Chapter 6, Problem 6.17NP

Interpretation Introduction

Interpretation : The value of ${\mu }_{O_2}^{mixture}(298.15\text{ K, 1 bar)}$ for oxygen in air should be calculated.

Concept Introduction :

The chemical potential of the icomponent in a mixture can be calculated as follows:

  ${\mu }_i^{mixture}(T,P)={\mu }_{O_2}^0(T)+RT\ln \frac{P}{P_0}+RT\ln x_i$

Here,

  $\begin{array}{l}{\mu }_i^{mixture}\text{ - chemical potential of i in a mixture}\\ {\mu }_{O_2}^0\text{ - chemical potential of pure i}\\ \text{R - universal gas constant}\\ \text{T - temperature}\\ \text{P - pressure}\\ x_i\text{ - mole fraction of i}\end{array}$