Chapter 4, Problem 4D.4ST

Interpretation Introduction

Interpretation:

The amount of partial pressure of methane is required to dissolve 21 mg of methane in 100 g of benzene at ${25}^{o}C$ has to be calculated.

Concept introduction:

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

Partial pressure of a gas in terms of its mole fraction and total pressure is,

`            ${\text{P}}_{\text{A}}\text{=}{\text{χ}}_{\text{A}}\text{×}{\text{P}}_{\text{TOTAL}}$

Henry’s law states that the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

$\text{p}={\text{K}}_{\text{H}}\text{x}\boxed{\begin{array}{l}\text{where,}\\ {\text{K}}_{\text{H}}\text{:}\text{Henry's}\text{constant}\\ \text{x}\text{:}\text{mole}\text{fraction}\\ \text{p}\text{:}\text{partial}\text{pressure}\end{array}}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$