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 ňö symbol but now provides “\textdegree“. As such, please be patient and do not blame me for all editor faults! ūüôā 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 ‚ȧ r ‚ȧ 2.3 millimeters

(ii) Angle range: 0 ‚ȧ \theta ‚ȧ 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

[9] Wikipedia. Baghdad. en.wikipedia.org[online]. 2013. Available from: http://en.wikipedia.org/wiki/Baghdad#Geography_and_climate

[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

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\textdegreeC 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\%