# Chapter 24: Fundamentals of Mass Transfer. Example 1

Book: Welty, James R.; Wicks, Charles E.; Wilson, Robert E. Fundamentals of Momentum, Heat, and Mass Transfer, third edition. New York: John Wiley and Sons. 1984.

Chapter 24: Fundamentals of Mass Transfer, Page 471

Note: LaTex causes certain parts of the text to be raised, italicized, and larger. It is not to the power. I believe the “power” will be recognizable once I point the latter out.

Example 1

The composition of air is often given in terms of only the two principle species in the gas mixture

oxygen, $O_2, y_{O_2} = 0.21$

nitrogen, $N_2, y_{N_2} = 0.79$

Determine the mass fraction of both oxygen and nitrogen and the mean molecular weight of the air when it is maintained at 25$\textdegree$C (298 K) and 1 ATM $(1.013 X 10^5 Pa)$. The molecular weight of oxygen is 0.032 kg/mol and of nitrogen is 0.028 kg/mol.

As a basis for our calculations, consider 1 mole of the gas mixture;

moles of oxygen present = (1 mol) $(y_{O_2})$ = (1 mol)(0.21) = 0.21 mol oxygen

mass of oxygen present = (0.21 mol) $(M_{O_2})$ = (0.21 mol) $(0.032 \frac{ kg}{mol})$ = 0.00672 kg oxygen

moles of nitrogen present = (1 mol) $(y_{N_2})$ = (1 mol)(0.79) = 0.79 mol nitrogen

mass of nitrogen present = (0.79 mol) $(M_{N_2})$ = (0.79) $(0.028 \frac{kg}{mol})$ = 0.0221 kg nitrogen

total mass present = kg oxygen + kg nitrogen = 0.00672 kg oxygen + 0.0221 kg nitrogen = 0.0288 kg

$w_{O_2} = \frac{0.00672 kg}{0.0288 kg} = 0.23$

$w_{N_2} = \frac{0.0221 kg}{0.0288 kg} = 0.77$

Since 1 mole of the gas mixture has a mass of 0.0288 kg, the mean molecular weight of the air must be 0.0288: $M_{air}$ = 0.0288 kg/mol. When one takes into account the other constituents that are present in air, the mean molecular weight of air is often rounded off to $0.029 \frac{kg}{mol}$.

This problem could also be solved using the ideal gas law, PV = nRT. At ideal conditions, 0$\textdegree$C or 273 K and 1 atm or $1.013 X 10^5 Pa$ pressure, the gas constant is evaluated to be

Remember, we want to calculate the mass fraction of oxygen and nitrogen, and the mean molecular weight of air. Which to calculate first is the key

PV = nRT

R = $\frac{PV}{nT}$

n = 1 mole; T = 273 K; P = $1.015 X 10^5 Pa$; V = $22.4 m^3$

R = $\frac{1.013 X 10^5 Pa x 22.4 m^3}{1 kg mol x 273 K}$ = $8.314 X 10^3 \frac{Pa m^3}{(kg mol) K}$

R = $8.314 X 10^3 \frac{(Pa) (m^3)}{(kg mol) (K)} X \frac{1 kg mole}{1000 moles}$ = $8.314 \frac{(Pa)(m^3)}{(mol) (K)}$

The of volume the gas mixture, at 298 K, is

V= $\frac{nRT}{P} = \frac{(1 mol) (8.314 \frac{Pa m^3}{mol K})(298 K)}{1.013 X 10^5 Pa} = 0.0245 (m^3)$

The concentrations are (1 mole basis)

$c_{O_2} = \frac{moles O_2}{volume of gas mixture} = \frac{0.21 mol}{0.0245 m^3} = 8.57 \frac{mol O_2}{m^3}$

$c_{N_2} = \frac{moles N_2}{volume of gas mixture} = \frac{0.79 mol}{0.0245 m^3} = 32.3 \frac{mol N_2}{m^3}$

$c = \sum_{i=1}^{n} c_i = 8.57 \frac{mol O_2}{m^3} + 32.3 \frac{mol N_2}{m^3} = 40.9 \frac{mol}{m^3}$

The total density, $\rho$ is

mass of $O_2$ = $(mol O_2) (Molecular Weight O_2) = (0.21 mol O_2)(0.032 \frac{kg}{mol}) = 0.00672 kg O_2$

mass of $N_2$ = $(mol N_2)(Molecular Weight N_2) = (0.79 mol N_2)(0.028 \frac{kg}{mol}) = 0.0221 kg N_2$

total mass = 0.00672 kg + 0.0221 kg = 0.0288 kg

total density = $\rho = \frac{total mass}{total volume} = \frac{0.0288 kg}{0.0245 m^3} = 1.180 \frac{kg}{m^3}$

and the mean molecular weight of the mixture is

$M = \frac{\rho}{c} = \frac{1.180\frac{kg}{m^3}}{40.9 \frac{mol}{m^3}} = 0.288 \frac{kg}{mol}$

As a side note, density, $\rho_i$, can be used to calculate mass percent of $O_2$ and $N_2$

$\rho_{O_2} = (c_{O_2})(M_{O_2}) = (8.57 \frac{mol O_2}{m^3})(0.032 \frac{kg}{mol})= 0.27 \frac{kg O_2}{m^3}$

$w_{O_2} = \frac{\rho_{O_2}}{\rho_total} = \frac{0.27 \frac{kg O_2}{m^3}}{1.180 \frac{kg}{m^3}} = 0.23$

$\rho_{N_2} = (c_{N_2})(M_{N_2}) = (32.3 \frac{mol N_2}{m^3})(0.028 \frac{kg N_2}{mol N_2} = 0.904 \frac{kg N_2}{m^3}$

$w_{N_2} = \frac{\rho_{N_2}}{\rho_total} = \frac{0.904 \frac{kg N_2}{m^3}}{1.180 \frac{kg}{m^3}}= 0.77$