Chapter 14, Problem 14.1P

Assuming air to be composed exclusively ${\text{O}}_{\text{2}}$ and ${\text{N}}_{\text{2}}$ , with their partial pressures in the ratio 0.21:0.79, what arc their mass fractions?

To determine

Mass fraction of each species in the mixture.

Answer to Problem 14.1P

Mass fractions of ${\text{O}}_{\text{2}}$ and ${\text{N}}_{\text{2}}$ ,

  ${\text{m}}_{{\text{O}}_{\text{2}}}=0.233$

  ${\text{m}}_{{\text{N}}_{\text{2}}}=0.767$

Explanation of Solution

  $\left[\begin{array}{l}{M}_{{\text{O}}_{\text{2}}}=32\text{kg/kmol}\\ M_{{\text{N}}_{\text{2}}}=28\text{kg/kmol}\end{array}\right]$

According to mass fraction definition,

  ${\text{m}}_{\text{i}}=\frac{{\rho }_{\text{i}}}{\rho }=\frac{{\rho }_{\text{i}}}{{\displaystyle \sum {\rho }_{\text{i}}}}$

With

  ${\rho }_{\text{i}}=\frac{{\text{p}}_{\text{i}}}{{\text{R}}_{\text{i}}\text{T}}=\frac{{\text{p}}_{\text{i}}}{\left(\Re /M_{\text{i}}\right)\text{T}}=\frac{M_{\text{i}}{\text{p}}_{\text{i}}}{\Re \text{T}}$

Thus,

  ${\text{m}}_{\text{i}}=\frac{\frac{M_{\text{i}}{\text{p}}_{\text{i}}}{\Re \text{T}}}{{\displaystyle \sum \frac{M_{\text{i}}{\text{p}}_{\text{i}}}{\Re \text{T}}}}$

Now,

Cancelling the terms and dividing both numerator and denominator by the total pressure $\text{p}$ ,

  ${\text{m}}_{\text{i}}=\frac{M_{\text{i}}{\text{x}}_{\text{i}}}{{\displaystyle \sum M_{\text{i}}{\text{x}}_{\text{i}}}}$

Then,

Mole fractions,

  ${\text{x}}_{{\text{O}}_{\text{2}}}=\frac{{\text{p}}_{{\text{O}}_{\text{2}}}}{\text{p}}$

  ${\text{x}}_{{\text{O}}_{\text{2}}}=\frac{0.21}{0.21+0.79}=0.21$

And

  ${\text{x}}_{{\text{N}}_{\text{2}}}=\frac{{\text{p}}_{{\text{N}}_{\text{2}}}}{\text{p}}$

  ${\text{x}}_{{\text{N}}_{\text{2}}}=\frac{0.79}{0.21+0.79}=0.79$

Thus,

Mass fractions,

  ${\text{m}}_{{\text{O}}_{\text{2}}}=\frac{M_{{\text{O}}_{\text{2}}}{\text{x}}_{{\text{O}}_{\text{2}}}}{M_{{\text{O}}_{\text{2}}}{\text{x}}_{{\text{O}}_{\text{2}}}+M_{{\text{N}}_{\text{2}}}{\text{x}}_{{\text{N}}_{\text{2}}}}$

  ${\text{m}}_{{\text{O}}_{\text{2}}}=\frac{32\times 0.21}{32\times 0.21+28\times 0.79}$

  ${\text{m}}_{{\text{O}}_{\text{2}}}=0.233$

And

  ${\text{m}}_{{\text{N}}_{\text{2}}}=1-{\text{m}}_{{\text{O}}_{\text{2}}}$

  ${\text{m}}_{{\text{N}}_{\text{2}}}=1-0.233$

  ${\text{m}}_{{\text{N}}_{\text{2}}}=0.767$