# Numerical Method Reviews

Hello,

This note will be a collection of my personally created numerical method reviews. I will be reviewing topics from two books[B1;B2].I have uploaded my PDFs to scribd because I have had hit and miss experience with LaTeX in WordPress. Also, the PDFs can be accessed, downloaded, and printed easily.

I have chosen books [B2; B5] because they are used in courses at MIT that are freely available online. Since I am using Octave, I found a great book for that environment too[B1]. I plan to study both books.

Purpose

Most books have an initial chapter that discusses pertinent topics to be successful for the rest of the book. The topics are briefly discussed, and the authors [B1;B2] really expect the reader to have full knowledge or gain full knowledge of the material before proceeding.

I am entering a learning phase and a review phase. I am reviewing calculus [B3] and algebra[B4], but I am learning linear algebra[B5]. This note will be a place that I categorize my reviews.

I am writing these reviews with LaTeX, so that takes time as well. Still, LaTeX lets me easily write in the language of mathematics once I learn it well.

Numerical Methods Using MATLAB Reviews[B2]

Quote: It is assumed that the reader is familiar with the notation and subject matter covered in undergraduate calculus sequence. This should have included topics of limits, continuity, differentiation, integration, sequences, and series. Throughout the book we refer to the following results.”[B2]

Quote: “Example 1 is about as easy a limit proof can get; most limit proofs require a little more algebraic and logical ingenuity. (sic: read basic knowledge should be well understood.) The reader who finds “δ – ε ” discussions hard going should not become discouraged; the concepts and techniques are intrinsically difficult. In fact, a precise understanding of limits evaded the finest mathematical minds for centuries.”[B3]”[B2,1]

Reviews For Numerical Methods Using MATLAB[B2]:

[B2,1] Harding, Chris. (2015, Dec. 21). Review of Limits. Retrieved (2015, Dec. 21). Available from: https://www.scribd.com/doc/293834966/Review-of-Limits

Scientific Computing With MATLAB And Octave[B1]

Quote: “In this book we will systematically use elementary mathematical concepts which the reader should know already, yet he or she might not recall them immediately.

We will therefore use this chapter to refresh them and we will condense notions which are typical of courses in Calculus, Linear Algebra and Geometry, yet rephrasing them in a way that is suitable for use in Scientific Computing.”[B1]

Reviews For Scientific Computing with MATLAB and Octave[B1]:

[B1,1] Harding, Chris. (2015, Dec. 26). Chapter 1, What Can’t Be Ignored: Section 1.1 To Addition And Multiplication Review Of Matrices.(Octave) Retrieved (2015, Dec. 26). Scribd[online]. Available from: https://www.scribd.com/doc/294069551/Chapter-1-What-Can-t-Be-Ignored-1-Section-1-1-To-Addition-And-Multiplication-Review-Of-Matrices

Books:

[B1] Quarteroni, Alfio; Saleri, Fausto; Gervasio, Paola. Scientific Computing with MATLAB and Octave. Fourth Edition. Retrieved (2015, Dec. 18). Shannon[online]

[B2] Mathews, John H.; Fink, Kurtis, D. (1999). Numerical Methods Using Matlab. (Third Edition). Prentice Hall

[B3] Anton, Howard. Calculus with Analytic Geometry. Fifth Edition.

[B3a] Barker, William H.; Ward, James E. (1995). The Calculus Companion to Accompany Calculus with Analytic Geometry. Fifth Edition by Howard Anton. John Wiley

[B4] Grossman, Stanley I. (1992) Algebra and Trigonometry, Second Edition. Saunders College

[B5] Strang, Gilbert. (2009). Introduction to Linear Algebra. South Asian Edition. Cambridge University Press India Pvt. Ltd.

# Limits Review: A Rigorous Approach

Hello,

Dr. Anton–Calculus Textbook–Quote:
“Example 1 is about as easy a limit proof can get; most limit proofs require a little more algebraic and logical ingenuity. (sic: read basic knowledge should be well understood.) The reader who finds “δ – ε” discussions hard going should not become discouraged; the concepts and techniques are intrinsically difficult. In fact, a precise understanding of limits evaded the finest mathematical minds for centuries.”[3]”[1]

Purpose

I have hopes to learn some Numerical Methods[2] to allow me to contribute to PLOTS (http://publiclab.org). Numerical methods are  often used by engineers to mathematically solve problems that do not have “exact” solutions. In truth, they can be used to solve problems that have exact solutions where the solution is quite difficult as well.

Also, the studying of math and engineering help me mentally by providing the psychotherapy methods of mindfulness and compensatory cognitive training for my schizoaffective disorder (bipolar type).

Here, I have provided my first LaTeX created PDF on this topic– LaTeX is great for writing mathematics and is free–that reviews limits and absolute values. I am writing such documents because it is likely that I will lose focus or be hospitalized since that has happened in the past. I can then read a document I made to provide an understanding to move forward. In this case, one of my numerical methods Books started with a vague and abstract definition of the rigorous limits approach[1;2], and I wanted a better understanding[1].

I plan to do similar PDFs for other “review” topics in Section 1.1[2]. By the way, I picked[2] because MIT has a free course that uses the same textbook.

In my reference[1], I also provide a great book that covers the use of Octave and MATLAB when doing numerical methods.

Great Video and Special Note

This MIT video on the Rigorous approach has great information of absolute values, etc. In fact, I suggest the reader pay particular attention, within the video[3], of the power of a simple absolute value property of “Triangle inequality”

$|a + b| \leq |a| + |b|$

that is discussed during the problem solving near video time of 35:12 to end at 46 minutes.

The professor is being kind. To be good, one must have an amazing memory and an amazing grasp of the fundamentals of mathematics. Really, I believe one should obviously begin this grasp in each math class, but, if one is intelligent enough to grasp future fundamentals ahead of the curve, then he or she can go back and review past sections as they pertain to the current set of problems. I am not that person, but hope to gain enough of a “working” knowledge that I can be competent and useful in the world of PLOTS. With that said, all the video[Embedded; 3] is important and a good test of one’s knowledge on this topic[1].

References:

[1] Harding, Chris. (2015, Dec. 21). Review of Limits. Retrieved (2015, Dec. 21). Available from: https://www.scribd.com/doc/293834966/Review-of-Limits; Google URL: goo.gl/vd5XAe

[2] Mathews, John H.; Fink, Kurtis, D. (1999). Numerical Methods Using Matlab. (Third Edition). Prentice Hall

[3] MIT OpenCourseWare. (2011, May 05). Unit I: Lec 5 | MIT Calculus Revisted: Single Variable Calculus, Unit !: Lecture 5: A More Rigorous Approach to Limits. Retrieved (2015, Dec. 28). youtube[online]. Available from: https://youtu.be/9tYUmwvLyIA?list=PLLEGnvlQMshG_TaCu75GpVOeXHiVe0JP2

# An investment opportunity : For those interested in OxiGene (OXGN)–a developer of novel anti-cancer drugs

Disclosure

First, I own OxiGene (OXGN) stock. Also, I have two STEM degrees: chemical engineering and biological sciences, so I have some understanding of the science behind OxiGene drugs. As such, I am not an expert, but I can provide a somewhat educated evaluation.

History

OxiGene once traded at near $8,000 per share,$5,000 per share, $3,000 per share, and from$1,000 per share to the current, December 26, 2013, near \$2.60 per share[6]. It is a small cap company with no current product on the market[3;7]. As such, its stock dropped over the years because of the recession and negative Earnings Per Share, EPS. What should be noted is that small-cap stocks are the usual winners of a recession recovery[4]. Also, the splits make the share price less.

Clinical Phases, Orphan Drugs, and Pipeline

With that said, Zybrestat, its leading drug, was in phase 3 for Anaplastic Thyroid Cancer (ATC)[12] and designated an orphan drug by the United States (US)[14;15], but, according to information from an OxiGene source, Zybrestat will not currently enter continued phase III trials for  ATC in the USA because of funding. As such, I assume OxiGene will not be marketing Zybrestat for ATC in the USA. Still, according to FDA website, OxiGene’s combretastatin drugs have been designated Orphan drugs for “Treatment of anaplastic thyroid cancer, medullary thyroid cancer, and stage IV papillary or follicular thyroid cancer”, “Treatment of ovarian cancer”, and “Treatment of acute myelogenous leukemia.”[16, search: combretastatin]. US Orphan drug status guarantees OxiGene exclusive marketing rights for seven years and various significant tax breaks[15]. According to the US Food and Drug Administration, Phase 3 can take from 1 to 4 years[13]. At the OxiGene website, a graphic indicates that the Zybrestat Phase 3 for ATC is half way completed, but I received a phone call from an OxiGene representative that said phase 3 for ATC will not continue in the USA due to FDA requirements for a large population of patients, near  300, in a future phase III trial. The FDA required number of subjects is, at this time, to costly for OxiGene. The representative did say that OxiGene is still planning to continue USA clinical trials for ovarian cancer. Once Zybrestat is approved by the FDA, assuming phase 3 trials continue to go well[5], a physician, oncologist as an example, can prescribe Zybrestat for other cancers, which is called “off-label” use[2], in ethical situations. In truth, I assume FDA mandates on ATC clinical trials could be an indication of future FDA mandate on ovarian cancer trials. OxiGene is also expecting early marketing of Zybrestat in the European Union, EU. Also, Zybrestat has been designated an EU orphan drug for Anaplastic Thyroid Cancer, ATC, and ovarian cancer. The OxiGene representative that contacted me also said OxiGene is still planning to market Zybrestat in the European Union once approved. Eventually, I have hope the stock will stabilize at a new price significantly greater than where I bought it.

As an example, current and past phase I thru III drug trials of OxiGene drugs, Zybrestat and OXi4503, show promise with Anaplastic Thyroid Cancer (ATC)[12], Ovarian Cancer[12], Acute Myeloid Leukemia (AML)[10;11], myelodysplastic syndrome (MDS)[10;11], and other abnormal vascular diseases such as macular degeneration[12]. Zybrestat has shown success against Pancreatic Neuroendocrine Tumor Model as well[9], and Zybrestat has had mixed results with non-small cell lung cancer[12]. This is important because investment companies look at “how many” people a drug will serve to earn profits per share. Remember, a physician can prescribe for “off-label use,” but the additional prescriptions might or might not affect the overall share price. Finally, Zybrestat has a good safety profile so far[5]. According to reference [5], Zybretsat did show some myocardial ischemia, but the OxiGene representative told me the ischemia problem has been eliminated by prescribing a drug that helps with hypertension. As the reader may know, the US FDA denies the approval of many drugs because of heart issues, but I cannot definitively say this is why the US FDA required near 300 subjects for a Zybrestat phase III trial for anaplastic thyroid cancer, ATC. As mentioned previously, an OxiGene representative told me that OxiGene will not currently be completing ATC phase 3 clinical trials in the USA because of funding.

Research Pipeline

OxIGene also has an effective research pipeline. One of OxiGene’s drugs in the research pipeline is benzosuberenes[1], which have shown success for prostate, ovarian, and non-small cell lung cancers[1] in the pico-molar range. OxiGene owns the rights to all of Baylor Universities studies on benzosuberenes. If you notice, the Division of Cancer Treatment and Diagnosis, National Cancer Institute at Frederick, National Institutes of Health, Frederick, Maryland, United States was apart of the following journal article too[1]. In truth, I do not believe the US Government wants OxiGene to fail. I believe the science behind OxiGene’s drugs is strong[1;3;5;7;8;9;10;11;12]. For a nice video of Zybrestat in action within a biologically active cancer tumor[8].

Low Stock Price

Although OxiGene has performed a reverse stock-split, which indicates a distressed company that is possibly facing NASDAQ delisting if the stock does not increase, I believe the strong science and importance of cancer therapy make the risky company a good buy. Still, I am not a financial analyst. With that said, I would ask a potential financial analyst to strongly consider the science behind OxiGene’s drugs when making a decision.

References:

[1] Tanpure, Rajendra P; George, Clinton S; Sriram, Madhavi; Strecker, Tracy E; Tidmore, Justin K; Hamel, Ernest; Charlton-Sevcik, Amanda K; Chaplin, David J; Trawick, Mary Lynn; Pinney, Kevin G. An Amino-Benzosuberene Analogue That Inhibits Tubulin Assembly and Demonstrates Remarkable Cytotoxicity, Jun 2012. Medchemcomm[online]. 2012. vol. 3(6). pp. 720-724. Available from: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3682772/ PMCID: PMC3682772; NIHMSID: NIHMS386131; doi: 10.1039/C2MD00318J

[2] Cancer.org. Off-label Drug Use: What is off-label drug use? cancer.org[online]. 2013. Available From: http://www.cancer.org/treatment/treatmentsandsideeffects/treatmenttypes/chemotherapy/off-label-drug-use

[3] Murphy, Bryan. Yes, This Time Really is Different for OxiGENE (OXGN), Nov 6, 2013. smallcapnetwork.com[online]. 2013. Available from: http://www.smallcapnetwork.com/Yes-This-Time-Really-is-Different-for-OXiGENE-OXGN/s/via/1789/article/view/p/mid/3/id/454/

[4] Consumer Reports Money Advisor. Post-recession investing: No matter which kind of economic recovery we get, diversification still wins, Oct 2009. consumerreports.org[online]. 2013. Available from: http://www.consumerreports.org/cro/money/personal-investing/post-recession-investing/overview/index.htm

[5] Kanthou, Chryso; Tozer, Gillian M. Microtubule depolymerizing vascular disrupting agents: novel therapeutic agents for oncology and other pathologies, Jun 2009. Int J Exp Pathol[online]. 2009. vol. 90(3). pp. 284-294. Available from: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2697551/ PMCID: PMC2697551; doi: 10.1111/j.1365-2613.2009.00651.x

[6] NASDAQ. OxiGene, Inc (OXGN) Interactive Chart Analysis. nasdaq.com[online]. 2013. Available from: http://www.nasdaq.com/symbol/oxgn/interactive-chart

[7] Oxigene. Clinical Trails. oxigene.com[online]. 2013. Available from: http://www.oxigene.com/pipeline/clinical_trials/

[8] OxiGene. How Our Drugs Work. How do ZYBRESTAT (fosbretabulin) and OXI4503 work? oxigene.com[online]. 2013. Available from: http://oxigene.com/our_science/how_our_drugs_work/

[9] Wall Street Journal. OxiGene Presents Preclinical Data Demonstrating Significant Antitumor Activity of Zybrestat ® in Pancreatic Neuroendocrine Tumor Model, Oct. 21, 2013. online.wsj.com[onlne]. 2013. Available from: http://online.wsj.com/article/PR-CO-20131021-905345.html

[10] Turner, Daniel BS; Gonzalez, Andres BS; Pettiford, Leslie RN; Meacham, Amy MS; Wise, Elizabeth BS; Bosse, Raphael C BS; Chaplin, Dai PhD; Hsu, Jack W. MD; Brown, Randy A. MD; Hiemenz, John W. MD; Norkin, Maxim MD, PhD; Wingard, John R. MD; Cogle, Christopher R. MD. 1463 A Phase I Study Of The Vascular Disrupting Combretastatin, OXi4503, In Patients With Relapsed and Refractory Acute Myeloid Leukemia (AML) and Myelodysplastic Syndromes (MDS), Dec 7, 2013. American Society of Hematology[online]. 2013. Available from: https://ash.confex.com/ash/2013/webprogram/Paper65534.html

[11] OxiGene. Pipeline. OXi4503 in Onocology. oxigene.com[online]. 2013. Available from: http://www.oxigene.com/pipeline/OXi4503_in_oncology/

[12] Oxigene. Pipeline. oxigene.com[online]. 2013. Available from: http://www.oxigene.com/pipeline/

[13] US Food and Drug Administration. Step 3 Clinical Research. patientnetwork.fda.gov[online]. 2013. Available from: http://www.patientnetwork.fda.gov/learn-how-drugs-devices-get-approved/drug-development-process/step-3-clinical-research

[14] Landes Bioscience. Oxigene breathes life into cancer treatments. Cancer Biology & Therapy[online]. vol 6(8). Available from: https://www.landesbioscience.com/journals/cbt/article/5015/

[16] US Food and Drug Administration. Search Orphan Drug Designations and Approvals. accessdata.fda.gov[online]. 2013. Available from: http://www.accessdata.fda.gov/scripts/opdlisting/oopd/

# 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