# Chapter 24: Fundamentals of Mass Transfer. Example 3

Example 3

Reevaluate the diffusion coefficient of carbon dioxide in air at 20$\textdegree$C and atmospheric pressure using the Fuller, Schettler, and Giddings equation and compare the new value with the one reported in example 2.

The equation is

$D_{AB} = \frac{0.001T^{1.75}(\frac{1}{M_A}\frac{1}{M_B})^{1/2}}{P[(\sum \nu)_A^{1/3} + (\sum \nu)_B^{1/3}]^2}$

Molecular weight carbon dioxide

$M_A = M_{CO_2} = M_C + 2 M_{O} = 12 + 2(16) = 44$

Molecular weight Air

$M_B = M_{Air}; 21\% \frac{mole O_2}{moles} O_2; 79\% N_2 \frac{mole N_2}{moles}$

1 mole basis

$0.21 \frac{mole O_2}{moles}(1 mole) = 0.21 mole O_2$

$0.79 \frac{mole N_2}{moles}(1 mole) = 0.79 mole N_2$

Total mass

$mass_{O_2} = 0.21 mole O_2 (M_{O_2}) = 0.21 mole O_2(32 \frac{g O_2}{mole O_2}) = 6.72 g O_2$

$mass_{N_2} = 0.79 mole N_2 (M{_{N_2}} = 0.79 mole N_2(28 \frac{g N_2}{mole N_2} = 22.1 g N_2$

Total Mass = 6.7 + 22.1 = 29 grams

Molecular Weight of Air = $\frac{29 grams}{1 mole}= 29 \frac{g}{mole}$

$\nu$ terms from book. For carbon dioxide and Air, the terms were already calculated (Page 491) and included as “Simple Molecules”.

$(\sum \nu)_A = (\sum \nu)_{CO_2 } = \nu_{CO_2} = 26.9$

$(\sum \nu)_B = (\sum \nu)_{Air} = \nu_{Air} = 20.1$

We have all the needed variables to use the Fuller, Schettler, and Giddings equation

$D_{AB} = \frac{0.001T^{1.75} (\frac{1}{M_{CO_2}}+\frac{1}{M_{Air}})^{1/2}}{P[(\nu_{CO_2})^{1/3} + (\nu_{Air})^{1/3}]^{2}} = \frac{0.00143(293 K)^{1.75}(\frac{1}{44} + \frac{1}{29})^{1/2}}{1 atm[(26.9)^{1/3} + (20.1)^{1/3}]^{2}} =$

$D_{AB} = \frac{7.1}{32.7} = 0.151 \frac{cm^2}{s}$

Compared to Hirschfelder, Bird, and Spoz at 20$\textdegree$C and 1 atm

$D_{AB} = \frac{0.001858T^{3/2}[\frac{1}{M_A} + \frac{1}{M_B}]^{1/2}}{P \delta_{AB}^2 \Omega_D} = 0.147 \frac{cm^2}{s}$

Compared to the Experimental at 20$\textdegree$C and 1 atm

$D_{AB_{T_2, P_2}} = \frac{P_1}{P_2}(\frac{T_2}{T_1})^{3/2}\frac{\Omega_{D|T1}}{\Omega_{D|T2}} = 0.155 \frac{cm^2}{s}$

Percent Difference from Predicted Experimental

Precent Difference = $\frac{0.151 - 0.155}{0.155} x 100 = -2.58\%$