Question

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

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Hint: Use Henry's law to solve this question. Its formula is-
${{P}_{gas}}={{k}_{H}}{{\chi }_{gas}}$
The total pressure is $10\operatorname{atm}$ which includes both nitrogen and oxygen, so find their individual partial pressure.

We will use Henry’s law to solve this question. This law gives the relationship between the partial pressure of a particular gas above a solution and the amount of that gas dissolved in that solution. According to Henry, the partial pressure is directly proportional to the amount of gas dissolved in the solution. The proportionality constant is known as Henry’s constant and is abbreviated as ${{k}_{H}}$. The equation is as follows:
${{P}_{gas}}={{k}_{H}}{{\chi }_{gas}}$
Where,${{P}_{gas}}$ is the partial pressure of the gas above the solution and ${{\chi }_{gas}}$ is the mole fraction of gas dissolved inside the solution.
Let’s find out the partial pressures of the gases mentioned in the question-
- Partial pressure of oxygen (${{P}_{{{O}_{2}}}}$)
Taking the total pressure as $10\operatorname{atm}$ which is also equal to$7600\operatorname{mm}\operatorname{Hg}$ (because $1\operatorname{atm}=760\operatorname{mm}Hg$) and the volume percentage of oxygen is$20%$, its partial pressure is-
$\dfrac{20}{100}\times 7600 = 1520\operatorname{mm}Hg$
- Partial pressure of nitrogen (${{P}_{{{N}_{2}}}}$)
Similarly, taking the volume percentage of nitrogen as $79%$, its partial pressure is-
$\dfrac{79}{100}\times 7600=6004\operatorname{mm}Hg$
We can now calculate the solubility of each of the gases individually as-
- solubility of oxygen
\begin{align} & {{P}_{{{O}_{2}}}}={{k}_{H}}{{\chi }_{{{O}_{2}}}} \\ & \Rightarrow {{\chi }_{{{O}_{2}}}}=\dfrac{{{P}_{{{O}_{2}}}}}{{{k}_{H}}} \\ \end{align}
Putting the respective values, we get
$\Rightarrow {{\chi }_{{{O}_{2}}}}=\dfrac{1520}{3.30\times {{10}^{7}}}=4.60\times {{10}^{-5}}$
- Solubility of nitrogen
\begin{align} & {{P}_{{{N}_{2}}}}={{k}_{H}}{{\chi }_{{{N}_{2}}}} \\ & \Rightarrow {{\chi }_{{{N}_{2}}}}=\dfrac{{{P}_{{{N}_{2}}}}}{{{k}_{H}}} \\ \end{align}
Putting the respective values, we get
$\Rightarrow {{\chi }_{{{O}_{2}}}}=\dfrac{6004}{6.51\times {{10}^{7}}}=9.22\times {{10}^{-5}}$
So, the mole fractions of oxygen and nitrogen dissolved inside water are $4.60\times {{10}^{-5}}$ and $9.22\times {{10}^{-5}}$ respectively.

Note: Convert all the units used into the unit which is asked by the question. As in the above case, we converted the unit of “atm” into that of “mmHg”.
Mole fraction is a dimensionless quantity because it is a ratio between the same units.