Question

When does Raoult's law become a special case of Henry's law?

Answer 1

Raoult's law:

• "At any given temperature, the partial pressure of any volatile liquid in a solution is equal to the product of the component's vapour pressure in its pure state and its mole fraction in the solution."

Henry's law:

• The law was first proposed in$1803$ by William Henry, an English physician and scientist.
• The weight of a gas dissolved by a liquid is proportional to the pressure of the gas on the liquid, according to Henry's law.
• It only applies to dilute solutions and low gas pressures.

Special case:

• When ${\mathrm{K}}_{\mathrm{H}}$ equals ${{\mathrm{p}}_1}^0$ in Henry's law, Raoult's law becomes a special instance of Henry's law.
• According to Raoult’s law:

$\mathrm{p}={\mathrm{p}}^0\mathrm{x}$

where $\mathrm{p}$ is the partial pressure, $\mathrm{x}$ is the mole fraction and ${\mathrm{p}}^0$ is the vapour pressure of the pure component.

• According to Henry’s law:

$\mathrm{p}={\mathrm{K}}_{\mathrm{H}}\mathrm{x}$

Where $\mathrm{p}$ is the partial pressure, $\mathrm{x}$ is the mole fraction and ${\mathrm{K}}_{\mathrm{H}}$ is the proportionality constant (Henry’s constant)

• We can see that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution by comparing the above equations. As a result, where ${\mathrm{K}}_{\mathrm{H}}={\mathrm{p}}^0$, Raoult's law is a particular instance of Henry's law.