Chapter 14, Problem 93A

Interpretation Introduction

Interpretation:

The pressure and the solubility have to be calculated using Henry’s Law.

Concept introduction:

Solubility: Solubility can be expressed as maximum amount of solute that can dissolve in a known amount of solvent at a given temperature and pressure.

Henry’s Law: In liquid, the solubility of a gas is directly proportional to its pressure above the liquid at given temperature.

Answer to Problem 93A

Solubility ( $\mathrm{gm}/litre$ )	Pressure (kPa)
2.9	25.08
3.7	32
4.51	39

Explanation of Solution

Given information:

$S_1=2.9\mathrm{gm}/litre$

$S_2=3.7\mathrm{gm}/litre$

$P_2=32.0\mathrm{kPa}$

The expression of Henry’s Law is $\frac{S_1}{P_1}=\frac{S_2}{P_2}$

The solubility of a gas is S₁ at P₁ pressure and S₂ at pressure P₂.

From the above table, we will use first two values in Henry’s equation:

$\begin{array}{c}\frac{S_1}{P_1}=\frac{S_2}{P_2}\\ P_1=\frac{S_1\times P_2}{S_2}\\ P_1=\frac{2.9gm/litre\times 32kPa}{3.7gm/litre}\\ =25.08kPa\end{array}$

From the above table, we will usenext two values in Henry’s equation:

$\begin{array}{c}\frac{S_1}{P_1}=\frac{S_2}{P_2}\\ S_2=\frac{S_1\times P_2}{P_1}\\ S_2=\frac{3.7gm/litre\times 39kPa}{32kPa}\\ S_2=4.51gm/litre\end{array}$