 10$ atm (or 7600 mm) directly in Henry’s Law for each gas.

Why it’s wrong:

Henry’s Law uses the partial pressure of the gas above the liquid, not the total pressure.

How to avoid:

Calculate partial pressures from the volume percentages (which equal mole fractions in the gas phase):

• $P_{\text{O}_2} = 0.20 \times 7600 = 1520$ mm Hg
• $P_{\text{N}_2} = 0.79 \times 7600 = 6004$ mm Hg

(Note: The remaining 1% is other gases — ignore for this problem.)

3. Confusing mole fraction with mass or volume

The Mistake:

Thinking the answer is a percentage by volume or mass in water.

Why it’s wrong:

Henry’s Law gives mole fraction ($x$) of the gas dissolved in water — a dimensionless number, not a percentage.

How to avoid:

Write Henry’s Law clearly:

The result is a small decimal (e.g., $4.61 \times 10^{-5}$). That’s the mole fraction.

4. Misinterpreting “composition of these gases in water”

The Mistake:

Stopping after finding individual mole fractions, without expressing the relative composition.

Why it’s wrong:

The question asks for composition — often meaning the ratio or relative amounts of O₂ and N₂ dissolved.

How to avoid:

After finding $x_{\text{O}_2}$ and $x_{\text{N}_2}$, state the relative composition:

$x_{\text{N}_2} = 9.22 \times 10^{-5}$ vs $x_{\text{O}_2} = 4.61 \times 10^{-5}$ — nitrogen dissolves about twice as much as oxygen in absolute amount, because its partial pressure is roughly 4 times higher. Oxygen is the more soluble gas only per unit of partial pressure (its $K_H$ is smaller) — which is why the dissolved N₂ : O₂ ratio (≈ 2 : 1) is much richer in oxygen than the ≈ 4 : 1 ratio in the air above.

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