# Diffusivity: Chapman and Enskog Versus Hirschfelder Equation when Compared to Experimental Value at 25 Degree C and 1 Atm, and Non-polar Versus Brokaw Polar Method

Title: Diffusivity: Chapman and Enskog Versus Hirschfelder Equation when Compared to Experimental Value at 25$\textdegree$C and 1 Atm, and Non-polar Versus Brokaw Polar Method

Chapman and Enskog Equation

$D_{AB} = \frac{3}{16} \frac{(\frac{4 \pi \kappa T}{M_{AB}})^{1/2}}{n \pi \sigma_{AB}^{2} \Omega_D}f_D$

When $f_D$ is unity and n is expressed by the ideal gas law

$D_{AB} = \frac{0.0026 T^{3/2}}{PM_{AB}^{1/2} \sigma_{AB}^{2}\Omega_D}$

Hirschfelder, Bird, and Spotz Equation

$D_{AB} = \frac{0.001858 T^{3/2}[(\frac{1}{M_A}) + (\frac{1}{M_B})]^{1/2}}{P \sigma_{AB}^{2} \Omega_D}$

Non-polar Comparison

There are suggested correction factors for polar compounds. Since water is a polar compound, I will use these factors in a later comparison. I am doing a non-polar comparison because I was surprised with the closeness of the Hirschfelder equation previously when using non-polar factors.

Hirschfelder, Bird, and Spotz equation[2]

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

$M_A = M_{H_2O} = 18 \frac{g}{mol}; M_B = M_{air} = 29 \frac{g}{mol}$

T = 298 K; P = 1 atm

Need $\sigma_{AB} \ and \ \Omega_D$. For comparative purposes, will use tabular values from [1].

Water: $\sigma_A = 2.641 \AA; \frac{\epsilon_A}{\kappa} = 809.1 K$

Air: $3.711 \AA; \frac{\epsilon_B}{\kappa} = 78.6 K$

$\sigma_{AB} = \frac{\sigma_A + \sigma_B}{2} = \frac{2.641 + 3.711}{2} = 3.18 \AA$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{\frac{\epsilon_A}{\kappa} \frac{\epsilon_B}{\kappa}} = \sqrt{(809.1)(78.6)} = 252.2 K$

Need $T^* = \frac{\kappa T}{\epsilon_{AB}}$ to calculate $\Omega_D$

Neufeld, et al.: $\Omega_D = \frac{A}{(T^*)^{B}} + \frac{C}{\exp{DT^*}} + \frac{E}{\exp{FT^*}} + \frac{G}{\exp{HT^*}}$

The constants

A;B;C;D;E;F;G;H will be placed in the Neufeld, et. al. equation

First calculate $T^*$

$\frac{\epsilon_{AB}}{\kappa T} = \frac{252.2 K}{298 K} = 0.846$

$T^{*} = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.846} = 1.18$

$\Omega_D = \frac{A}{(T*)^B} + \frac{C}{\exp{DT^*}} + \frac{E}{\exp{FT^*}} + \frac{G}{\exp{HT^*}} =$

$\frac{1.06036}{1.18^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(1.18))}} + \frac{1.03587}{\exp{((1.52996)(1.18))}} + \frac{1.76474}{((3.89411)(1.18))} = \Omega_D$

$\Omega_D = 1.33$

Back to Hirschfelder equation

$D_{AB} = \frac{0.001858 T^{3/2}[\frac{1}{M_A} + \frac{1}{M_B}]^{1/2}}{P \sigma_{AB}^{2} \Omega_D} = \frac{0.001858 (298)^{3/2}[\frac{1}{18} + \frac{1}{29}]^{1/2}}{1 (3.18)^{2} 1.33} =$

$D_{AB} = \frac{2.87}{13.45} = 0.213 \frac{cm^2}{s}$

Compared to experimental value from [2] at 25$\textdegree$C and 1 atm

Diffusivity of water in air: $D_{AB} = 0.260 \frac{cm^2}{s}$

Chapman and Enskog equation

$D_{AB} = \frac{0.00266 T^{3/2}}{P M_{AB}^{1/2} \sigma_{AB}^2 \Omega_D}$

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1} = 2[\frac{1}{18} + \frac{1}{29}]^{-1} = 22.21$

T = 298 K; P = 1 atm; $\sigma_{AB} = 3.18 \AA; \ \Omega_D = 1.33$

$D_{AB} = \frac{0.00266 (298)^{3/2}}{1 (22.21)^{1/2} (3.18)^2 (1.33)} = \frac{13.68}{63.38} =$

$D_{AB} = 0.216 \frac{cm^2}{s}$

Compared to Hirschfelder equation and experimental value

Hirschfelder[2]: $D_{AB} = 0.213 \frac{cm^2}{s}$

Percent Difference:

$\frac{Hirschfelder - Experimental}{Experimental} x 100 = \frac{0.213 - 0.260}{0.260} x 100 = -18.1\%$

Experimental: $D_{AB} = 0.260 \frac{cm^2}{s}$

Chapman[1]: $D_{AB} = 0.216 \frac{cm^2}{s}$

Percent Difference:

$\frac{Chapman - Experimental}{Experimental} x 100 = \frac{0.216 - 0.260}{0.260} x 100 = -16.9\%$

Polar Molecule Correction Comparison

Sadly, I have discovered that most empirical correlations lack sufficient data to estimate the diffusivity of many compounds. As an example, I, as a 1991 Gulf War veteran, desire to calculate the diffusivity of sarin in air. I have discovered that most empirical correlations do not take the phosphorus atom into consideration. Also, the Brokaw relationships for polar gases have correction equations that consider polar diffusing through polar.

In my analysis, I will first use the Brokaw method and only consider the polar molecule of water since air is non-polar. I will highlight the potential error of using this method by “error” when I use the correction equations. To be specific, $\Omega_{D_{H_2O}}$ will be calculated based on the new variable $\delta_{H_2O}$ instead of a $\delta_{AB}$ of two polar species. I hope to see if I can use the Brokaw method to calculate the diffusivity of sarin in air since the correction factors used in the Brokaw method include phosphorus and fluorine. I wish I could have found a phosphorus “diffusion volume increment”, $\nu$, but I could not find one. I did find a fluorine[1], Nitrogen, Sulfur, Iodine, etc and might use either Nitrogen or Sulfur in the Fuller, et al. equation as an estimate when I calculate the diffusivity of sarin in air.

Since both equations gave approximately the same value during the non-polar comparison, I will use the Chapman-Enskog equation. Once again, the temperature and pressure are 25$\textdegree$C and 1 atmosphere.

Chapman-Enskog equation[1]

$D_{AB} = \frac{0.00266 T^{3/2}}{PM_{AB}^{1/2}\sigma_{AB}^2 \Omega_D}$

T = 298 K; P = 1 atm

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1} = 2[\frac{1}{18} + \frac{1}{29}]^{-1} = 22.21$

Brokaw Method

Neufield, et al. Relation:

$\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{(DT^*)}} + \frac{E}{\exp{(FT^*)}} + \frac{G}{\exp{(HT^*)}}$

$T^* = \frac{\kappa T}{\epsilon_{AB}}$

$\Omega_D = \Omega_D(Neufield) + \frac{0.19 \delta_{AB}}{T^*}$

Possible error: Since air is non-polar, I will only be using the delta of water, $\delta_A$ in the above equation.

Changed equation: $\Omega_D = \Omega_D(Neufield) + \frac{0.19 \delta_A}{T^*}$

For Water (A)

$\frac{\epsilon_A}{\kappa} = 1.18(1 + 1.3\delta_A^2)T_b; T_b = normal \ boiling \ point (1 atm), K = 373 K$

$\delta_A = \frac{1.94 x 10^3 \mu_p^2}{V_b T_b}; V_b = Molar \ Volume \ at \ T_b; \mu_p = Dipole \ moment, D$

Calculation by Le Bas method[1]

$V_b = 2(H) + 1(O) = 2(3.7) + (7.4) = 14.8 \frac{cm^3}{mol}$

Percent Difference: $\frac{Calculated - Experimental}{Experimental} = \frac{14.8 - 18.8}{18.8} = -21.3\%$

Note: ChemSpider provides a “Molar Volume” of $18.045 cm^3$ for water. Although I assume the latter was calculated at normal boiling point, I am not certain. Still, the value is extremely close to the experimental value in reference [1], $18.8 \frac{cm^3}{mol}$. For this reason, I will use the Chemspider value for water. Why? ChemSpider also provides a Molar Volume value for Sarin. If ChemSpider responds by email that the value was not calculated at boiling point, I will reconsider. Still, the percent difference of ChemSpider for water Molar Volume when compared to experimental[1] is:

ChemSpider Percent Difference:

$\frac{ChemSpider - Experimental}{Experimental}x100 = \frac{18.045 - 18.8}{18.8}x 100 = -4.02\%$

Delta: $\delta_{A} = \frac{1.94x10^3 \mu_p}{V_bT_b}$

$\mu_p = dipole \ moment, \ debyes$

$V_b = liquid \ molar \ volume \ at \ normal \ boiling \ point, \ \frac{cm^3}{mole}$

$T_b = normal \ boiling \ point \ (1 atm), \ K$

$\delta_{A} = \frac{1.94 x 10^3(1.855)^2}{(18.045)(373)} = 0.992$

$\frac{\epsilon_A}{\kappa} = 1.18(1 + 1.3 \delta_A^2)T_b = 1.18(1 + 1.3(0.992)^2)373 = 1003 \ K$

$\sigma_A = (\frac{1.585V_b}{(1 + 1.3\delta_A^2)})^{1/3} = (\frac{1.585(18.045)}{(1 + 1.3(0.992)^2)})^{1/3} = 2.32 \AA$

Need: $T^* = \frac{\kappa T}{\epsilon_{AB}}$

Water: $\sigma_A = 2.32 \AA; \frac{\epsilon_A}{\kappa} = 1003 K$

From [1]: Air: $\sigma_B = 3.711 \AA; \frac{\epsilon_B}{\kappa} = 78.6 K$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{\frac{\epsilon_A}{\kappa}\frac{\epsilon_B}{\kappa}} = \sqrt{(1003.0)(78.6)} = 280.8$

$\frac{\epsilon_{AB}}{\kappa T} = \frac{280.8}{298.0} = 0.942$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.942} = 1.06$

Neufield, et al. Relation

$\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}} =$

$\Omega_D = \frac{1.06036}{(1.06)^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(1.06))}} + \frac{1.03587}{\exp{((1.52996)(1.06))}} + \frac{1.76474}{\exp{((3.89411)(1.06))}} =$

$\Omega_D = 1.4$

Brokaw relation for polar molecules

$\Omega_D = \Omega_D(Neufield) + \frac{0.19 \delta_{AB}^2}{T^*}$

Changed for water only: $\Omega_D = \Omega_D (Neufield) + \frac{0.19 \delta_A^2}{T^*}$

$\Omega_D = 1.4 + \frac{(0.19)(0.992)^2}{1.06} = 1.57$

Check Chapman and Enskog equation

$D_{AB} = \frac{0.00266 T^{3/2}}{P M_{AB}^{1/2} \sigma_{AB}^2 \Omega_D}$

T = 298 K; P = 1 atm; $M_{AB} = 22.21 \ and \ \Omega_D = 1.57$

Need: $\sigma_{AB}^2$

From Brokaw relation

$\sigma_{AB} = \sqrt{\sigma_A \sigma_B} = \sqrt{(2.32)(3.711)} = 2.93 \AA$

Chapman and Enskog equation for diffusivity of polar water in non-polar air

$D_{AB} = \frac{0.00266 (298)^{3/2}}{1(22.21)(2.93)^2(1.57)} = \frac{13.68}{63.52} = 0.215 \frac{cm^2}{s}$

Non-polar versus polar comparisons

Using: $D_{AB} = \frac{0.00266 T^{3/2}}{PM_{AB}^{1/2} \sigma_{AB}^2 \Omega_D}$ and Brokaw relationships for polar corrections

Chapman and Enskog non-polar molecule[1]: $D_{AB} = 0.216 \frac{cm^2}{s}$

Chapman and Enskog with Brokaw relationships for polar molecules[1]: $D_{AB} = 0.215 \frac{cm^2}{s}$

Hirschfelder nonpolar[2]: $D_{AB} = 0.213 \frac{cm^2}{s}$

The experimental value[2]: $D_{AB} = 0.260 \frac{cm^2}{s}$

Note: All the ‘calculated” values are quite close. As such, I assume I can, when needed, use Chapman and Enskog and the Brokaw method to calculate the diffusivity of a polar molecule of sarin in non-polar air.

References

[1] Poling, Bruce E.; Prausnitz, John M.; O’Connell, John P. (2001) The Properties of Gases and Liquids, Fifth Edition. New York: Mcgraw-Hill.

[2] Welty, James R.; Wicks, Charles E.; Wilson, Robert E. (1984) Fundamentals of Momentum, Heat, and Mass Transfer, third edition. New York: John Wiley & Sons.

[3] Harding, Byron. Chapter 24: Fundamentals of Mass Transfer. Diffusivity of water in Air at 20 Degrees Celsius and 1 Atmosphere, December 2012. chrisbharding.wordpress.com[online]. 2012. Available from: https://chrisbharding.wordpress.com/2012/12/28/chapter-24-fundamentals-of-mass-transfer-diffusivity-of-water-in-air-at-25-degrees-celsius/

[4] ChemSpider. The free chemical database. chemspider.com[online]. 2013. Available from: http://www.chemspider.com/

[5] Gregory, J.K.; Clary, D.C.; Liu, K.; Brown, M.G.; Saykally, R.J. The Water Dipole Moment in Water Clusters, February 1997. science[online]. vol. 275. pp. 814. Available from: http://www.cchem.berkeley.edu/rjsgrp/publications/papers/1997/187_gregory_1997.pdf

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## 3 thoughts on “Diffusivity: Chapman and Enskog Versus Hirschfelder Equation when Compared to Experimental Value at 25 Degree C and 1 Atm, and Non-polar Versus Brokaw Polar Method”

1. Karla |

hi, i have the same problem with phosphorus, i would like to calculate the diffusion molecular volume for chlorpyrifos and diazinon (both with a phosphorus atom in their molecule) to explain why these are not include perfectly within cyclodextrin, If anyone knows how to board my problem I would really appreciate it. Thanks!

• Hello,

I am sorry for the late response. I have been busy. I am sure you already figured out the problem.