Chapter 16.1, Problem 1SP

Interpretation Introduction

Interpretation: To calculate the solubility of a gas in water when pressure increases.

Concept introduction: Henry's law is a law used to determine the solubility of gases in a liquid. This law states that the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of the gas.

At constant temperature, Henry's law is described as:

  $\frac{S\text{1}}{P\text{1}}=\frac{S\text{2}}{P\text{2}}$

Answer to Problem 1SP

The solubility when pressure increases to $288\text{ KPa}$ is $0.443\frac{g}{L}$ .

Explanation of Solution

To determine the solubility of a gas in the liquid, Henry's law is used. According to Henry's law, at constant temperature the relation between solubility and pressure is as follows:

  $\frac{S\text{1}}{P\text{1}}=\frac{S\text{2}}{P\text{2}}$

Where:

• S₁ is the solubility at an initial pressure.
• S₂ is the solubility at final pressure.
• P₁ is the initial pressure.
• P₂ is the final pressure.

The data given in the question is as follows:

• Solubility of gas in liquid ( S₁ ) is $0.16\frac{g}{L}$ .
• Pressure( P₁ ) is $104\text{ KPa}$ .
• Pressure increases to $288\text{ KPa}$ which is P₂ .

Put all the values in the above formula to get the value of new solubility ( S₂ ).

  $\frac{0.16\frac{g}{L}}{104\text{ KPa}}=\frac{S\text{2}}{288\text{ KPa}}$

  $\frac{0.16\frac{g}{L}}{104\text{ KPa}}\times 288\text{ KPa}=S\text{2}$

  $S\text{2}=0.443\frac{g}{L}$

Therefore, the new solubility is $0.443\frac{g}{L}$ .

Conclusion

The solubility when pressure increases to $288\text{ KPa}$ is $0.443\frac{g}{L}$ .