# 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

[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. Example 2

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

Example 2

Evaluate the diffusion coefficient of carbon dioxide in air at 20$\textdegree$C and atmospheric pressure. Compare this value with the experimental value reported in Appendix Table J.1.

Will be using the following diffusivity equation:

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

We have temperature and pressure. We can calculate the molecular weights via a periodic chart. $\delta$ and $\Omega$ can be obtained from Tables K.1 and K.2.

From K.2 of the appendix values $\delta$ and $\frac{\epsilon}{\kappa}$ are obtained:

Carbon dioxide: $\delta$ in $\AA$, 3.996 and $\frac{\epsilon_{CO_2}}{\kappa}$ in K, 190

Air: $\delta$ in $\AA$, 3.617 and $\frac{\epsilon_{N_2}}{\kappa}$ in K, 97

$\delta_{AB} = \frac{(\delta_A + \delta_B)}{2} = \frac{(3.996 \AA + 3.617 \AA)}{2} = 3.806 \AA$

$\frac{\epsilon_{AB}}{\kappa}= \sqrt{(\frac{\epsilon_{A}}{\kappa})(\frac{\epsilon_B}{\kappa}}) = \sqrt{(190)(97)} = 136$

T = 20 + 273 = 293 K

P = 1 atm

$\frac{\epsilon_{AB}}{\kappa T} = \frac{136}{293} = 0.463$

$\frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.463} = 2.16$

$\Omega_D (Table K.1) = 1.047$

This value was obtained by interpolation

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

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

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

From Table K.1

$y_i = \Omega$ and $x_i = \frac{\kappa T}{\epsilon_{AB}}$

and

$x = 2.16$

Interpolate

$y = \frac{(1.041-1.057)}{(2.20-2.10)}(2.16 - 2.10) + 1.057 = 1.047 = \Omega_D$

We have all variable except molecular weights. Considering the most prevalent gasses:

$M_{CO_2} = M_C + 2M_O = 12 + 2(16) = 12 + 32 = 44$

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

Now, we have all the information needed to calculate the diffusivity of $CO_2$ in air when using:

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

$D_{AB} = \frac{0.001858(293^{3/2})(\frac{1}{44} + \frac{1}{29})^{1/2}}{1 atm(3.806 \AA)^2 (1.047)} = 0.147 \frac{cm^2}{s}$

Now, we want to compare to the experimental value that is reported in Table J.1

$T, K = 273, D_{AB}P \frac{cm^2 atm}{s} = \frac{0.136 \frac{cm^2 atm}{s}}{1 atm} = 0.136 \frac{cm^2}{s}$

Since the value is reported at 273 K, must use a conversion equation to compare at 293 K

$\frac{D_{AB,T_1}}{D_{AB,T_2}} = (\frac{T_1}{T_2})^{3/2}(\frac{\Omega_{D,T_2}}{\Omega_{D,T_1}})$

at $T_1 = 293 K$ and $\Omega_{D, T_1} = 1.047$

at $T_2 = 273 K$ and $\Omega_{D, T_2} = ?$ from Table Table K.1

$\frac{\epsilon_AB}{\kappa}\frac{1}{T_2} = 136\frac{1}{273} = 0.498$

$\frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.498} = 2.01$

Once again, interpolation of Table K.1 is needed

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

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

$x = 2.01$

$y = \frac{1.057 -1.075}{2.10-2.00}(2.01-2.00) + 1.075 = 1.074 = \Omega_{D, T_2}$

Since we have all the values for the conversion equation

$D_{AB, 293} = (\frac{293}{273})^{3/2}(\frac{1.074}{1.047})(0.136) = 0.155 \frac{cm^2}{s}$

The diffusivity of carbon dioxide in air

Calculated: $0.147 \frac{cm^2}{s}$ and Corrected Experimental: $0.155 \frac{cm^2}{s}$

Percent Difference

$\frac{Calculated - Corrected Experimental}{Corrected Experimental} x 100= \frac{0.147 - 0.155}{0.155} x 100=5.16\%$

# 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 $N_2$

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

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

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

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