# Drop Evaporation at 10 degrees C, 1 atmosphere, and 71% relative humidity (January in Iraq and near Baghdad): Sarin (Nerve Agent) versus Water

Note: Sadly, I have noticed that the code of LaTex changes in WordPress. As an example, the text “\textdegree” use to provide the ˚ symbol but now provides “$\textdegree$“. As such, please be patient and do not blame me for all editor faults! 🙂 It truly is an experiment in progress and I am dependent upon LaTex and WordPress consistency.

Title: Drop Evaporation at 10 degrees C, 1 atmosphere, and 71% relative humidity (January in Iraq and near Baghdad): Sarin (Nerve Agent) versus Water

Conclusion: The molar flux of water is greater than sarin. As such, I assume the evaporation of water is greater than sarin. The latter is supported by a relative volatility (water:sarin) that is 12.6 at the specified conditions. Also, the boiling point of sarin is greater than water.

1991 Gulf War Illness

Before I continue, I would like the reader to know that more than 250,000 United States 1991 Gulf War veterans are suffering from 1991 Gulf War Illnesses. The illness can be psychologically and medically debilitating. For more information and to provide support, please please read the  December 2012 scientific journal articles that connect chemical weapons to potential cause of illnesses[7;8]. Also, I wrote a post about differing hypotheses and 1991 Gulf War Illness[17].

Actual mathematical properties of a potential  drop

Equation: $z = 1-\frac{1}{8}(x^2 + y^2)$

The base: $y = \sqrt{2.3^2 - x^2}$

The base radius: 2.3 millimeters; The height: 1 millimeter

Drop volume: Double integration in polar coordinates

$Volume = \iint_R z \ \mathrm{d}A = \iint_R f(x,y) \ \mathrm{d}A = \iint_R f(rcos(\theta), rsin(\theta)) \ r \ \mathrm{d}r \ \mathrm{d}\theta$

In polar coordinates

$r^2 = x^2 + y^2$

$z = 1 - \frac{1}{8}(x^2 + y^2) = 1 - \frac{1}{8}(r^2)$

$Volume =\iint_R (1 - \frac{1}{8}(x^2 + y^2)) \mathrm{d}A = \iint_R (1 - \frac{1}{8}(r^2)) \ r\mathrm{d}r \ \mathrm{d}\theta$

R is a unit disk in the xy plane and one reason I can use polar coordinates.

(i) For fixed $\theta$, r range: 0 ≤ r ≤ 2.3 millimeters

(ii) Angle range: 0 ≤ $\theta$ ≤ 2$\pi$

$Volume = \int_0^{2\pi} \int_0^{2.3} (1-\frac{1}{8}(r^2)) \ r\mathrm{d}r \ \mathrm{d}\theta$

From TI-92:

$Volume = \int_0^{2\pi} [\frac{-(r^2-8)^2}{32}]_{r=0}^{r=2.3} \ \mathrm{d}\theta = \int_0^{2\pi}(1.77) \ \mathrm{d}\theta$

$Volume = \int_0^{2\pi}(1.77) \ \mathrm{d}\theta = [1.77\theta_{\theta = final} - 1.77\theta_{\theta = initial}]_0^{2\pi} = (1.77(2\pi) - 1.77(0)) =$

$Drop \ volume = 11.12 \ mm^3$

Convert to cubic centimeters for calculations

$\frac{1 \ cm}{10 \ mm} \ and \ \frac{1^3 \ cm^3}{10^3 \ mm^3} = \frac{1 \ cm^3}{1000 \ mm^3}$

$Drop \ volume = 11.12 \ mm^3(\frac{1 \ cm^3}{1000 \ mm^3}) = 0.011 \ cm^3$

Density of fluids

Sarin[12-14]: ChemSpider: 1.07; Noblis: 1.096 at 20 deg C; WISER: 1.0887 at 25 deg C

Note: Difficult finding density data on sarin. As such, will assume the density changes little between above values and 10 deg Celsius.

Sarin average: $Density \ average =\frac{(1.07+1.096+1.0887)}{3} = 1.09 = 1.1 \ \frac{g}{cm^3}$

Water at 10 deg C[3;15]: Perry’s: 999.699; Engineering Tool Box: 999.7

Water average: $Density \ average = \frac{(999.699 + 999.7)}{2} = 999.699 \frac{kg}{m^3}$

Conversion: $(999.699 \frac{kg}{m^3}) (\frac{1 m^3}{100^3cm^3})(\frac{1000 g}{1 kg}) = 1.0 \frac{g}{cm^3}$

Water average: $Density \ average = 1.0 \ \frac{g}{cm^3}$

Evaporation mass: Drop Volume x density

Sarin: $Mass = 0.011cm^3(1.1 \frac{g}{cm^3}) = 0.012 \ grams$

Water: $Mass = 0.011cm^3(1.0 \frac{g}{cm^3}) = 0.011 \ grams$

Evaporation moles: Mass divide by molecular weight

Sarin: $Moles_{C_4H_{10}FO_2P} = \frac{0.012 \ grams}{(\frac{140.1 \ grams}{mole})} = 8.6 x 10^{-5} \ moles$

Water: $Moles_{H_2O} = \frac{0.011 \ grams}{\frac{18 \ grams}{mole}} = 6.11 x 10^{-4} \ moles$

Mass transfer: Evaporation

Sarin

The moles of sarin evaporated per square centimeter per unit time may be expressed by[1]

$N_{A,z} = \frac{cD_{AB}}{(z_2-z_1)} \frac{(y_{A1} - y_{A2})}{y_{B,lm}}$

Total molar concentration, c

$PV = nRT; c =\frac{n}{V} = \frac{P}{RT} = \frac{cm^3}{mol}$

The gas constant “R” will be calculated at standard temperature and pressure, “STP”

$Temperature = 273 K; Pressure = 1 atm; Molar \ volume: \ \frac{L}{mol} = 22.4 \frac{L}{mol}$

Conversion: $22.4 \frac{L}{mol}(\frac{1000 \ cm^3}{1\ Liters}) = 2.24x10^4 \ \frac{cm^3}{mol}$

$R = \frac{PV}{nT} = \frac{(1 \ atm)(2.24x10^4\frac{cm^3}{mol})}{273 \ K} = 82.05 \frac{atm \ cm^3}{mol \ K}$

$c = \frac{moles}{cm^3} = \frac{P}{RT} = \frac{1 \ atm}{(82.05 \frac{atm cm^3}{mol K})(283 \ K)} = 4.31x10^{-5} \ \frac{mol}{cm^3}$

Sarin diffusivity in air at 10 deg Celsius and 1 atmosphere[16]

$D_{AB} = 0.070 \frac{cm^2}{s}$

Assume the gas film

$(z_2 - z_1) = 0.5 cm$

Mole fraction Sarin

$y_{A1} = \frac{p_{A1}}{P_{total}}; y_{A2} = 0$

From[13a]:

Sarin vapor pressure:

$log \ p_A(Torr) = 9.4(\pm 0.1) - \frac{2700 (\pm 40) }{T(K)} \ from \ 0 \ to \ 147 \ deg \ C$

$log \ p_A(Torr) = 9.4 - \frac{2700}{283} = -0.1406; 10^{log \ p_A} = 10^{-0.1406} = 0.723 \ Torr$

Conversion: $0.723 \ Torr(\frac{1 \ atm}{760 \ Torr}) = 9.51x10^{-4} \ atm$

$y_{A1} = \frac{9.51x10^{-4} \ atm}{1 \ atm} = 9.51x10{-4}$

Assume no sarin in the air at a distance away from drop, $y_{A2} = 0$

For a binary system

$y_{B1} = 1 - y_{A1} = 1 - 9.51x10^{-4} = 0.9991;y_{B2} = 1 - y_{A2} = 1 - 0 = 1$

$y_{B,lm} = \frac{(y_{B2} - y_{B1})}{ln(\frac{y_{B2}}{y_{B1}})} = \frac{(1-0.9991)}{ln(\frac{1}{0.9991})} = \frac{9.0x10^{-4}}{9.52x10^{-4}} = 0.946$

The sarin flux

$N_{A,z} = \frac{cD_{AB}}{(z_2-z_1)}\frac{(y_{A1}-y_{A2})}{y_{B,lm}} = \frac{(4.31x10^{-5})(0.070)}{0.5}\frac{(9.51x10^{-4} - 0)}{0.946} =2.18x10^{-5} \frac{mol}{cm^2 \ hr}$

Water

The moles of water evaporated per square centimeter per unit time may be expressed by[1]

$N_{A,z} = \frac{cD_{AB}}{(z_{2}-z_{1})}\frac{(y_{A1}- y_{A2})}{y_{B,lm}}$

Total molar concentration, c

$PV = nRT; c = \frac{n}{V} = \frac{P}{RT} = \frac{cm^3}{mol}$

As before, the gas constant “R” will be calculated at standard temperature and pressure, “STP”

$Temperature = 273K; Pressure = 1atm; Molar \ volume= \frac{L}{mol} = 22.4\frac{L}{mol}$

Conversion: $22.4 \frac{L}{mol}(\frac{1000 \ cm^3}{1 \ Liters}) = 2.24x10^4 \frac{cm^3}{mol}$

$R = \frac{PV}{nT} = \frac{(1 \ atm)(2.24x10^4\frac{cm^3}{mol})}{273 \ K} = 82.05 \frac{atm \ cm^3}{mol \ K}$

$c = \frac{moles}{cm^3} = \frac{P}{RT} = \frac{1 \ atm}{(82.05 \frac{atm \ cm^3}{mol \ K})(283 \ K)} = 4.31x10^{-5} \ \frac{mol}{cm^3}$

Water diffusivity in air at 10 deg Celsius and 1 atmosphere[16]

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

Assume the gas film

$(z_2-z_1) = 0.5 \ cm$

Mole fraction of water

$y_{A1} = \frac{p_{A1}}{P_{total}}; y_{A2}= \frac{p_{A2}}{P_{total}}$

From[4]:

Water vapor pressure:

$log_{10} \ P_{vp} = A - \frac{B}{T + C - 273.15}$

Constants A, B, C[Appendix A;4], T in kelvins, and pressure is in bar

$log_{10} \ P_{vp} = 5.11564 - \frac{1687.537}{283+230.17-273.15} = -1.91518$

$P_{vp} = 10^{-191518} = 0.0122 \ bars$

Conversion: $\frac{1 \ atm}{1.01325 bars}(0.0122 \ bars) = 0.012 \ atm; \frac{760 \ mmHg}{1 \ atm}(0.012 \ atm) = 9.11 mmHg$

$y_{A1} = \frac{p_{A1}}{P_{total}} = \frac{0.012 \ atm}{1 atm} = 0.012$

From[2] and relative humidity of 71% (January weather in Iraq)[9]

Partial pressure of water in flowing stream

Relative humidity[2]:

$s_r(h_r) = \frac{p_{v}}{p_v^*(T)}x 100\% = 71\%$

At 283 K, previous equation gave: $p_v^* = 0.012 \ atm$

$\frac{71\%}{100}(0.012 \ atm) = p_v = p_{A2} = 0.0085 \ atm$

$y_{A2} = \frac{p_{A2}}{P_{total}} = \frac{0.0085 \ atm}{1 \ atm} = 0.0085$

For a binary system

$y_{B1} = 1 - y_{A1} = 1 - 0.012 = 0.988; y_{B2} = 1 - y_{A2} = 1 - 0.0085 = 0.992$

$y_{B,lm} = \frac{(y_{B2} - y_{B1})}{ln(\frac{y_{B2}}{y_{B1}})} = \frac{(0.992 - 0.988)}{ln(\frac{0.992}{0.988})} = 0.990$

Molar flux of water

$N_{A,z} = \frac{cD_{AB}}{(z_2-z_1)} \frac{(y_{A1} - y_{A2})}{y_{B,lm}} = \frac{(4.31x10^{-5})(0.193)}{0.5} \frac{(0.012 - 0.0085)}{0.990} = 5.88x10^{-8} \ \frac{mol}{cm^2 \ s}$

Conversion:

$N_{A,z} = 5.88x10^{-8} \frac{mol}{cm^2 \ s}\frac{3600 \ s}{1 \ hr} = 2.12x10^{-4} \ \frac{mol}{cm^2 \ hr}$

Molar Flux: Sarin versus water comparison

Sarin: $N_{A,z} = 2.18x10^{-5} \ \frac{mol}{cm^2 \ hr}$

Water: $N_{A,z} = 2.12x10^{-4} \ \frac{mol}{cm^2 \ hr}$

Ratio: $\frac{Water}{Sarin} = \frac{2.12x10^{-4}}{2.18x10^{-5}} = 9.71$

Although the above is a simple evaluation based on “diffusion through a stagnant gas film”[1] and not the most rigorous, the ratio makes since because the ratio of vapor pressures at 10 deg Celsius, “relative volatility”[18], is

$\alpha_{water-sarin} = \frac{p_{H_2O}}{p_{C_4H_{10}FO_2P}} = \frac{0.012 \ atm}{9.51x10^{-4} \ atm} = 12.6$

Per US Department of Energy[19]

“The evaporation of a liquid depends upon its vapor pressure — the higher the vapor pressure at a given temperature the faster the evaporation — other condition being equal.

The higher/lower the boiling point the less/more readily will a liquid evaporate.”[19]

The boiling points are:

Sarin[14]: 147 deg Celsius; Water[15a]: 100 deg Celsius

Conclusion:

The evaporation of water is greater than the evaporation of sarin.

References:

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

[2] Felder, Richard M; Rousseau, Ronald W. (1986) Elementary Principles of Chemical Processes, Second Edition. New York: John Wiley & Sons.

[3] Perry, Robert H; Green, Don W. (1997) Perry’s Chemical Engineers’ Handbook, Seventh Edition. New York. McGraw-Hill.

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

[5] Anton, Howard. Calculus with Analytic Geometry, Fifth Edition. New York: John Wiley & Sons.

[6] Barker, William H; Ward, James E. (1995) The Calculus Companion. Calculus: Howard Anton, Fifth Edition.

[7] 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

[8] 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] Harding, Byron. Diffusivity of Water versus Sarin (Nerve Agent) in Air at 10 Degrees Celsius (50 Degrees Fahrenheit) and 1 Atmosphere, January 2013. chrisbharding.wordpress.com[online]. 2013. Available from: https://chrisbharding.wordpress.com/2013/01/07/test/

[11] Removed

[12] ChemSpider. The free chemical database. Sarin. chemspider.com[online]. 2013. Available from: http://www.chemspider.com/Chemical-Structure.7583.html

[13] Noblis. Chemistry of GB (Sarin). noblis.org[online]. 2013. Available from: http://www.noblis.org/MissionAreas/nsi/ChemistryofLethalChemicalWarfareAgents/Pages/Sarin.aspx

[13a] Noblis. Parameters for Evaluation of the Fate, Transport, and Environmental Impacts of Chemical Agents in Marine Environments. noblis.org[online]. 2012. Available from: http://pubs.acs.org/doi/pdf/10.1021/cr0780098

[14] Wireless Information System for Emergency Responders. WISER. Sarin, CAS RN: 107-44-8. webwiser.nlm.nih.gov[online]. 2013. Available from: http://webwiser.nlm.nih.gov/getSubstanceData.do?substanceID=151&displaySubstanceName=Sarin&UNNAID=&STCCID=&selectedDataMenuItemID=30

[15] The Engineering ToolBox. Water-Density and Specific Weight. engineeringtoolbox.com[online]. 2013. Available from: http://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html

[15a] The Engineering Toolbox. engineeringtoolbox.com[online]. 2013. Available from: http://www.engineeringtoolbox.com/

[16] Harding, Byron. Diffusivity of Water versus Sarin (Nerve Agent) in Air at 10 Degrees Celsius (50 Degrees Fahrenheit) and 1 Atmosphere, January 2013. chrisbharding.wordpress.com[online]. 2013. Available from: https://chrisbharding.wordpress.com/2013/01/07/test/

[17] 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

[18] Chopey, Nicholas P. (1994). Handbook of Chemical Engineering Calculations, Second Edition. Boston Massachusetts: Mc Graw Hill.

[19] US Department of Energy. Newton: Ask A Scientist.Evaporation and Vapor Pressure. newton.dep.anl.gov[online]. 2012. Available from: http://www.newton.dep.anl.gov/askasci/phy00/phy00130.htm

# 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

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[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

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