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Q1.39:
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.

Solution :

Step 1: Determination of Partial Pressures of Gases

According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is the product of its mole fraction in the gas phase and the total pressure of the mixture [Per Dalton's Law: $P_i = y_i \times P_{total}$].

Given:

  • Total Pressure ($P_{total}$) = $10 \text{ atm} = 10 \times 760 \text{ mm Hg} = 7600 \text{ mm Hg}$
  • Volume percentage of $O_2$ = $20\% \implies y_{O_2} = 0.20$
  • Volume percentage of $N_2$ = $79\% \implies y_{N_2} = 0.79$

Calculating Partial Pressures ($P_i$):

  • $P_{O_2} = 0.20 \times 7600 \text{ mm Hg} = 1520 \text{ mm Hg}$
  • $P_{N_2} = 0.79 \times 7600 \text{ mm Hg} = 6004 \text{ mm Hg}$

Step 2: Applying Henry's Law to Calculate Mole Fraction in Water

Henry's Law states that the partial pressure of a gas is proportional to its mole fraction in the solution: $P_i = K_H \times x_i$, where $K_H$ is the Henry's Law constant and $x_i$ is the mole fraction of the gas in the liquid phase [Per Henry's Law].

Rearranging for $x_i$: $x_i = \frac{P_i}{K_H}$

Step 3: Numerical Computation of Mole Fractions

Given $K_H(O_2) = 3.30 \times 10^7 \text{ mm Hg}$ and $K_H(N_2) = 6.51 \times 10^7 \text{ mm Hg}$:

Gas Partial Pressure ($P_i$) Henry's Constant ($K_H$) Mole Fraction ($x_i$)
Oxygen ($O_2$) $1520 \text{ mm Hg}$ $3.30 \times 10^7 \text{ mm Hg}$ $\frac{1520}{3.30 \times 10^7} \approx 4.61 \times 10^{-5}$
Nitrogen ($N_2$) $6004 \text{ mm Hg}$ $6.51 \times 10^7 \text{ mm Hg}$ $\frac{6004}{6.51 \times 10^7} \approx 9.22 \times 10^{-5}$

Graphical Representation of Solubility Equilibrium

Partial Pressure (P) Mole Fraction (x) $N_2$ Slope $O_2$ Slope

Final Solution: The composition of gases in water is $x_{O_2} = 4.61 \times 10^{-5}$ and $x_{N_2} = 9.22 \times 10^{-5}$.


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