# Diffusivity of Water versus Sarin (Nerve Agent) in Air at 10 Degrees Celsius (50 Degrees Fahrenheit) and 1 Atmosphere

Diffusivity of Water versus Sarin in Air at 10 Degrees Celsius (50 Degrees Fahrenheit) and 1 Atmosphere[see bottom of post]

1991 Gulf War veterans are suffering from 1991 Gulf War Illness[3;References]. Scientific research suggests the combination of experimental medication, pyridostigmine bromide as an example, over use of pesticides, chemical weapon-sarin as an example-destruction at plants and football sized bunkers, oil fires, etc as the potential cause[6-9].

Dr. Robert Haley, MD, UT SouthWestern Medical Center, and Intelligence Analyst James Tuite have reported how 1991 Gulf War veterans might have been contaminated with chemical weapons prior to the ground war, “Desert Storm”[9]. In fact, their work provides data proving that sophisticated equipment detected chemical weapons in Saudi Arabia prior to the ground war[9a]. It is also hypothesized that the “toxic cocktail” has caused autonomic dysfunction, nerve death, and brain death[9-14].

As a 1991 Gulf War veteran, I have been affected. I am also a chemical engineer with a degree in biological sciences. Like most educated, I have lost much of my knowledge in chemical engineering and biological sciences, but I can, if I find a good example, still “plug and chug” by using “tested and trusted” equations, which is advised anyhow. 🙂 Here, I compare the diffusivity of sarin vapor and water vapor in air by using Chapman and Enskog equation with Brokaw relations for polar gases correction. I have shown that the equation can be used when considering the diffusivity of polar in a non-polar matrix[19]. After performing the latter calculation, I noticed that reference [1] also suggests Brokaw relations to be used for diffusivity of one polar gas molecule in a non-polar matrix[1].

I will be comparing the diffusivity of polar sarin = A in non-polar air = B at 10$\textdegree$C and 1 atmosphere. I chose 10$\textdegree$C because I discovered data, possibly experimental, that stated that 90% volume of 1 mm sarin drop on a non-absorbable surface at 10$\textdegree$C evaporated in 0.24 hours[17].

Equations

Chapman and Enskog Equation[1]. Reference [1] reports that this equation has a “Average absolute error” of 7.9% when used without Brokaw relations. The range is from 0% to 25%. The authors[1] did not provide an average for Browkaw relations but do provide specific absolute error values. When I averaged the Brokaw values[1], I obtained a 10.9% average absolute error with a range from 0% to 33%.

Chapman and Enskog Equation[1]

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

Neufield, et al. Equation

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

Polar Gases: Brokaw Relations

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

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

$\delta = \frac{1.94x10^3 \mu_p^2}{V_bT_b}$

$\mu_p = dipole \ moment, \ debyes$

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

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

$\frac{epsilon}{\kappa} = 1.18(1 + 1.3\delta^2)T_b$

$\sigma = (\frac{1.585V_b}{1 + 1.3 \delta^2})^{1/3}$

$\delta_{AB} = (\delta_A \delta_B)^{1/2}$

$\frac{\epsilon_{AB}}{\kappa} = (\frac{\epsilon_A}{\kappa} \frac{\epsilon_B}{\kappa})^{1/2}$

$\sigma_{AB} = (\sigma_A \sigma_B)^{1/2}$

When $f_D$ is chosen as unity and “n” is expressed by the ideal-gas law, the Chapman-Enskog Equation

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

Brokaw Diffusivity: Water in Air at 10$\textdegree$C and 1 Atmosphere

Molecular Weight

Water:

$M_A = M_{H_2O} = 2(MW_H) + 1(MW_O) = 2(1.008) + 1(16.00) = 18 \frac{g}{mol}$

Air: 1 mole basis

$21\% \frac{molO_2}{mol} \ O_2 \ and \ 79\% \frac{molN_2}{mol}\ N_2$

$Moles \ O_2 = 0.21 \frac{molO_2}{mole}(1 \ mol) = 0.21 \ molO_2; Moles \ N_2 = 0.79 \frac{molN_2}{mol} (1 \ mol) = 0.79 \ molN_2$

Grams oxygen:

$0.21 \ molO_2(MW_{O_2}) = 0.21 molO_2(\frac{32 \ grams \ O_2}{mol \ O_2}) = 6.72 \ grams \ O_2$

Grams nitrogen:

$0.79 \ molN_2(MW_{N_2}) = 0.21 molO_2(\frac{ 28 \ grams \ N_2}{mol \ O_2}) = 22.12 \ grams \ N_2$

Air: $M_B = M_{air} = \frac{(6.72 + 22.12)}{1mol} = 28.8 \frac{g}{mol}$

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1} = 2[\frac{1}{18} + \frac{1}{28.8}]^{-1} = 22.2$

Need: $\sigma; \delta; \Omega_D$

Note: I will only be calculating a delta value for water because air is non-polar[1;19].

$\delta_A = \delta_{H_2O} = \frac{1.94x10^3 \mu_p^2}{V_bT_b}$

From [16]: $V_b \frac{cm^3}{mol} = 18.045 \frac{cm^3}{mol}$

From [20]: $\mu_{p_{H_2O}} = 1.855$

$T_b = 373 K$

$\delta_{A_{H_2O}} = \frac{1.94x10^3(1.855)^2}{(18.045)(373)} = \frac{6.68x10^3}{6.73x10^3} = 0.992$

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

$\sigma_A = (\frac{1.585V_b}{1 + 1.3\delta_A^2})^{1/3} = (\frac{1.585 (18.045)}{1 + 1.3(0.992)^2})^{1/3} = (12.55)^{1/3} = 2.32 \AA$

Need $T^*$ to calculate $\Omega_D$

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

$\frac{\epsilon_{AB}}{\kappa} = (\frac{\epsilon_A}{\kappa} \frac{\epsilon_B}{\kappa})^{1/2}$

Water: $\frac{\epsilon_A}{\kappa} = 1003 K; Air[1, Appendix B]:78.6 K$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{\frac{\epsilon_A}{\kappa} \frac{\epsilon_B}{\kappa}} = \sqrt{(1003 K)(78.6 K)} = 280.8 K$

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

$\frac{\epsilon_{AB}}{\kappa T} = \frac{280.8 K}{283 K} = 0.992$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.992} = 1.01$

Neufield, et al.:

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

$\Omega_D = \frac{1.06036}{1.01^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(1.01))}} + \frac{1.03587}{\exp{((1.52996)(1.01))}} + \frac{1.76474}{\exp{((3.89411)(1.01))}} =$

$\Omega_D = 1.43$

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

$\Omega_D = 1.43 + \frac{0.19(0.992)^2}{1.01} = 1.62$

Need $\sigma_{AB} = \sqrt{\sigma_A \sigma_B}$

Water: 2.32 $\AA$; Air (Appendix B[1]): 3.711 $\AA$

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

Diffusivity: Polar water in non-polar air at 10$\textdegree$C and 1 atmosphere

$D_{AB} = \frac{0.00266 T^{3/2}}{PM_{AB}^{1/2} \sigma_{AB}^2 \Omega_D} = \frac{0.00266 (283)^{3/2}}{1 (22.2)^{1/2} (2.93)^2 (1.62)} = \frac{12.66}{65.53} =$

$D_{AB} = 0.193 \frac{cm^2}{s}$

Brokaw Diffusivity of Sarin in Air at 10$\textdegree$C and 1 Atmosphere

Molecular Weight

Sarin, $C_4H_{10}FO_2P$:

$M_A = M_{C_4H_{10}FO_2P} = 4(MW_C) + 10(MW_H) + 1(MW_F) + 2(MW_O) + 1(MW_P) =$

$M_{C_4H_{10}FO_2P} = 4(12.01) + 10(1.008) + 1(19.00) + 2(16.00) + 1(30.97) = 140.1 \frac{g}{mol}$

Air: 1 mole basis

$21\% \frac{molO_2}{mol} \ and \ 79\% \frac{molN_2}{mol}$

0.21 $\frac{molO_2}{mol}$(1 mol) = 0.21 mol oxygen gas; 0.79 $\frac{molN_2}{mol}$(1 mol) = 0.79 mol nitrogen gas

Grams oxygen:

$0.21 (molO_2)(32 \frac{gO_2}{molO_2}) = 6.72 grams \ O_2$

Grams nitrogen:

$0.79 (molN_2)(28 \frac{gN_2}{molN_2}) = 22.1 grams \ N_2$

Air: $M_B = M_{air} = \frac{(6.72 + 22.12)}{1 mol} = 28.8 \frac{g}{mol}$

$M_{AB} = 2[\frac{1}{140.1} + \frac{1}{29.0}]^{-1} = 48.1$

Need: $\delta; \sigma; \Omega_D$

Note: I will only be calculating the delta value for the polar gas sarin because air is non-polar[1;19].

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

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

$\frac{\epsilon_i}{\kappa} = 1.18(1 + 1.3\delta_i^2)T_b$

Sarin: $\frac{\epsilon_A}{\kappa} = 1.18(1 + 1.3(0.418)^2)(420) = 608.2 K$

$\frac{\epsilon_{AB}}{\kappa} = (\frac{\epsilon_A}{\kappa} \frac{\epsilon_B}{\kappa})^{1/2}$

Sarin: $\frac{\epsilon_A}{\kappa} = 608.2 K$

Air[Appendix B;1]: $\frac{\epsilon_B}{\kappa} = 78.6 K$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{\frac{\epsilon_A}{\kappa} \frac{\epsilon_B}{\kappa}} = \sqrt{(608.2)(78.6)} = 216.1 K$

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

$\frac{\epsilon_{AB}}{\kappa T} = \frac{216.1}{283} = 0.764$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.764} =1.31$

$\delta_A = \frac{1.94x10^3 \mu_p^2}{V_bT_b}$

$\mu_p =$ dipole moment, debyes

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

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

Sarin[18;16a]: $\delta_A = \frac{1.94x10^3(3.44)^2}{(130.9)(420)} = 0.418$

$\sigma_i = (\frac{1.585V_b}{1 + 1.3\delta_i^2})^{1/3}$

Sarin[16a]: $\sigma_A = (\frac{1.585(130.9)}{1 + 1.3(0.418)^2})^{1/3} = 5.5 \AA$

$\sigma_{AB} = (\sigma_A \sigma_B)^{1/2}$

Sarin: $\sigma_A = 5.5 \AA$

Air[Appendix B;1}: $\sigma_B = 3.711 \AA$

$\sigma_{AB} = \sqrt{(\sigma_A)(\sigma_B)}= \sqrt{(5.5)(3.711)} = 4.52 \AA$

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

$\Omega_D = \frac{1.06036}{(1.31)^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(1.31))}} + \frac{1.03587}{\exp{((1.52996)(1.31))}} + \frac{1.76474}{\exp{((3.89411)(1.31))}} =$

$\Omega_D = 1.24$

$\Omega_D = \Omega_D(Neufield) + \frac{0.19 \delta_A^2}{T^*} = 1.24 + \frac{0.19(0.418)^2}{1.31} = 1.27$

Chapman-Enskog equation after polar correction

Diffusivity of Sarin in Air:

$D_{AB} = \frac{0.00266T^{3/2}}{PM_{AB}^{1/2}\sigma_{AB}^2 \Omega_D} = \frac{0.00266(283)^{3/2}}{1(48.1)^{1/2}(4.52)^2(1.27)} = \frac{12.66}{180.0} = 0.070 \frac{cm^2}{s}$

Diffusivity Comparison in Air: Water Versus Sarin in Descending Order

Water: $D_{AB} = 0.193 \frac{cm^2}{sec}$

Sarin: $D_{AB} = 0.070 \frac{cm^2}{sec}$

Diffusivity Ratio: $\frac{Water}{Sarin} = \frac{0.193}{0.070} = 2.74$

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. 1991 Gulf War Illnesses and Differing Hypotheses: Nerve and Brain Death Versus Stress, December 2012. gather.com[online] 2012. Available from: http://www.gather.com/viewArticle.action?articleId=281474981824775

[4] Removed

[4a] Removed

[6] National Academies Press. Institute of Medicine. Committee on Gulf War and Health: Health Effects of Serving in the Gulf War, Update 2009. Board on Health of Select Populations. Gulf War and Health, Volume 8. nap.edu[online]. 2010. pp. 320. Available from: http://www.nap.edu/catalog.php?record_id=12835 ISBN-10: 0-309-14921-5; ISBN-13: 978-0-309-14921-1

[7] Research Advisory Committee on Gulf War Veterans’ Illnesses. Gulf War Illness and Health of Gulf War Veterans. Scientific Findings and Recommendations, 2008. va.gov[online]. 2012. Available from: http://www.va.gov/RAC-GWVI/docs/Committee_Documents/GWIandHealthofGWVeterans_RAC-GWVIReport_2008.pdf

[8] Research Advisory Committee on Gulf War Veterans’ Illnesses. Research Advisory Committee on Gulf War Veterans’ Illnesses Findings and Recommendation, June 2012. va.gov[online]. 2012. Available from: http://www.va.gov/RAC-GWVI/docs/Committee_Documents/CommitteeDocJune2012.pdf

[9] Kennedy, Kelly. Study: Wind blew deadly gas to U.S. troops in Gulf War, December 2012. ustoday.com[online]. 2012. Available from: http://www.usatoday.com/story/news/world/2012/12/13/sarin-gas-gulf-war-veterans/1766835/

[9a] Haley, Robert W.; Tuite, James J. Meteorological and Intelligence Evidence of Long-Distance Transit of Chemical Weapons Fallout from Bombing Early in the 1991 Persian Gulf War, December 2012. karger.com[online]. 2012. vol. 40. pp. 160-177. Available from: http://content.karger.com/ProdukteDB/produkte.asp?Aktion=ShowFulltext&ArtikelNr=345123&Ausgabe=257603&ProduktNr=224263 DOI: 10.1159/000345123

[9b] Haley, Robert W.; Tuite, James J. Epidemiologic Evidence of Health Effects from Long-Distance Transit of Chemical Weapons Fallout from Bombing Early in the 1991 Persian Gulf War, December 2012. karger.com[online]. vol. 40. pp. 178-189. Available from: http://content.karger.com/ProdukteDB/produkte.asp?Aktion=ShowFulltext&ArtikelNr=345124&Ausgabe=257603&ProduktNr=224263 DOI: 10.1159/000345124

[10] Oswal, DP; Garrett, TL; Morris, M; Kucot, JB. Low-dose sarin exposure produces long term changes in brain neurochemistry of mice. Neurochem Res[online]. 2013. vol. 1. pp. 108-116. Available from: http://www.ncbi.nlm.nih.gov/pubmed/23054072 doi: 10.1007/s11064-012-0896-9

[11] Shewale, SV.; Anstadt, MP; Horenziak, M; Izu, B.; Morgan, EE.; Lucot, JB.; Morris, M. Sarin causes autonomic imbalance and cardiomyopathy: an important issue for military and civilian health, July 2012. J. Cardiovasc Pharmacol.[online]. 2012. vol 60(1). pp. 76-87. Available from: http://www.ncbi.nlm.nih.gov/pubmed/22549449 doi: 10.1097/FJC.0b013e3182580b75

[12] DTIC. Online Information for the Defense Community.Chan, Victor T; Soto, Armando; Wagner, Jessica A; Watts, Brandy S.; Walters, Amy D.; Hill, Tiffany M. Mechanisms of Organophosphates (OP) Injury: Sarin-Induced Hippocampal Gene Expression Changes and Pathway Perturbation, Jan 2012. dtic.mil[online]. 2012. Available from: http://www.dtic.mil/docs/citations/ADA560343

[13] Medical News Today. Low-Level Exposure to Organophosphate Pesticides Damage Brain and Nervous System, December 2012. medicalnewstoday.com[online]. 2012. Available from: http://www.medicalnewstoday.com/releases/253534.php

[14] Fulco, Carolyn E; Liverman, Catharyn T.; Sox, Harold C. National Academy Press. Committee on Health Effects Associated with Exposures During the Gulf War. Gulf War and Health: Volume 1. Depleted Uranium, Sarin, Pysidostigmine Bromide, Vaccines, 2000. Effects of Long-Term Exposure to Organophosphate Pesticides in Humans. nap.edu[online]. 2012. Available from: http://www.nap.edu/openbook.php?record_id=9953&page=R1

[15] NCBI.PubChem. Sarin-Compound Summary (CID 7871). pubmed.ncbi.nlm.nih.gov[online]. 2012. Available from: http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=7871

[16] ChemSpider. The free chemical database. Water. chemspider.com[online]. 2013. Available from: http://www.chemspider.com/Chemical-Structure.937.html?rid=01a81689-c122-434f-a0a1-b4e6e3ca8109

[16a] ChemSpider. The free chemical database. Sarin (isopropyl methylphosphonofluoridate). chemspider.com[online]. 2013. Available from: http://www.chemspider.com/Chemical-Structure.7583.html?rid=8885b92c-43db-4dbf-a9fd-280d32df0450

[17] US National Library of Medicine. WISER: Wireless Information System for Emergencey Responders. Sarin, CAS RN: 107-44-8. Volatilization. webwiser.nlm.nih.gov[online]. 2012. Available from: http://webwiser.nlm.nih.gov/getSubstanceData.do;jsessionid=E6C28B95977867F872631D36CDD61D42?substanceID=151&displaySubstanceName=Sarin&UNNAID=&STCCID=&selectedDataMenuItemID=81

[18] Lee, Ming-Tsung; Vishnyakov, Aleksey; Gor, Gennady Yo.; Neimark, Alexander V. Interactions of Phosphororganic Agents with Water and Components of Polyelectrolyte Membranes, October 2011. J. Physical Chemistry[online]. 2012. Available from: http://www.princeton.edu/~ggor/Gor_JPCB_2011.pdf

[19] Harding, Byron. 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, January 2013. Available from: https://chrisbharding.wordpress.com/2013/01/04/chapman-and-enskog-versus-hirschfelder-equation-and-compared-to-experimental-value-at-25c-and-1-atm/

[20] 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

# Temporary Divergence: Diffusivity: That Smell-Methyl Mercaptan (Methanethiol), Odorless Natural Gas, Odorless Propane, and Even Flatulence

Temporary Divergence: Diffusivity: That Smell-Methyl Mercaptan (Methanethiol), Odorless Natural Gas, Odorless Propane, and Even Flatulence

Lynyrd Skynyrd. That Smell. youtube.com[online]. 2013. Available from: http://youtu.be/ZDB-yswOrzc

Diffusivity of Methyl Mercaptan Versus Methane and Propane

Methyl mercaptan[3-6], “methanethiol”, is the byproduct of many natural processes. Flatulence is one example[7]. Because of its odor threshold, 1 ppb has been reported[4], methanethiol is also added to odorless natural gas, methane, and odorless propane for detection purposes. Apparently, it is used as a communication warning system in mining operations as well[4].

In this blog post, I will be comparing the diffusivity of the polar chemical methanethiol to the non-polar chemicals methane and propane in air. I have heard reports that the diffusivity of methanethiol is significantly greater than methane and propane. See bottom of post for diffusivities.Since reference[1] has tabular values for methane and propane in appendix B, I will use the tabular values and the Chapman-Enskog equation to calculate diffusivity values for methane and propane. For methanethiol, I will use Fuller, et al equation and tabular values for the atoms making up methanethiol, $CH_3-SH$

Chapman-Enskog Equation. From reference[1], the average absolute error of this “theoretical equation” is 7.9%

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

If $f_D$ is chosen as unity and “n” expressed by ideal-gas law

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

For Non-polar gases: Methane and Propane

Methane $CH_4$ in air at 25$\textdegree$C and 1 atmosphere (atm)

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

$M_A = M_{CH_4} = 1(MW_C) + 4(MW_H) = 1(12.01) + 4(1.008) = 16.0 \frac{g}{mol}$

$M_B = M_{air} = 29.0 \frac{g}{mole}$

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1} = 2[\frac{1}{16.0} + \frac{1}{29.0}]^{-1} = 20.6$

Need $\Omega_D$

Neufield, et al.: $\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}}$

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

From appendix B[1]

Methane: $\sigma = 3.758 \AA; \frac{\epsilon_A}{\kappa} = 148.6 K$

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

$\sigma_{AB} = \frac{\sigma_A + \sigma_B}{2} = \frac{3.758 + 3.711}{2} = 3.735 \AA$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{(148.6)(78.6)} = 108.1 K$

$\frac{\epsilon_{AB}}{\kappa T} = \frac{108.1 K}{298 K} = 0.363$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.363} = 2.76$

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

$\Omega_D =\frac{1.06036}{(2.76)^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(2.76))}} + \frac{1.03587}{\exp{((1.52996)(2.76))}} + \frac{1.76474}{\exp{((3.89411)(2.76))}} =$

$\Omega_D = 0.972$

Diffusivity of Methane in Air:

$D_{AB} = \frac{0.00266 T^{3/2}}{P M_{AB}^{1/2} \sigma_{AB}^2 \Omega_D} = \frac{0.00266 (298)^{3/2}}{1(20.6)^{1/2} (3.735)^2 (0.972)} = \frac{13.68}{61.54} = 0.222 \frac{cm^2}{s}$

Propane $CH_3CH_2CH_3$ in air at 25$\textdegree$C and 1 atmosphere (atm)

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

$M_A = M_{CH_3CH_2CH_3} = 3(MW_C) + 8(MW_H) = 3(12.01) + 8(1.008) = 44.1 \frac{g}{mol}$

$M_B = M_{air} = 29 \frac{g}{mol}$

$M_{AB} = 2[\frac{1}{44.1} + \frac{1}{29}]^{-1} = 35.0$

Need $\Omega_D$

Neufield, et al.: $\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}}$

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

From appendix B[1]

Propane: $\sigma_A = 5.118 \AA; \frac{\epsilon_A}{\kappa} = 237.1 K$

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

$\sigma_{AB} = \frac{\sigma_A + \sigma_B}{2} = \frac{5.118 + 3.711}{2} = 4.42 \AA$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{(\frac{\epsilon_A}{\kappa})(\frac{\epsilon_B}{\kappa})} = \sqrt{(237.1)(78.6)} = 136.5 K$

$\frac{\epsilon_{AB}}{\kappa T} = \frac{136.5}{298} = 0.458$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.458} = 2.18$

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

$\Omega_D = \frac{1.06036}{(2.18)^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(2.18))}} + \frac{1.03587}{\exp{((1.52996)(2.18))}} + \frac{1.76474}{\exp{((3.89411)(2.18))}} =$

$\Omega_D = 1.05$

Diffusivity of Propane in Air:

$D_{AB} = \frac{0.00266T^{3/2}}{PM_{AB}^{1/2} \sigma_{AB}^2 \Omega_D} = \frac{0.00266(298)^{3/2}}{1 (35.0)^{1/2} (4.42)^2 (1.05)} = \frac{13.68}{121.36} =$

$D_{AB} = 0.113 \frac{cm^2}{s}$

For polar molecule $CH_3-SH$, will use Fuller, et al. equation. From reference[1], the absolute relative error of this equation is 5.4%. Authors report an average absolute error of about 4% when using

Fuller, et al.

$D_{AB} = \frac{0.00143 T^{1.75}}{PM_{AB}^{1/2}[(\sum \nu)_A^{1/3} + (\sum \nu)_B^{1/3}]^2}$

T = 25$\textdegree$C = 298 K; P = 1 atm

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

$M_A = M_{CH_3SH} = 1(C) + 4(H) + 1(S) = 1(12.01) + 4(1.008) + 1(32.07) = 48.11 \frac{g}{mol}$

$M_B = M_{air} = 29 \frac{g}{mol}$

$M_{AB} = 2[\frac{1}{48.11} + \frac{1}{29}]^{-1} = 18.1$

Summation of “Atomic and Structural Diffusion Volume Increments from table 11-1[1]

$(\sum \nu)_A = (\sum \nu)_{CH_3SH} = 1 (C) + 4(H) + 1(S) = 1(15.9) + 4(2.31) + 1(22.9) = 40.04$

$(\sum \nu)_B = (\sum \nu)_{air} = 19.7$

$D_{AB} = \frac{0.00143T^{1.75}}{PM_{AB}^{1/2}[(\sum \nu)_A^{1/3} + (\sum \nu)_B^{1/3}]^2} = \frac{0.00143(298)^{1.75}}{1(18.1)^{1/2}[(40.0)_A^{1/3} + (19.7)_B^{1/3}]^2} =$

Diffusivity of methanethiol in air

$D_{AB} = \frac{30.56}{159.38} = 0.192 \frac{cm^2}{s}$

Diffusivities in air in decreasing order

Methane: $D_{AB} = 0.222 \frac{cm^2}{s}$

Methanethiol: $D_{AB} = 0.192 \frac{cm^2}{s}$

Propane: $D_{AB} = 0.113 \frac{cm^2}{s}$

At a detection threshold of 1 part per billion (ppb) and the above diffusivities, one might detect methanethiol prior to experiencing propane. In truth, there is an equation that takes “mixture” into account but I do not know the percent mixture of each component[2].

Equation for mixture

$D_{1-mixture} = \frac{1}{\frac{z_2}{D_{1-2}} + \frac{z_3}{D_{1-3}} + .... + \frac{z_n}{D_{1-n}}}$

$z_n$ is the mole fraction of component “n” in the gas mixture evaluated on a component-1-free basis

$z_2 = \frac{y_2}{y_2 + y_3 + ... + y_n}$

ihatemyhate. Friends Selection – Ross Flirts. youtube.com[online]. 2013. Available from: http://youtu.be/kH5JhYsfNMA

• Wait until 2nd attempt

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] ScienceBlogs. Molecule of the day. Methanethiol (They put that in, you know), March 2009. scienceblogs.com[online]. 2013. Available from: http://scienceblogs.com/moleculeoftheday/2009/03/18/methanethiol-they-put-that-in/

[4] Wikipedia. Methanethiol. Also known as methyl mercaptan. en.wikipedia.org[online]. 2013. Available from: http://en.wikipedia.org/wiki/Methanethiol

[5] NCBI.PubChem Substance. Methanethiol-Substance Summary (SID 3699). Also known as Methylmercaptan (CAS: 74-93-1). pubchem.ncbi.nlm.nih.gov[online]. 2013. Available from: http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=3699

[6] US National Institute of Standards and Technology. NIST. Methanethiol. webbook.nist.gov[online]. 2013. Available from: http://webbook.nist.gov/cgi/cbook.cgi?ID=74-93-1&Units=SI

[7] Wikipedia. Flatulence. en.wikipedia.org[online]. 2013. Available from: http://en.wikipedia.org/wiki/Flatulence

# 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

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

# Chapter 24: Fundamentals of Mass Transfer. Diffusivity of Water in Air at 20 Degrees Celsius and 1 Atmosphere

Diffusivity of Water in Air at 1 atmosphere and 20$\textdegree$C. A = $H_2O$ and B = Air.

Diffusivity equation. Assuming nonpolar and nonreacting molecules

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

$T = 20 + 273.15 = 293 K ; P = 1 atm ; M_A = 2 M_H+ M_O= 2(1) + 16 = 18;$

$M_B = \%O_2(M_{O_2}) + \%N_2(M_{N_2}) = 0.21(32) + 0.79(28) = 29$

Need $\delta_{AB}$ and $\Omega_D$, From Table K.2 of the appendix, values are obtained for water and Air

Water: $\delta_A, \AA: 2.649 \AA$ and $\frac{\epsilon_A}{\kappa}, K: 356 K$

Air: $\delta_B, \AA: 3.617 \AA$ and $\frac{\epsilon_A}{\kappa}, K: 97 K$

$\delta_{AB}$

$\delta_{AB} = \frac{\delta_A + \delta_B}{2} = \frac{2.649 + 3.617}{2} = 3.13 \AA$

Now I need $\Omega_D$, which can be found on Table K.1. To find $\Omega_D$, need to calculate $\frac{\kappa T}{\epsilon_{AB}}$:

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{(\frac{\epsilon_{A}}{\kappa})(\frac{\epsilon_{B}}{\kappa})} = \sqrt{(356)(97)} = 186$

T = 20 + 273 = 293 K

P = 1 atm

$\frac{\epsilon_{AB}}{\kappa T} = \frac{186}{293} = 0.634$

For Table K. 1, need the inverse

$\frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.634} = 1.58$

Must interpolate:

$\frac{y - y_0}{x-x_0} = \frac{y_1 - y_0}{x_1-x_0}$

$y = \frac{y_1 - y_0}{x_1 - x_0}(x-x_0) + y_0$

$\Omega_D = y; x = 1.58$

$y = \frac{1.167 - 1.182}{1.60 - 1.55}(1.58 - 1.55) + 1.182 = 1.173 = \Omega_D$

Now the diffusivity equation and diffusion of water vapor in air at 20$\textdegree$C and 1 atm

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

$D_{AB} = \frac{0.001858(293)^{3/2}(\frac{1}{18} + \frac{1}{29})^{1/2}}{1(3.13)^{2}(1.173)} = 0.243 \frac{cm^{2}}{s}$

Compare to the experimental value from Table J.1 of book

$D_{AB}P \frac{cm^{2}atm}{s} = 0.260 \frac{cm^{2}atm}{sec}$ at T = 298 K

Need to find the experimental diffusivity value at pressure of 1 atm and 298 K

$\frac{D_{AB}P}{P} = \frac{0.260\frac{cm^{2}P}{s}}{1 atm} = 0.260 \frac{cm^{2}}{s}$

Now, we will use an equation to predict the experimental diffusivity at 20$\textdegree$C. The following equation can be used for pressures below 25 atmospheres

$\frac{D_{AB, T_2, P_2}}{D_{AB, T_1, P_1}} = (\frac{P_1}{P_2}) (\frac{T_2}{T_1})^{3/2} \frac{\Omega_{D|T_1}}{\Omega_{D|T_2}}; T_2 = 293K; T_1 = 298 K$

Need $\Omega_{D|T_1}$

$\frac{\epsilon_{AB}}{\kappa} = 186; \frac{\epsilon_{AB}}{\kappa T} = \frac{186}{298} = 0.624; \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.624} = 1.60$

No need to interpolate Table K.1 since 1.60

$\Omega_{D|T_1} = 1.167$

Diffusion coefficient prediction

$D_{AB, T_2, P_2} = (\frac{P_1}{P_2})(\frac{T_2}{T_1})^{3/2}\frac{\Omega_{D|T_1}}{\Omega_{D|T_2}} = (\frac{1 atm}{1 atm}) (\frac{293}{298})^{3/2}(\frac{1.167}{1.173})(0.260) = 0.252 \frac{cm^2}{s}$

Percent difference

Percent Difference = $\frac{Calculated - Corrected Experimental}{Corrected Experimental} x 100 = \frac{0.243 - 0.252}{0.252} x 100 = -3.57\%$

There is nice agreement between calculated and predicted experimental water in air diffusivity values. The final diffusivity values for water in air at 20$\textdegree$C and 1 atm are

$D_{Calc} = 0.243 \frac{cm^2}{s}$ and $D_{Experimental} = 0.252 \frac{cm^2}{s}$

Summary

Nonpolar Hirschfelder, Bird, and Spotz Equation

$D_{AB} = \frac{0.001858 T^{3/2}[(\frac{1}{M_A}) + (\frac{1}{M_B})]^{1/2}}{P \sigma_{AB}^{2} \Omega_D} = 0.243 \frac{cm^{2}}{s}$

Experimental Value from Table J.1

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

Predicted Value

$D_{AB_{T_2,P_2}} = D_{AB_{T_1, P_1}} (\frac{P_1}{P_2})(\frac{T_2}{T_1})^{3/2}\frac{\Omega_{D|T_1}}{\Omega_{D|T_2}} = 0.252 \frac{cm^2}{s}$

References:

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