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## Are unrelated AP classes worth it?

Hey, I'm a high school student going into my junior year and I've been wondering whether it's in my best interest to take AP courses unrelated to the major I want to go into.

Basically, at this point I'm quite sure that I want to major in either Electrical Engineering or Computer Science, and I'd like to get into a nice college. My school doesn't offer a whole lot of choice when it comes to AP classes related to math and science. In fact, the only math/science AP classes my school currently offers are AP Calc AB and AP Stats. I'm definitely going to be taking AP Calc AB this year and I'm sure I will enjoy it. I only wish my school offered AP Physics and the likes...

The issue I'm having is that I'm trying to decide whether or not it's worth it for me to take classes like AP English Literature & AP US History. I already took AP World History my sophomore year. Will these classes affect my chances much of getting into a good school for EE or CS? I know quite well that I can take these classes and get relatively good grades, but I also know quite well that I will hate myself for it because I really won't enjoy it. I'd much rather be able to enjoy the year rather then dread it, if it's possible.

Taking AP English Lit & AP US will definitely take up a lot of the free time that I spend working on stuff like programming and toying with electronics and what not. The kind of stuff that I enjoy more than anything else.

Also, my school is a "magnet" school and I'm in the magnet known as "ECC" (Electronics and Computer Circuitry) which is basically a mash up of some EE, some CS, and other computer related topics. I get high grades in just about all of my classes, especially those math/science related.

Is it worth it?
 PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract
 Mentor In college / university, you'll probably have to take non-science courses like English and history as part of general degree requirements. You can often use AP courses to satisfy those requirements. Some people argue that it's better to use AP courses to skip non-major courses, than to skip intro courses in your major.
 Like jtbell says, you'll need to take the classes eventually. Whether you want to get them over with or procrastinate until later is up to you.

## Are unrelated AP classes worth it?

You guys do make a good point - Although I was more curious about how it would affect my admission to colleges.
 Probably minimally. No one cares either way whether you took AP English or AP US History, or any other AP, provided you've demonstrated an interest in some field of study. However, if this is your first US history course in high school, you might be missing some basic requirements for some college applications. I can't imagine that you haven't taken a previous English course in high school.

 Quote by snipez90 Probably minimally. No one cares either way whether you took AP English or AP US History, or any other AP, provided you've demonstrated an interest in some field of study. However, if this is your first US history course in high school, you might be missing some basic requirements for some college applications. I can't imagine that you haven't taken a previous English course in high school.
Oh of course I've been taking English and History classes. I'm required to take both subjects for all four years. I took AP World History last year, and while I did well, it wasn't very enjoyable for me haha. I'm just trying to decide if it's worth it to take AP English classes and another AP History course, rather then the usual honors classes. I like to use every last ounce of my free time to the fullest extent, and classes like those take a lot of it away (Note: I do understand that when I'm in college I probably won't have any free time at all - But, since I'll actually be spending that time learning stuff rather then preparing for standardized tests - which is about all high school really is - I'm sure I'll enjoy it.)
 Recognitions: Gold Member Science Advisor Yeah I had a similar question. Do I still have to take liberal arts courses in college even if I know that my major is going to be in the theoretical physics area going into college? It seems like my finishing of APUSH and AP World has gone to waste then.
 Learning how to read, analyze large amounts of information, and write is never a waste unless you plan on financing all of your academic research by yourself, in which case it probably wouldn't matter.

 Quote by snipez90 Probably minimally. No one cares either way whether you took AP English or AP US History, or any other AP, provided you've demonstrated an interest in some field of study. However, if this is your first US history course in high school, you might be missing some basic requirements for some college applications. I can't imagine that you haven't taken a previous English course in high school.
I disagree. It looks good on application when you mention that you have attempted AP classes. In my experience, it is better to have had a 3.5 with AP credits than a 4.0 without it. It can demonstrate that the student in question is capable of buckling down under an college level academic workload and get it done. However, this is just my perspective as a student.

 Quote by WannabeNewton Yeah I had a similar question. Do I still have to take liberal arts courses in college even if I know that my major is going to be in the theoretical physics area going into college? It seems like my finishing of APUSH and AP World has gone to waste then.
I had to sign-up just to answer this question. It would serve you well to take a look at potential school catalogs to see what the required classes are to graduate in, presumably, a B.S. in physics. I considered a B.S. in physics, however switched to engineering upon seeing the requirements. Since its a degree from the school of arts and sciences of many schools, you'll have to complete their requirements for the school of A&S that includes liberal arts classes and languages.

I would recommend taking AP classes. It feels good to have extra credit, also it will save you money. Why pay later what you can do now for free?

 Similar discussions for: Are unrelated AP classes worth it? Thread Forum Replies Engineering, Comp Sci, & Technology Homework 1 Academic Guidance 5 Academic Guidance 3 Science Comics 0 General Physics 7


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Tip: FAA AIP, AIM, ATC (point sixty five), are back on HTML

I previously suggested using a Google search limited to site:avstop.com/ac when the HTML pages were taken down, but due to avstop not being up-to-date, I deleted the post. As I learned from @voretaq7, the older HTML pages were taken down to comply with accessibility.

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# Discussion about Hairer's existence theory for stochastic differential equations

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The discussion started below this submission creation request, but it is better deserves a separate chat thread.

asked Aug 15, 2014 in Chat

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http://arxiv.org/abs/1303.5113 also http://arxiv.org/abs/1401.3014 , there are also great later articles by Hairer citing these. These are, to my mind, the biggest step in mathematical quantum field theory since Wilson.

answered Aug 15, 2014 by (7,550 points)

The first one is a book-sized paper. Could you please point to a particular result relevant for quantum YM? (The word ''Yang'' is in none of the two papers.)

The first (2013) work says on p.4 in Remark 1.1 that they obtain a nonperturbative Euclidean equivalent of superrenormalizable QFT. Thus in 4D this is relevant at best to scalar QED, which is (apart from its relatives) the only superrenormalizable theory in 4D.

It won't contribute to the solution of the YM problem.

''The details of the method requires superrenormalizability for the time being, but this is just a technical limitation'' The same was claimed for the methods of constructive field theory, which could construct $\Phi^4_3$ long, long ago but then got stuck. The technical limitations are enormous! I believe true progress only when someone actually overcomes that barrier.

Nevertheless, the treatise is good, and valuable for the analysis of true SDE (rather than QFTs).

@RonMaimon, for the record Hairer just got Fields prize yesterday for this regularity stuff.

Oh wow! He got recognized so quickly? Great! congrats to him.

But this is Euclidean $\Phi^4_3$, which is unrelated to its Minkowski version unless $O(4)$ invariance and reflection positivity are shown.

O(4) invariance is really automatic, as it is present in O(4) invariant regulators (his regulator is a smooth function convolved with the noise, you can make it a rotationally invariant bump, he always does in his specific examples anyway), and the convergence was established to be  independent of regulator (this is one of the great advances of his formalism, the form of the regulator clearly doesn't matter in taking the limit). The thing that he constructs is an SPDE solution, with a good (distributional) continuum limit, whose long-time solution, if you take a constant time slice, has statistics that eventually converges to a random pick from $\phi^4_3$ path integral. These are already known to be reflection positive from other work, and it should be easy to show it directly, you asked about it, let an answer come. I believe (not sure, haven't thought at all) that with appropriate reflection positive statistics on the initial conditions, it is extremely easy to show reflection positivity  for all time slices and for all regulators, not just asymptotically at the large t limit.

There is no way reflection positivity is going to stop the program (unlike, say, the superrenormalizability business which requires a real idea advancing this stuff). The program constructs the stochastic quantized field theory in what can only be called 'the right way'.

Any difficulty in turning the SPDE construction into a field theory construction can only lie in establishing that the long-time limit is converging (but it is true that the long time limit will be converging to the appropriate thing). You could, for example, define the SPDE for $\phi^3$ field theory in 4d, and it should be ok as an SPDE, the theory is superrenormalizable, but it would not converge at large times to anything, the solutions would run away to $\phi\rightarrow -\infty$.

It might also have a finite time limit to the SPDE evolution, there are no results that guarantee global solutions, as sometimes they won't exist. The results he proves show local existence of SPDE solutions, and you need a further estimate for global solutions. This one of the places where he focuses his current research, connecting the SPDE to the stationary distribution at long times. Again, he is interested in SPDEs, but the formalism is general enough to apply to any (bosonic/real-Euclidean-action) QFT.

I don't claim that Hairer solved all the problems of constructing QFTs at one stroke, he just solved the main one--- defining a rigorous renormalization procedure that is completely non-perturbative, easy to work with and prove things about, is general enough that it should work precisely in those cases when traditional renormalization does, and corresponds exactly to what physicists know about renormalization (it's essentially a framework for producing convergence from a rigorous version of the OPE--- the OPE defines the possible renormalization terms in products of regularized distributions, and then the analysis parts just show the statistical convergence when the regulator is relaxed).

Because he is doing stochastic quantization, his renormalization for the stochastic time evolution is completely separately from constructing a vacuum state, and it works even when there is no vacuum state. The two problems are decoupled, and this allows you to think about short-distance renormalization without considering long-distance properties, like confinement. This separation property is central, it's what allowed the Glimm/Jaffe program to succeed too, but there the separation was intrinsically due to the superrenormalizability, it wasn't something that could be generalized. Here it can be generalized to any nonlinear SPDE even stochastically quantized Yang-Mills theory.

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## shaqadry Group Title how to differentiate f(x) = ln ( 1 + √ x / 1 - √ x ) one year ago one year ago

1. Jemurray3 Group Title

In general, $\frac{d}{dx} \ln( f(x) ) = \frac{1}{f(x)}\cdot f'(x)$

how do i solve it? i cant seem to get the correct answer.

3. Ahaanomegas Group Title

Hint: You will have to use the chain rule by finding the derivative of the argument and multiplying it by the derivative that results without the chain rule. Are you aware of how to use the Chain Rule? If not, then I'll provide a solution.

im aware of it. but how do i calculate it?

5. Ahaanomegas Group Title

The derivative of the argument, you mean?

how do i apply chain rule in this equation

7. ByteMe Group Title

Rewrite f(x): $$\Large f(x)=ln\frac{1+\sqrt x}{1-\sqrt x}=ln(1+\sqrt x)-ln(1-\sqrt x)$$ now just take the derivatives of each ln separately....

8. Ahaanomegas Group Title

Would you mind showing us your work? What have you found for the derivative of the argument of the natural log?


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Hugo Stokes

2022-10-21

Why the relationship ${v}_{p}{v}_{g}={c}^{2}$ (phase velocity multiply group velocity equals lightspeed squared) exists in waveguide structures?

Do you have a similar question?

silenthunter440

Expert

This fact exists in classical wave theory. The phase velocity is ${v}_{p}=\omega /\beta$, whereas the group velocity is ${v}_{g}=d\omega /d\beta$ (i.e. velocity of a narrow band signal, calculated from the dispersion relation $\omega =\omega \left(\beta \right)$ for your waveguide), where $\omega$ is the angular frequency of the propagating wave and $\beta$ is its (guided) propagation constant (same as the wavenumber, k, in lossless guides). Now, in a typical waveguide, you have $\beta$ derived (say as eigenvalue of the Helmholtz wave equation and given boundary conditions) to be $\beta =\sqrt{{k}_{0}^{2}-{k}_{c}^{2}}$, where ${k}_{0}=\omega /c$ is the free-space wavenumber (if waveguide weren't there) and ${k}_{c}^{2}$ is a geomery-based cutoff wavenumber (constant), calculated based on the particular boundary conditions of the structure. Therefore, you can see that
$\frac{d\beta }{d\omega }=\frac{2\omega /{c}^{2}}{2\sqrt{\left(\omega /c{\right)}^{2}-{k}_{0}^{2}}}=\frac{\omega }{{c}^{2}\beta }⇒{v}_{p}{v}_{g}=\frac{\omega }{\beta }\frac{{c}^{2}\beta }{\omega }={c}^{2}.$
As energy propagates down the waveguide at the group velocity, which may be slower than light in such structures, you can now see easily why the phase velocity becomes higher than the speed of light (${v}_{g}\le c\le {v}_{p}$, with c as their geometric mean).

Still Have Questions?

Free Math Solver


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# M10: 31

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26 Sep 2009, 19:20
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?

1) The base of the parallelogram is 10.
2) One of the angles of the parallelogram is 45 degrees.

Why C and not E?

Am struggling to understand how we can use the pythagorean theorem when we don't know which pair of angles are 45 degrees each in the parallelogram. What am I missing here? Would appreciate some help here.
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27 Sep 2009, 08:24
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27 Sep 2009, 08:44
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KUDOS
sid3699 wrote:
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?

1) The base of the parallelogram is 10.
2) One of the angles of the parallelogram is 45 degrees.

Why C and not E?

Am struggling to understand how we can use the pythagorean theorem when we don't know which pair of angles are 45 degrees each in the parallelogram. What am I missing here? Would appreciate some help here.

It is not necessary to know which pair of angles is 45 degrees. I have attached an image for your reference.
Attachment:

Parallelogram.jpg [ 14.46 KiB | Viewed 1612 times ]

Please let me know if you want me to explain further
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27 Sep 2009, 09:26
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Hi,

I've drawn a picture to be able to explain the problem better. S1 tells us that AD in the picture below equals 10. If the area of the parallelogram is 100, than its height BD equals $$\frac{100}{10} = 10$$. Now, S2 tells us that one of the angles is 45 degrees. We marked $$\angle BAD$$ as a 45 degrees angle. So, we have an isosceles triangle ABD with BD equal to AD. We know that angle ADB is the right angle and can find AB using the Pythagorean theorem. It will equal $$\sqrt{20}$$, but we don't even have to find the length of the side or calculate the perimeter if we already know we have enough info. You can save some time on these DS question this way.

Did it help? Hope it did .
Attachments

m10-31.png [ 1.83 KiB | Viewed 1606 times ]

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27 Sep 2009, 14:52
Awesome. Thanks very much.
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30 Sep 2011, 08:41
I got this question wrong for a stupid reason (thought "quadrilateral" instead of "parallelogram"). Later it clicked, but it took longer since I got confused reading this piece in the explanation:

Statement (1) by itself is insufficient. We know the base and the height of the parallelogram; however, the height can equal the side of the parallelogram (in case of a rectangle) or it can exceed it.

Statement (2) by itself is insufficient. Nothing can be established about the length of the base of the parallelogram.

Statements (1) and (2) combined are sufficient. The length of the side can be found by the Pythagorean theorem.

As I read it, it was difficult to visualize the height exceeding a side. Am I reading it correctly? How would you correct the bold part?
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04 Oct 2011, 05:25
sid3699 wrote:
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?

1) The base of the parallelogram is 10.
2) One of the angles of the parallelogram is 45 degrees.

Why C and not E?

Am struggling to understand how we can use the pythagorean theorem when we don't know which pair of angles are 45 degrees each in the parallelogram. What am I missing here? Would appreciate some help here.

Area of parallelogram = base * height = 100. perimeter = 2 (base + side)

St1: base = 10, so height = 10. side is unknown still. not sufficient

St 2: angle = 45. so the triangle formed by base (part) and height is an isosceles triangle 90-45-45. sqrt(2):1:1. so the height and base part should be equal, but unknown.

Both: from St1 base = 10 and height = 10, through st2 side = 10* sqrt(2). As we know side and base, perimeter could be calculated.

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28 Nov 2011, 00:34
Why not B?

Since you know that one of the angles is 45 degrees, you can tell that that the parallelogram is made of two 45/45/90 triangles. This would lead you to knowing that the height = the base.

This would lead you to knowing that:

100 = height * base = height * height = base * base

So therefore, height and base must be equal to 10.

And this leads you to knowing that the perimeter must be equal to 20 + 20 * root(2).

Am I missing something or making any leaps of logic that I shouldn't be?

Thanks!
Re: M10: 31   [#permalink] 28 Nov 2011, 00:34
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The UvA-LINKER will give you a range of other options to find the full text of a publication (including a direct link to the full-text if it is located on another database on the internet).
De UvA-LINKER biedt mogelijkheden om een publicatie elders te vinden (inclusief een directe link naar de publicatie online als deze beschikbaar is in een database op het internet).

# Zoekresultaten

Zoekopdracht: faculteit: "UvA" en publicatiejaar: "2000"

 Auteurs Tom H. Koornwinder, N. Touhami Titel QDYBE: some explicit formulas for exchange matrix and related objects in case of sl(2), q=1 Uitgever s.n. Jaar 2000 Faculteit UvA Instituut/afd. UvA: Universiteitsbibliotheek Trefwoorden Quantum dynamical Yang-Baxter equation, sl(2), intertwining operator, fusion matrix, exchange matrix, weighted trace function, hypergeometric functions, Racah polynomials, Gegenbauer functions Samenvatting This mainly tutorial paper is intended as a somewhat larger example for parts of the theory exposed in the Lectures on the dynamical Yang-Baxter equations by P. Etingof and O. Schiffmann, math.QA/9908064. We explicitly compute the matrix entries of the intertwining operator, fusion matrix and exchange matrix associated to sl(2,R) for q=1. We also consider the weighted trace function and the dual Macdonald-Ruijsenaars equation for this particular case. As expected, the matrix entries of the exchange matrix turn out to be Racah polynomials. However, the road to their explicit formula via the fusion matrix is quick, and it also yields an interesting way to derive their orthogonality relations.SR4580320520032003De rechtspositie van werknemers bij surseance van betaling en faillissement31uva/uba/fdr/hsiDeventerKluwer9013006558Monografieën sociaal rechtArbeidsrechtelijke aspecten van reorganisatie1198491020270;1;02;21122000;;;04;01;01-03-1997Verburg;;L.G.;Zondag ;;W.A.;4590220520002000Het reïntegratieplan bij verzoeken om ontbinding van de arbeidsovereenkomst van zieke werknemers101uva/uba/fdr/hsiDen HaagElzevier9057495856Onderzoek uitgevoerd in opdracht van het Ministerie van Sociale Zaken en Werkgelegenheid door IVA Tilburg en het Hugo Sinzheimer Instituut1199801020270;1;02;21122000;;;04;01;01-03-19971001505;1;01;21122000;;;08;01;01-03-1984Marcelissen;;F.;;Pas;van de;I.;;4600320520002000Ethnic and religious state-neutrality: utopia or myth?uva/uba/fmg/aissrMellenModes of belonging and politics of recognition1200591000114;1;01;25002100;25011200;;01;01;01-10-1976Sicakkan;;H.;Lithman;;Y.;4610320520002000Immigrationuva/uba/fmg/aissrLondonRoutledge0415234417Routledge ECPR studies in European political scienceInternational distributive justice: cosmopolitanism and its critics1201621000114;1;01;25002100;25011200;;01;01;01-10-1976Caney;;S.;Lehning;;P.;4620320519981998Ethnizität, Rassismus und Klassen: post-Marxism and beyondAS 2473096125uva/uba/fmg/aissrHamburgArgument-Verlag3886192474Argument Sonderbände, N.F.Die Wiederentdeckung der Klassen1200421000114;1;01;25002100;25011200;;01;01;01-10-1976Bader;;V.M.;prif. dr.Benschop;;A.;dr.Krätke;;M.;4630821419191919Credit-based accept-zero sampling schemes for the control of outgoing qualityuva/uba/fnwi/kvi114546Baillie;;David H.;;Klaassen;;Chris A.J.;;A general procedure is presented for switching between accept-zero attributes or variables sampling plans to provide acceptance sampling schemes with a specified limit on the (suitably defined) average outgoing quality (AOQ). The switching procedure is based on credit, defined as the total number of items accepted since the last rejection. The limit on the AOQ is maintained by mandatory 100 % inspection of lots that are rejected when the credit is zero, with acceptance of all inspected items found to be conforming in such lots. The general procedure is developed and the application of the procedure to three situations is considered. Denoting the smallest value in a sample by $X_{(1)}$, a single, lower specification limit by $L$ and a constant by $c$, the method is applied to: - sampling by attributes, with acceptance criterion $X_{(1)}\geq L;$ - sampling by variables from a normal distribution with unknown process mean $\mu$ and known process standard deviation $\sigma$, with acceptance criterion $X_{(1)}\geq L+c\sigma;$ - sampling by variables from a normal distribution with unknown process mean $\mu$ and unknown process standard deviation $\sigma$, with acceptance criterion $X_{(1)}\geq L+cs$, where $s$ is the sample standard deviation. Finally it is shown that the guarantee on the upper limit to the AOQ remains valid when all rejected lots are submitted to 100 % inspection.SR4640821419191919Zero Information in the Two-Sample Mixed Proportional Hazards Modeluva/uba/fnwi/kvi114547Klaassen;;Chris A.J.;;Lenstra;;Andries J.;;The mixed proportional hazards model generalizes the Cox model by incorporating a random effect. In the case of two sampl Soort document Rapport Download document/1340 Document finder


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## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 15th 2021

for completeness, to go with U(ℋ), for the moment mainly in order to record references, such as:

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeSep 19th 2021
• (edited Sep 19th 2021)

added the statement that $PU(\mathcal{H})$ is well-pointed, which is made explicit on p. 23 of

but follows by a more general result due to Dardalat and Pennig.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeSep 19th 2021
• (edited Sep 19th 2021)

added statement that $S^1 \to \mathrm{U}(\mathcal{H}) \to PU(\mathcal{H})$ is a fiber bundle. Am copying this statement also to U(H) and to coset space coprojection admitting local sections

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeOct 30th 2021

• Ellen Maycock Parker, The Brauer Group of Graded Continuous Trace $C^\ast$-Algebras, Transactions of the American Mathematical Society 308 1 (1988) (jstor:2000953)
• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeDec 13th 2021

for the graded version of $PU(\mathcal{H})$ I have added pointer also to

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeDec 13th 2021

These authors also point (in addition to the reference in #4) to

• E. Parker, Graded continuous trace $C^\ast$-algebras and duality, Operator algebras and topology (Craiova, 1989), 130–145, Pitman Res. Notes Math. Ser., 270.

but I haven’t yet found any online trace of this article/book (I find the 1983 version of the series, but not the one from 1989).


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# Tag Info

10

The integrand is periodic modulo $\pi$ in each variable, so it suffices to integrate each variable over $[0, \pi]$ and replace the constant factor by $2^{-7}$. If we were to apply a change of variables (e.g. set $x = \cos(s_1)$ and similarly for the other five variables), we would have an integral of a piecewise-algebraic function which thus belongs to ...

6

Nemo's representation of $F(\eta)$ in terms of a hypergeometric function can be evaluated without difficulty for large $\eta$: $$F(\eta)=\frac{\sqrt{3} {\eta}^2 \Gamma \left(\frac{2}{3}\right) \; _1F_2\left(\frac{2}{3};\frac{4}{3},\frac{5}{3};-\frac{{\eta}^3}{27}\right)}{6\pi }-\frac{12 {\eta} \; _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{{\eta}^3}... 5 Some buzzwords that should lead to some non-textbook examples: In the fields of uncertainty quantification, statistical inverse problems or Bayesian inference one wants, for example, compute conditional expectations for posterior distributions. The domain of integration has as many dimensions as the the quantity of interest has degrees of freedom, and this ... 4 For small n Monte Carlo integration is not needed. For n up to 100 see Kolmogorov-Smirnov Tests when Parameters are Estimated with Applications to Tests of Exponentiality and Tests on Spacings (table 3). These exact results were used to construct the Graphs for Use with the Lilliefors Test for Normal and Exponential Distributions. There are also ... 4 For every square matrix C, let r(C) denote its spectral value. We say that a complex number \lambda is a dominant eigenvalue of C if \lambda is the only eigenvalue of C with modulus r(C). a semisimple eigenvalue of C if it is an eigenvalue of C and its algebraic multiplicity coincides with its geometric multiplicity. Theorem. The ... 3 The answers are yes and yes. Consider the Hermite Gauss functions : \psi_n(x)=e^{-x^2/2}H_n(x) , we have two properties:$$\psi_n''(x)+(2n+1-x^2)\psi_n(x)=0 $$and$$\psi_{n+1}=\psi_n'(x) +x\psi_n(x) $$For n large, we take a lenght L>0 which is very small (for example L=1/\log{n}). On any [x_0-L/2,x_0+L/2]\subset ]- \sqrt{2n},\sqrt{2n}[, \... 3 Do you know the Riemann zeta function ? \displaystyle\frac{\log \zeta(\sigma+2i\pi \xi)}{\sigma+2i\pi \xi} is the Fourier transform of the prime counting function \displaystyle J(e^u) e^{-\sigma u} = e^{-\sigma u}\sum_{p^k \le e^u} \frac{1}{k}=e^{-\sigma u}\sum_{k\ge 1} \frac{\pi(e^{u/k})}{k} 3 It looks to me like you are searching for what are called monomial cubature rules on the simplex in the literature on numerical integration. As an example, a 4th-degree monomial in 2-D (x,y) would be any of: x^4, x^3y, x^2y^2, xy^3, y^4. A 4th-degree monomial cubature rule will exactly integrate any of those, or any linear combination of them. There are ... 2 Here are a few papers that discuss high-dimensional Monte Carlo integrals, together with quotes from Math Reviews. MR2719643 (2011i:65038) Griebel, Michael; Holtz, Markus; Dimension-wise integration of high-dimensional functions with applications to finance, J. Complexity 26 (2010), no. 5, 455–489. "In addition to error bounds, the authors also ... 2 I must premise that I am not a specialist in numerical analysis, therefore I may be not right when talking about more popular methods in this field pertaining the solution of ITEs. Said that, however, I think I can be of some help. Could you recommend me any articles or book with a brief overview of some methods (maybe classical one), please? Perhaps a ... 2 The key here is the simple change-of-interval/rescaling formula, found e.g. at the link in the OP, according to which $$T_n(f)(x)=T_{n,[0,x]}(f)=x\sum_1^n w_i f(xx_i), \tag{*}$$ where the w_i's and x_i's are such that T_n(f)(1)=T_{n,[0,1]}(f)=\sum_1^n w_i f(x_i) for all f. From here, one immediately has the affirmative ... 1 You want the relation between \phi(x) evaluated at a point x on the boundary and the derivative \partial\phi/\partial n evaulated at the same point x, which is given by$$\phi(x)=\frac{\partial\phi(x)}{\partial n}\int_{-\Delta s/2}^{\Delta s/2}\ln |s|\,ds=[\ln(\Delta s/2)-1]\Delta s\frac{\partial\phi(x)}{\partial n}.$$Hence G_{ii}=[\ln(\Delta s/2)-... 1 Following Michael Renardy's suggestion, write g=\phi\psi, and write$$ g(x)=g(0)+x\cdot G(x) , $$where G is a smooth function. Then the integral becomes$$ \int_\Omega fg = g(0)\int_\Omega f + \sum_i\int_\Omega \frac{x_i}{|x|}G_i(x)d^3x . $$The integrands in the latter integrals are bounded, and the first integral can be written as$$ \int_\Omega f = \...

1

It is not hard to show, assuming that $b>0$, that the smooth solutions of the Euler-Lagrange equation are given by the formula $$y(x) = \frac{a}{b} - c_1^2 - \left(\frac{x-c_0}{2c_1}\right)^2 < \frac{a}{b}$$ where $c_0$ and $c_1\not=0$ are constants. For example, see my answer to the MO question Riemannian surfaces with an explicit distance function?...

1

Under the assumption that your solution is smooth enough, you can use the calculus of variations to turn your minimization problem into a differential equation. The differential equation is then given by the Euler-Lagrange equation, and can be solved by standard methods for numerical simulation of differential equations.

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## Definition

The addition operation in the domain of integers $\Z$ is written $+$.

We have that the set of integers is the Inverse Completion of Natural Numbers.

Thus it follows that elements of $\Z$ are the isomorphic images of the elements of equivalence classes of $\N \times \N$ where two tuples are equivalent if the difference between the two elements of each tuple is the same.

Thus addition can be formally defined on $\Z$ as the operation induced on those equivalence classes as specified in the definition of integers.

That is, the integers being defined as all the difference congruence classes, integer addition can be defined directly as the operation induced by natural number addition on these congruence classes:

$\forall \tuple {a, b}, \tuple {c, d} \in \N \times \N: \eqclass {a, b} \boxminus + \eqclass {c, d} \boxminus = \eqclass {a + c, b + d} \boxminus$


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## Precalculus (6th Edition) Blitzer

$\dfrac{1}{2}$
RECALL: $\log_b{b^x} = x.$ Note that $9=\sqrt{81} = 81^{\frac{1}{2}}.$ Thus, the given expression can be written as $\log_{81}{81^{\frac{1}{2}}}.$ Use the rule above to obtain: $\dfrac{1}{2}.$


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a) Ni(CO)4 and NiCl42- are diamagnetic; and [Ni(CN)4] 2- is paramagnetic. 13) Amongst following, the lowest degree of paramagnetism per mole of This is incorrect. dsp2 - square planar - wherein. J. inorg nucl. sp sp2 sp3 sp3d sp3d2. Ni(CO)4 is .........? dsp2 - square planar - wherein. dz 2. the compound at 298 K will be shown by: (new) Click here to see 3d Interactive Solved Question paper, BEST CSIR NET - GATE - SET Study Material The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Note that in the term 'dsp2', 'd' precedes 'sp2' indicating that the d-orbital used in hybridization comes from a lower (inner) shell than the 's' and 'p' orbitals. If the complex ion involves âsp3â hybridization, it would have tetrahedral structure. The CO ligands will be opposite each other. For the formation of square planar structure by dsp2 hybridisation, two unpaired d-electrons are paired up due to energy made available by the approach of ligands, making one of the 3d orbitals empty. In dsp 2 hybridization, one d-orbital [which is d(x 2 ây 2)] is involved in hybridization with one sandtwo p-orbital.This leads to the square planar geometry. Ni forms octahedral, square planar and tetrahedral complexes in +2 oxidation state. The four dsp 2 hybrid orbitals adopt square planar geometry The complex ion [Ni(CN) 4 ] 2 - involves dsp 2 Hybridization If the compound undergoes reaction with strong ligand than dsp 2 , dsp 3 ,d 2 sp 3 hybridization occurs. 2) What are the strong field and weak field ligands? diamagnetic. dx 2-dy 2 and dz 2. Structure and basic properties. Donor atoms are situated on the x and y -axes of the square planar structure.

What are the effects of sugar in cat food? why dsp2 hybridization is square planar and sp3 hybridization tetrahedral shape . 90o - Generally formed by strong field ligand... - â¦ 4. sp3. This is incorrect. chloride, [Ni(NH3)4]Cl2? Ni(CO)4 . 5) Is CO paramagnetic or diamagnetic.....? d2sp3 - inner complex. Question 44: Give the formula of each of the following coordination entities: (i) Co3 + ion is bound to one Cl-, one NH 3 molecule and two bidentate ethylene diamine (en) molecules. MathJax reference. Type of hybridization. 9) The theory that can completely explain the nature of bonding in Question. 4) What is the hybridization & structure of [CoCl4]2-? Can you explain this answer? Square-planar molecules dsp2 hybridization.docx - Square-planar molecules dsp2 hybridization For example the ions Pt2 and Cl\u2013 can form the, hybridization scheme, it helps to start with the neutral Pt atom, then imagine, it losing two electrons to become an ion, followed by grouping of the two, All of the four-coordinated molecules we have discussed so far have. & Online Coaching, BEST CSIR NET - GATE - IIT JAM Chemistry Study Material Tags: Class 12 , Chemistry , Coordination Compounds Asked by Srinivas 1 Answers. $\endgroup$ â Jan Mar 21 '16 at 1:26 Square planar. why dsp2 hybridization is square planar and sp3 hybridization tetrahedral shape . and is referred to as a high spin complex. square planar - Bond angle. The dsp 2 hybrid orbitals are inner orbital complexes in which the electrons get paired up due to the presence of a strong field ligand. b) [Ni(CN)4]2- and NiCl42- are diamagnetic; and Ni(CO)4 is paramagnetic. [Ni(CN)4]2- Magnetic nature: Diamagnetic (low spin). Ni(CO)4 = Ni + 4CO The valence shell electronic configuration of ground state Ni atom is 3d8 4s2. As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic The geometry of Ni(CO) 4 and Ni(PPh 3) 2 Cl 2 are (a) both square planar (b) tetrahedral and square planar, respectively (c) both tetrahedral (d) square planar and tetrahedral, respectively In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. What is the magnetic nature of this compound? Likewise, the square-planar ligand arrangement only has one nonbonding d-orbital. Hence [Fe(H2O)6]2+ is more paramagnetic. Hence the electronic configurations of metal ions coordinated to water are same as in isolated ions. Hybridization defects Hybridisation of s and p orbitals to form effective sp x hybrids requires that they have comparable radial extent. Square planar. Trigonal bipyramidal. 37, pp. In [Ni(DMG)2] the nickel is in the +2 oxidation state and to have a square planar geometry because of chelation the pairing of electrons takes place. The correct option is C. Related. c) Ni(CO)4 and [Ni(CN)4]2- are diamagnetic; and NiCl42- is paramagnetic. The geometry is prevalent for transition metal complexes with d 8 configuration. & Online Coaching, Click here to see 3d Interactive Solved Question paper. 1) How do you calculate the magnetic moment of ions of transition elements? Difference between dsp2 and sp3 hybrid orbitals. a)[Ni(CN)4]4- b)[Cu(NH3)4]2- c)[PtCI4]2- d)All of theseCorrect answer is option 'D'. But, it is possible if the two unpaired 3d-electrons are paired up due to the energy made available by the approach of ligands. The geometry of the orbital arrangement: Linear: Two electron groups involved resulting in sp hybridization, the angle between the orbitals is 180°. View Square-planar molecules dsp2 hybridization.docx from CHEMISTRY 101 at Colorado School of Mines. So, the dx^2-dy^2 orbital is used in dsp2 hybridization, along with the s, px and py orbitals to form a square planar geometry. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, â¦ Types of d orbitals. E) none of the above. 10) The complex ion (NiCl4)2â is tetrahedral. Whereas there is only one unpaired electron in Cu2+ and hence CuSO4.5H2O shows lowest degree of paramagnetism. * In presence of water, which is a weak field ligand, the configurations of metal ions in hydrated compounds reflect those in isolated gaseous ions i.e., no pairing of electrons is possible as the interaction with water molecules is weak. The correct order of bond angles (smallest first) The molecule having smallest bond angle is Which one of the â¦ Characteristics of dsp2 hybridization:... - Geometry. One electron in each C atom remains in an unhybridized p orbital * In presence of strong field CN- ions, all the electrons are paired up. This is incorrect. * The valence shell electronic configuration of ground state Ni atom is 3d8 4s2. Hence MnSO4.4H2O shows greater paramagnetic nature. This video is about dsp2 hybridization in a coordination spere e. g [Ni(CN)4]¯².it also explains the hybridization of compounds having square planar geometry. This includes Rh(I), Ir(I), Pd(II), Pt(II), and Au(III). * All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach Ni atom. Note: The charge on metal ions is equal to the charge on the complex since water is a neutral ligand. many unpaired electrons are there in the complex? sp3d2 - outer complex. and square planer respectively . [16] Identify The Geometry Of The Complex Ion If The Hybridization Is Dsp2. So, the dx^2-dy^2 orbital is used in dsp2 hybridization, along with the s, px and py orbitals to form a square planar geometry. As for sp2 hybridization, it is an attempt to explain trigonal planar electron pair geometry where the electron pairs are approximation 120 degrees apart. The four dsp 2 hybrid orbitals adopt square planar geometry The complex ion [Ni(CN) 4 ] 2 - involves dsp 2 Hybridization If the compound undergoes reaction with strong ligand than dsp 2 , dsp 3 ,d 2 sp 3 hybridization occurs. Thus [Ni(CN)4]2- is d) Ni(CO)4 is diamagnetic; [Ni(CN)4]2- and NiCl42- are paramagnetic. 12) Write the hybridisation and magnetic behaviour of the complex Distribution of hybrid orbitals in space. coordination number 4, hybridization dsp2. [Ni(CN) 4] 2-Hybridization:dsp 2 [Ni(CN) 4] 2- Shape: Square planar [Ni(CN) 4] 2- Magnetic nature: Diamagnetic (low spin) NiCl 4 2-= Ni 2+ + 4Cl-* Again in NiCl 4 2-, there is Ni 2+ ion, However, in presence of weak field Cl-ligands, NO pairing of d-electrons occurs. I mean how to identify whether sp3 or dsp2 Three electrons from each C atom are in sp 2 hybrids. * In [Ni(CN)4]2-, there is Ni2+ ion for which the electronic configuration in the valence shell is 3d8 4s0. * The outer shell electronic configurations of metal ions in the above complexes are shown below. * The paramagnetic nature of a compound is proportional to the number of unpaired electrons in it. Tetrahedral 4. dsp2. The dsp2 hybridisation in [Ni(CN)4]2- ion is shown below. This is called inner shell hybridization. * As mentioned in previous question, the electronic configuration of metal ions is not much affected by weak field ligand water. How In the previous unit on bonding, we saw terms such as 'sp3d'. (iii) Ni(CO) 4 has spb3 hybridization, tetrahedral shape, whereas [Ni(CN) 4]2- has dsp2 hybridization, therefore, it has a square planar shape. Option 3),square planar. are solved by group of students and teacher of Class 12, which is also the largest student community of Class 12. There will be one empty 4d-orbital, along with the 5s and two 5p orbitals, giving dsp2 hybridization which is square planar. In analyzing the bonding, it is a complex of Rh(I), a d 8 transition metal ion. 11) What is the magnetic moment of nickel ion in tetraammine nickel(ii) dx 2-y 2. * Again in NiCl42-, there is Ni2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. Formally, one may say that putting the d before s and p implies a lower-shell d-orbital. Now, depending upon the hybridization, there are two types of possible structure of Cu+ and Cu2+ ion are formed with co-ordinationnumber 4. By this, there is no unpaired electron and the complex would be diamagnetic. A square planar is a result of dsp 2 hybridisation where the inner d sub shell (d x2 y2 orbital) participates. [NiCl4]2- Magnetic nature: Paramagnetic (low spin). Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl- ligands in 5. sp3d. Copper Zinc Chromium Manganese Potassium 3.) This is correct. tetrahedral geometry. sp3 - dsp2 - The wilikinson catalyst is and the hybridisation and its shape are . Course Hero is not sponsored or endorsed by any college or university. 6. sp3d2 (nd orbitals are involved; outer orbital complex or high-spin or spin-free complex) Octahedral. * All the metal ions in the above compounds are divalent and their outer shell electronic configurations are shown below. and square planer respectively . Option 2),tetrahedral. sp3 - dsp2 - The wilikinson catalyst is and the hybridisation and its shape are . According to single crystal X-ray diffraction the compound adopts a slightly distorted square planar structure. The dsp 2 hybrid orbitals are inner orbital complexes in which the electrons get paired up due to the presence of a strong field ligand. Thus Ni(CO)4 is Printed in Greal Britain TRANSFORMATION OF SQUARE-PLANAR dsp2 CONFIGURATION OF Ni(II) ALDOXIMATES INTO OCTAHEDRAL sp3d2 CONFIGURATION BY REACTION WITH TIN(IV) CHLORIDE N. S. BIRADAR Department of Chemistry, Karnatak University. * Fe2+ ion has more number of unpaired electrons. sp 2. sp 3. sp 3 d. Chem., 1975, Vol. Thus [Ni(CN)4]2- is diamagnetic. The dsp2 hybridization favours a planar geometry in which the four hybridized orbitals lie in the equatorial xy plane, as in ICl4- or XeF4. Carbon-carbon triple bonds: sp hybridization in acetylene; Multiple bonds between unlike atoms; The nitrate ion; Conjugated Double Bonds; Benzene; Hybrids involving d orbitals. Option 1), trigonal bipyramidal. Option 3),square planar. 12) According to Valence Bond theory, in the square planar Ni(CN)42 orbital hybridization pattern is complex ion, the 1 A) dsp2 B) sp3 C) sp3 D) sp. are , respectively....... 8) The magnetic moment (spin only) of  NiCl42- The Questions and Answers of Which complex has square planar shape & dsp2- hybridization? My suggestion is never to use hybridisation approaches for transition metal complexes. Option 1), trigonal bipyramidal. Square Planar Tetrahedral Linear Octahedral Trigonal Planar 2.) Among the triatomic molecules/ions, B e C l 2 , N 3 â, N 2 O, N O 2 + , O 3 , S C l 2 , I C l 2 â, I 3 â and X e F 2 , the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is On the other hand, the notation with the d first as in dsp2 denotes the d-orbital with a lower principle quantum number is used, as â¦ The five basic shapes of hybridization are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. Mn2+ ion has more number of unpaired electrons. On the other hand, for the formation of square planar structure through the dsp2 hybridization, one of the 3d-orbital of nickel atom should be empty and available for hybridization. Square-planar molecules: dsp2 hybridization For example, the ions Pt2+ and Clâ can form the ion Tags: Class 12 , Chemistry , Coordination Compounds Asked by Srinivas 1 Answers. The empty 4d, 3s and two 4p orbitals undergo dsp2 hybridization to make bonds with CN- ligands in square planar geometry. d2sp3 - inner complex. It acquires stability through chelation and intramolecular hydrogen bonding. Pergamon Press. Option 2),tetrahedral. $\begingroup$ Square planar usually happens if there is a reason to keep two electrons (sometimes even one) in a $\mathrm{d_{z^2}}$ orbital for stabilisation reasons. Identify The Transition Metal That Is Used In Hemoglobin Synthesis In The Human Body. 6) The geometry and magnetic behaviour of the complex Ni(CO)4 is........?

This leaves us with two nitrogen p-orbitals which form two mutually perpendicular Ï bonds to the two atomic p orbitals on the carbon. 1. 6 1901-19t34. While 2p orbitals are on average less than 10% larger than 2s, in part attributable to the lack of a radial node in 2p orbitals, 3p orbitals which have one radial node, exceed the 3s orbitals by 20â33%. The dsp2 hybridization favours a planar geometry in which the four hybridized orbitals lie in the equatorial xy plane, as in ICl4- or XeF4. The empty 4d, 3s and two 4p orbitals undergo dsp2 hybridization to make bonds with CN- ligands in square planar geometry. The empty 4s and three 4p orbitals undergo sp3 hybridization and form bonds with CO ligands to give Ni(CO)4. Orbital Hybridization Figure 10.5 BONDS SHAPE HYBRID REMAIN 2 linear sp 2 pâs 3 trigonal sp 2 1 p planar 4 tetrahedral sp 3 none Compounds Containing Double Bonds Valence Bond Theory (Hybridization) C atom has four electrons. Ni(CO)4 Shape & Structure (geometry): Tetrahedral, Ni(CO)4 Magnetic nature: Diamagnetic (low spin). Again, if the complex involves âdsp2â hybridization, it would have square planar structure. What is the hybridization of the central atom in a molecule with a square-planar molecular geometry ? sp. sp3d2 - outer complex. This is incorrect. diamagnetic. Ballot Secrecy - is it a Voter's Privilege or a Voter's Obligation? Structures and Properties of Matter Review - Answers.pdf, SSCJ_4121_COORD_AND_ORGANO_CHEM_LECTURE_2.pptx, Unionville High School • CHEMISTRY SCH4U1, University of Pennsylvania • CHEMISTRY 101, Technical University of Mombasa • CHEMISTRY SPCI/02494, University of Southern California • CHEM 115. The hybridization of Ni in [Ni(DMG)2] is dsp2. The dsp2 hybrid orbitals are inner orbital complexes in which the electrons get paired up due to the presence of a strong field ligand. Identify The Geometry Around The Fe2+ Ion In The Hemoglobin Complex. This is correct. Square Planar Complexes. What is the hybridization of the central atom in a molecule with a square-planar molecular geometry ? Orbitals are involved ; outer orbital complex or high-spin or spin-free complex Octahedral... D-Orbitals, NiCl42- is paramagnetic 4p orbitals undergo sp3 hybridization and form bonds with CO to. What are the effects of sugar in cat food ligands approach Ni atom is 4s2! Up when strong field ligand... - â¦ square planar structure ion involves hybridization... Complex ) Octahedral by any college or university prevalent for transition metal ion +2 oxidation state 4s! 2- ion is shown below 3s and two 4p orbitals undergo dsp2 hybridization is planar... Of ions of transition elements: paramagnetic ( low spin ) one may say that putting the before. Co ligands to give Ni ( CN ) 4 is diamagnetic two types of possible structure of [ ]... A Voter 's Privilege or a Voter 's Privilege or a Voter Obligation! Paired up saw terms such as 'sp3d ' make bonds with Cl- in. The outer shell electronic configurations are shown below how many unpaired electrons in it this, is..., there are unpaired electrons are pushed into 3d orbitals and get paired up when strong field CN-,! Outer orbital complex or high-spin or spin-free complex ) Octahedral or a Voter 's Privilege or a Voter 's or. Dsp2 - the wilikinson catalyst is and the hybridisation and Magnetic behaviour of central... Tetrahedral Linear Octahedral Trigonal planar 2. involves âsp3â hybridization, it have. Slightly distorted square planar shape & dsp2- hybridization diffraction the compound adopts a distorted... Same as in isolated ions * All the metal ions in the complex ion if the complex ion if hybridization. Water are same as in isolated ions of bonding in Ni ( CO ) 4 2-. Identify whether sp3 or dsp2 why dsp2 hybridization to make bonds with CN- ligands in square planar and hybridization. There are unpaired electrons in it empty 4s and three 4p orbitals undergo dsp2 hybridization is square planar tetrahedral Octahedral! Electronic configuration of ground state Ni atom the d before s and p implies a lower-shell d-orbital where. Analyzing the bonding, we saw terms such as 'sp3d ' metal ion ( )! Are the strong field CO ligands approach Ni atom is 3d8 4s2 is... Unpaired electron and the complex: paramagnetic ( low spin ) Ni ( CN ) 4 = Ni 4CO! Endorsed by any college or university the central atom in a molecule with a square-planar molecular geometry 2- Magnetic:. Prevalent for transition metal complexes with d 8 transition metal complexes that completely! Mentioned in previous question, the electronic configurations of metal ions in the Human Body on metal ions is to! Ligands in square planar complexes lower-shell d-orbital to give Ni ( CO ) 4 ] 2- ion shown. In Cu2+ and hence CuSO4.5H2O shows lowest degree of paramagnetism of metal ions in the Human Body moment of of... Square-Planar molecules dsp2 hybridization.docx from Chemistry 101 at Colorado School of Mines we saw terms such as '... Degree of paramagnetism be diamagnetic would have square planar shape square planar hybridization dsp2 dsp2- hybridization (... Are paramagnetic a square-planar molecular geometry configurations are shown below a high complex... - dsp2 - the wilikinson catalyst is and the hybridisation and Magnetic behaviour of complex... Student community of Class 12, a d 8 transition metal that is Used in Hemoglobin in! P implies a lower-shell d-orbital hybridisation in [ Ni ( DMG ) 2 ] is dsp2 * the shell. ( d x2 y2 orbital ) participates the complex ion ( NiCl4 ) 2â tetrahedral! Transition elements high-spin or spin-free complex ) Octahedral the transition metal complexes in previous,. 4 and [ Ni ( CN ) 4 ] 2- ion is square planar hybridization dsp2 below dsp2-. Hero is not sponsored or endorsed by any college or university metal ions in the above complexes are shown.... Into 3d orbitals and get paired up due to the energy made available the! With a square-planar molecular geometry dsp 2 hybridisation where the inner d sub shell d...

## square planar hybridization dsp2

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# Maxwell-boltzman distrubution

If I have two particles that follows the maxwell velocity distribution:
$$\phi(v_i)dv_i=4 \pi v_i^2 \left ( \frac{m_iv_i}{2\pi kT} \right ) ^{3/2}e^{\frac{-m_iv_i^2}{2kT}}dv_i$$
Why is their combined distribution:
$$\phi(v)dv=4 \pi v^2 \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2}e^{\frac{-\mu v^2}{2kT}}dv$$
where mu is the reduced mass and v=v2-v1
I have these questions because I don't quite follow these derivations.
http://dissertations.ub.rug.nl/FILES/faculties/science/2007/a.matic/c2.pdf [Broken]
http://www.astro.psu.edu/users/rbc/a534/lec11.pdf
Namely, I not sure why the following holds:
$$\int_0^\infty \int_0^\infty \phi(v_1) \phi(v_2) v_1 v_2 \sigma dv_1 dv_2 = 4\pi \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2} \int_0^\infty v^3 \sigma e^{\frac{-\mu v^2}{2kT}}dv$$

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10.7 The Kinetic Molecular Theory of Gases

Learning Objective

1. To understand the significance of the kinetic molecular theory of gases.

The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.

A Molecular Description

The kinetic molecular theory of gasesA theory that describes, on the molecular level, why ideal gases behave the way they do. explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:

1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but these collisions are perfectly elastic; that is, they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.

Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8 "The Behavior of Real Gases", we explain how this theory must be modified to account for the behavior of real gases.

Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a force exerted by the gas on the walls, just as a bowling ball exerts a force on the pins it strikes. Anything that increases the frequency with which the molecules strike the walls or increases the momentum of the gas molecules (i.e., how hard they hit the walls) increases the pressure; anything that decreases that frequency or the momentum of the molecules decreases the pressure.

Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules.

Postulate 5 provides a molecular explanation for the temperature of a gas. Recall from Chapter 5 "Energy Changes in Chemical Reactions" that the kinetic energy of an object is given by $KE = 12mv2,$ where m is the mass of the object and v is its velocity, or speed. Postulate 5 refers to the average kinetic energy of the molecules of a gas, which can be represented as $KE¯,$ and states that at a given temperature, all gases have the same value of $KE¯.$ The average kinetic energy of the molecules of a gas is therefore

Equation 10.31

$K E ¯ = 1 2 m v 2 ¯$

where $v2¯$

is the average of the squares of the speeds of the particles. For n particles,

Equation 10.32

$v 2 ¯ = v 1 2 + v 2 2 + v 3 2 + ⋅ ⋅ ⋅ + v n 2 n$

The square root of $v2¯$ is the root mean square (rms) speed(vrms)The speed of a gas particle that has average kinetic energy.:

Equation 10.33

$v rms = v 1 2 + v 2 2 + v 3 2 + ⋅ ⋅ ⋅ + v n 2 n$

Compare this with the formula used to calculate the average speed:

Equation 10.34

$v ¯ = v 1 + v 2 + v 3 + ⋅ ⋅ ⋅ + v n n$

The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation 10.33 tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases.

Note the Pattern

At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed.

Example 15

The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed $(v¯),$ root mean square speed (vrms), and most probable speed (vp).

Given: particle speeds

Asked for: average speed $(v¯),$ root mean square speed (vrms), and most probable speed (vp)

Strategy:

Use Equation 10.34 to calculate the average speed and Equation 10.33 to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving.

Solution:

The average speed is the sum of the speeds divided by the number of particles:

The rms speed is the square root of the sum of the squared speeds divided by the number of particles:

The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence vp = 6.0 m/s. The vrms of the particles, which is related to the average kinetic energy, is greater than their average speed.

Exercise

Ten particles were found to have speeds of 0.1, 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0, and 6.0 m/s. Calculate their average speed $(v¯),$ root mean square speed (vrms), and most probable speed (vp).

Boltzmann Distributions

At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction, as you will learn in Chapter 14 "Chemical Kinetics". This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure 10.14 "The Distributions of Molecular Speeds for a Sample of Nitrogen Gas at Various Temperatures". Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure 10.14 "The Distributions of Molecular Speeds for a Sample of Nitrogen Gas at Various Temperatures" were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributionsA curve that shows the distribution of molecular speeds at a given temperature., after one of the other major figures responsible for the kinetic molecular theory of gases.

Figure 10.14 The Distributions of Molecular Speeds for a Sample of Nitrogen Gas at Various Temperatures

Increasing the temperature increases both the most probable speed (given at the peak of the curve) and the width of the curve.

The Relationships among Pressure, Volume, and Temperature

We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.

Pressure versus Volume

At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure.

Volume versus Temperature

Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, unless the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.

Pressure of Gas Mixtures

Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.

Example 16

The temperature of a 4.75 L container of N2 gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the

1. average kinetic energy of the N2 molecules?
2. rms speed of the N2 molecules?
3. average speed of the N2 molecules?
4. impact of each N2 molecule on the wall of the container during a collision with the wall?
5. total number of collisions per second of N2 molecules with the walls of the entire container?
6. number of collisions per second of N2 molecules with each square centimeter of the container wall?
7. pressure of the N2 gas?

Given: temperatures and volume

Asked for: effect of increase in temperature

Strategy:

Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas.

Solution:

1. Increasing the temperature increases the average kinetic energy of the N2 molecules.
2. An increase in average kinetic energy can be due only to an increase in the rms speed of the gas particles.
3. If the rms speed of the N2 molecules increases, the average speed also increases.
4. If, on average, the particles are moving faster, then they strike the container walls with more energy.
5. Because the particles are moving faster, they collide with the walls of the container more often per unit time.
6. The number of collisions per second of N2 molecules with each square centimeter of container wall increases because the total number of collisions has increased, but the volume occupied by the gas and hence the total area of the walls are unchanged.
7. The pressure exerted by the N2 gas increases when the temperature is increased at constant volume, as predicted by the ideal gas law.

Exercise

A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the

1. average kinetic energy of the He atoms?
2. rms speed of the He atoms?
3. average speed of the He atoms?
4. impact of each He atom on the wall of the container during a collision with the wall?
5. total number of collisions per second of He atoms with the walls of the entire container?
6. number of collisions per second of He atoms with each square centimeter of the container wall?
7. pressure of the He gas?

Answer: a. no change; b. no change; c. no change; d. no change; e. decreases; f. decreases; g. decreases

Diffusion and Effusion

As you have learned, the molecules of a gas are not stationary but in constant motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. Figure 10.15 "The Diffusion of Gaseous Molecules" shows white fumes of solid ammonium chloride (NH4Cl) forming when containers of aqueous ammonia and HCl are placed near each other, even with no draft to stir the air. This phenomenon suggests that NH3 and HCl molecules (as well as the more complex organic molecules responsible for the aromas of pizza and perfumes) move without assistance.

Figure 10.15 The Diffusion of Gaseous Molecules

When open containers of aqueous NH3 and HCl are placed near each other in a draft-free environment, molecules of the two substances diffuse, collide, and react to produce white fumes of solid ammonium chloride (NH4Cl).

DiffusionThe gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with a uniform composition. is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. As we shall see in Chapter 11 "Liquids", Chapter 12 "Solids", and Chapter 13 "Solutions", diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. We can describe the phenomenon shown in Figure 10.15 "The Diffusion of Gaseous Molecules" by saying that the molecules of HCl and NH3 are able to diffuse away from their containers, and that NH4Cl is formed where the two gases come into contact. Similarly, we say that a perfume or an aroma diffuses throughout a room or a house. The related process, effusionThe escape of a gas through a small (usually microscopic) opening into an evacuated space., is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.

The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship, $rate∝ 1M,$ is referred to as Graham’s lawA law that states that the rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass., after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses. If r is the effusion rate and M is the molar mass, then

Equation 10.35

$r 1 r 2 = M 2 M 1$

Although diffusion and effusion are different phenomena, the rate of diffusion is closely approximated using Equation 10.35; that is, if M1 < M2, then gas #1 will diffuse more rapidly than gas #2. This point is illustrated by the experiment shown in Figure 10.16 "A Simple Experiment to Measure the Relative Rates of the Diffusion of Two Gases", which is a more quantitative version of the case shown in Figure 10.15 "The Diffusion of Gaseous Molecules". The reaction is the same [NH3(aq) + HCl(aq) → NH4Cl(g)], but in this experiment, two cotton balls containing aqueous ammonia and HCl are placed along a meter stick in a draft-free environment, and the position at which the initial NH4Cl fumes appear is noted. The white cloud forms much nearer the HCl-containing ball than the NH3-containing ball. Because ammonia (M = 17.0 g/mol) diffuses much faster than HCl (M = 36.5 g/mol), the NH4Cl fumes form closer to HCl because the HCl molecules travel a shorter distance. The ratio of the distances traveled by NH3 and HCl in Figure 10.16 "A Simple Experiment to Measure the Relative Rates of the Diffusion of Two Gases" is about 1.7, in reasonable agreement with the ratio of 1.47 predicted by their molar masses [(36.5/17.0)1/2 = 1.47].

Figure 10.16 A Simple Experiment to Measure the Relative Rates of the Diffusion of Two Gases

Cotton balls containing aqueous NH3 (left) and HCl (right) are placed a measured distance apart in a draft-free environment, and the position at which white fumes of NH4Cl first appear is noted. The puff of white NH4Cl forms much closer to the HCl-containing ball than to the NH3-containing ball. The left edge of the white puff marks where the reaction was first observed. The position of the white puff (18.8 − 3.3 = 15.5 cm from the NH3, 28.0 − 18.8 = 9.2 cm from the HCl, giving a ratio of distances of 15.5/9.2 = 1.7) is approximately the location predicted by Graham’s law based on the square root of the inverse ratio of the molar masses of the reactants (1.47).

Heavy molecules effuse through a porous material more slowly than light molecules, as illustrated schematically in Figure 10.17 "The Relative Rates of Effusion of Two Gases with Different Masses" for ethylene oxide and helium. Helium (M = 4.00 g/mol) effuses much more rapidly than ethylene oxide (M = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon $294.00= 2.7$ times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Mylar balloons can retain their helium for days.

Figure 10.17 The Relative Rates of Effusion of Two Gases with Different Masses

The lighter He atoms (M = 4.00 g/mol) effuse through the small hole more rapidly than the heavier ethylene oxide (C2H4O) molecules (M = 44.0 g/mol), as predicted by Graham’s law.

Note the Pattern

At a given temperature, heavier molecules move more slowly than lighter molecules.

Example 17

During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of 235U. Naturally occurring uranium is only 0.720% 235U, whereas most of the rest (99.275%) is 238U, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound UF6 (boiling point = 56°C).

1. Calculate the ratio of the rates of effusion of 235UF6 and 238UF6 for a single step in which UF6 is allowed to pass through a porous barrier. (The atomic mass of 235U is 235.04, and the atomic mass of 238U is 238.05.)
2. If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the nth power. How many effusion steps are needed to obtain 99.0% pure 235UF6?

Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U

Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6

Strategy:

A Calculate the molar masses of 235UF6 and 238UF6, and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235UF6 after a single effusion step.

B Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required.

Solution:

1. A The first step is to calculate the molar mass of UF6 containing 235U and 238U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of 235UF6 is

234.04 + (6)(18.998) = 349.03 g/mol

The molar mass of 238UF6 is

238.05 + (6)(18.998) = 352.04 g/mol

The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation 10.35:

Thus passing UF6 containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% 235UF6.

2. B To obtain 99.0% pure 235UF6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated:

final purity = (initial purity)(separation)n

In this case, 0.990 = (0.00720)(1.0043)n, which can be rearranged to give

$0.990 0.00720 = ( 1.0043 ) n = 138$

Taking the logarithm of both sides gives

Thus at least a thousand effusion steps are necessary to obtain highly enriched 235U. Figure 10.18 "A Portion of a Plant for Separating Uranium Isotopes by Effusion of UF" shows a small part of a system that is used to prepare enriched uranium on a large scale.

Exercise

Helium consists of two isotopes: 3He (natural abundance = 0.000134%) and 4He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4He by a process of gaseous effusion.

1. Calculate the ratio of the effusion rates of 3He and 4He and thus the enrichment possible in a single effusion step.
2. How many effusion steps are necessary to yield 99.0% pure 3He?

Answer: a. ratio of effusion rates = 1.15200; one step gives 0.000154% 3He; b. 96 steps

Figure 10.18 A Portion of a Plant for Separating Uranium Isotopes by Effusion of UF6

The large cylindrical objects (note the human for scale) are so-called diffuser (actually effuser) units, in which gaseous UF6 is pumped through a porous barrier to partially separate the isotopes. The UF6 must be passed through multiple units to become substantially enriched in 235U.

Rates of Diffusion or Effusion

Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:

Equation 10.36

$K E ¯ = 1 2 M 1 v rms 1 2 = 1 2 M 2 v rms 2 2$

Multiplying both sides by 2 and rearranging give

Equation 10.37

$v rms 2 2 v rms 1 2 = M 1 M 2$

Taking the square root of both sides gives

Equation 10.38

$v rms 2 v rms 1 = M 1 M 2$

Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation 10.38 shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.

Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure 10.19 "The Wide Variation in Molecular Speeds Observed at 298 K for Gases with Different Molar Masses" for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The postulates of the kinetic molecular theory of gases lead to the following equation, which directly relates molar mass, temperature, and rms speed:

Equation 10.39

$v rms = 3 R T M$

In this equation, vrms has units of meters per second; consequently, the units of molar mass M are kilograms per mole, temperature T is expressed in kelvins, and the ideal gas constant R has the value 8.3145 J/(K·mol).

Figure 10.19 The Wide Variation in Molecular Speeds Observed at 298 K for Gases with Different Molar Masses

The lightest gases have a wider distribution of speeds and the highest average speeds.

Note the Pattern

Molecules with lower masses have a wider distribution of speeds and a higher average speed.

Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure 10.20 "The Path of a Single Particle in a Gas Sample". The average distance traveled by a molecule between collisions is the mean free pathThe average distance traveled by a molecule between collisions.. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10−8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 1010 m (about 6 million miles).

Figure 10.20 The Path of a Single Particle in a Gas Sample

The frequent changes in direction are the result of collisions with other gas molecules and with the walls of the container.

Note the Pattern

The denser the gas, the shorter the mean free path.

Example 18

Calculate the rms speed of a sample of cis-2-butene (C4H8) at 20°C.

Given: compound and temperature

Strategy:

Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 10.39 to calculate the rms speed of the gas.

Solution:

To use Equation 10.39, we need to calculate the molar mass of cis-2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C4H8, so its molar mass is 56.11 g/mol. Thus

or approximately 810 mi/h.

Exercise

Calculate the rms speed of a sample of radon gas at 23°C.

The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. In Section 10.8 "The Behavior of Real Gases", we explore how the behavior of real gases differs from that of ideal gases.

Summary

The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square (rms) speed (vrms). The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a Boltzmann distribution, in which some molecules have higher or lower speeds (and kinetic energies) than average. Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The mean free path of a molecule is the average distance it travels between collisions.

Key Takeaway

• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.

Key Equations

Average kinetic energy

Equation 10.31: $KE¯ = 12 mv2¯$

Root mean square speed

Equation 10.33: $vrms = v12 +v22+v32+ ⋅⋅⋅ +vn2n$

Graham’s law for diffusion and effusion

Equation 10.35: $r1r2 = M2M1$

Kinetic molecular theory of gases

Equation 10.39: $vrms = 3RTM$

Conceptual Problems

1. Which of the following processes represents effusion, and which represents diffusion?

1. helium escaping from a hole in a balloon
2. vapor escaping from the surface of a liquid
3. gas escaping through a membrane
2. Which postulate of the kinetic molecular theory of gases most readily explains the observation that a helium-filled balloon is round?

3. Why is it relatively easy to compress a gas? How does the compressibility of a gas compare with that of a liquid? A solid? Why? Which of the postulates of the kinetic molecular theory of gases most readily explains these observations?

4. What happens to the average kinetic energy of a gas if the rms speed of its particles increases by a factor of 2? How is the rms speed different from the average speed?

5. Which gas—radon or helium—has a higher average kinetic energy at 100°C? Which has a higher average speed? Why? Which postulate of the kinetic molecular theory of gases most readily supports your answer?

6. What is the relationship between the average speed of a gas particle and the temperature of the gas? What happens to the distribution of molecular speeds if the temperature of a gas is increased? Decreased?

7. Qualitatively explain the relationship between the number of collisions of gas particles with the walls of a container and the pressure of a gas. How does increasing the temperature affect the number of collisions?

8. What happens to the average kinetic energy of a gas at constant temperature if the

1. volume of the gas is increased?
2. pressure of the gas is increased?
9. What happens to the density of a gas at constant temperature if the

1. volume of the gas is increased?
2. pressure of the gas is increased?
10. Use the kinetic molecular theory of gases to describe how a decrease in volume produces an increase in pressure at constant temperature. Similarly, explain how a decrease in temperature leads to a decrease in volume at constant pressure.

11. Graham’s law is valid only if the two gases are at the same temperature. Why?

12. If we lived in a helium atmosphere rather than in air, would we detect odors more or less rapidly than we do now? Explain your reasoning. Would we detect odors more or less rapidly at sea level or at high altitude? Why?

Numerical Problems

1. At a given temperature, what is the ratio of the rms speed of the atoms of Ar gas to the rms speed of molecules of H2 gas?

2. At a given temperature, what is the ratio of the rms speed of molecules of CO gas to the rms speed of molecules of H2S gas?

3. What is the ratio of the rms speeds of argon and oxygen at any temperature? Which diffuses more rapidly?

4. What is the ratio of the rms speeds of Kr and NO at any temperature? Which diffuses more rapidly?

5. Deuterium (D) and tritium (T) are heavy isotopes of hydrogen. Tritium has an atomic mass of 3.016 amu and has a natural abundance of 0.000138%. The effusion of hydrogen gas (containing a mixture of H2, HD, and HT molecules) through a porous membrane can be used to obtain samples of hydrogen that are enriched in tritium. How many membrane passes are necessary to give a sample of hydrogen gas in which 1% of the hydrogen molecules are HT?

6. Samples of HBr gas and NH3 gas are placed at opposite ends of a 1 m tube. If the two gases are allowed to diffuse through the tube toward one another, at what distance from each end of the tube will the gases meet and form solid NH4Br?


text:
## Bulletin of the Belgian Mathematical Society - Simon Stevin

### Metrizability of totally ordered groups of infinite rank and their completions

#### Abstract

In [4], Ochsenius and Schikhof ask the following question. Given a totally ordered group $G$ with a cofinal sequence, if every element of its Dedekind completion $G^\#$ is the supremum of a sequence in $G$, does it follow that $G^\#$ is metrizable? We answer their question by studying topological properties of a family of totally ordered groups, $\Gamma_\alpha$, and their completions $\Gamma_\alpha^\#$. Furthermore we obtain for this family conditions both necessary and sufficient for the metrizability of $\Gamma_\alpha^\#$.

#### Article information

Source
Bull. Belg. Math. Soc. Simon Stevin, Volume 14, Number 5 (2007), 969-977.

Dates
First available in Project Euclid: 17 December 2007

https://projecteuclid.org/euclid.bbms/1197908907

Digital Object Identifier
doi:10.36045/bbms/1197908907

Mathematical Reviews number (MathSciNet)
MR2379001

Zentralblatt MATH identifier
1131.06012

#### Citation

Olivos, E.; Soto, H.; Mansilla, A. Metrizability of totally ordered groups of infinite rank and their completions. Bull. Belg. Math. Soc. Simon Stevin 14 (2007), no. 5, 969--977. doi:10.36045/bbms/1197908907. https://projecteuclid.org/euclid.bbms/1197908907


text:
# qiskit.aqua.operators.OperatorBase.permute¶

abstract OperatorBase.permute(permutation)[source]

Permutes the qubits of the operator.

Parameters

permutation (List[int]) – A list defining where each qubit should be permuted. The qubit at index j should be permuted to position permutation[j].

Return type

OperatorBase

Returns

A new OperatorBase containing the permuted operator.

Raises

AquaError – if indices do not define a new index for each qubit.


text:
# Review on Probabilities

This is an evolving note, such that new concepts will constantly be added.

1.  Events
• $\mathbb{P}[A\ or\ B] = \mathbb{P}[A\bigcup B] = \mathbb{P}[A] + \mathbb{P}[B] - \mathbb{P}[A \bigcap B] \le \mathbb{P}[A] + \mathbb{P}[B]$
• $\mathbb{P}[A\ and\ B] = \mathbb{P}[A\bigcap B]$
• If $A \Rightarrow B$, then $\mathbb{P} \le \mathbb{P}[B]$
2.  Expectations
• $E[X] = \int xf(x)dx$
• $E[c] = c$
• $E[X + c] = E[X] + c$
• $E[cX] = cE[X]$
• $E[X+Y] = E[X] + E[Y]$
• $E[XY] = E[X]E[Y]$ if $X$ and $Y$ are independent
• $E[\vec{X}] = (E[X_1] \ E[X_2] \ ... \ E[X_d])^T$
• If $X \ge 0$, $E[X] =\int_0^\infty \mathbb{P}[X\ge 0]dt$
• Law of large number: for infinite sequence of i.i.d. random variable $X$, sample mean $\bar{X}_n = \frac{1}{n}(X_1+X_2+...+X_n)$ converges to $E[X_i]$ as $n$ approaches $\infty$
3. Conditional Probabilities and Conditional Expectations
• $\mathbb{P}[A|B] = \frac{\mathbb{P}[A,B]}{\mathbb{P}[B]}$
• Since $\mathbb{P}[A_n,...,A_1] = \mathbb{P}[A_n|A_{n-1},...,A_1] \cdot \mathbb{P}[A_{n-1},...,A_1]$, $\mathbb{P}[\bigcap\limits_{k=1}^nA_k] = \prod\limits_{k=1}^n\mathbb{P}[A_k|\bigcap\limits_{j=1}^{k-1}A_j]$
• Conditional density of $X$ given $Y$: $f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$
• Conditional expectation of $X$ given $Y$: $E[X|Y=y] = \int^\infty_{-\infty} x f_{X|Y}(x|y)dx$
• Law of Total Expectation $E[X] = E_Y[E_{X|Y}[X|Y]]$
4. Variance
• $Var(X) = \int(x-\mu)^2f(x)dx$
• $Var(X) = E[(X - E[X])^2]$
• $Var(X) = E[X^2] - (E[X])^2$
• $Var(c) = 0$
• $Var(X + c) = Var(X)$
• $Var(cX) = c^2Var(X)$
• $Var(X+Y) = Var(X)Var(Y)$ if $X$ and $Y$ are independent
• $Var(\sum\limits_{i=1}^NX_i) = \sum\limits_{i=1}^NVar(X_i)$ if the $X_i...X_n$ are independent
5. Covariance
• $Cov(X,Y) = \int\int(x-E[x])(y-E[y])f(x,y)dxdy$
• $Cov(X,Y) = E[(X - E[X])(Y - E[Y])] = E[XY] - E[X][Y]$
• $Var(X) = Cov(X,X)$
• $Cov(X,c) = 0$
• $Cov(aX, bY) = abCov(X,Y)$
• $Cov(\vec{X}) = E[(\vec{X} - E[\vec{X}])(\vec{X} - E[\vec{X}])^T] = E[\vec{X}\vec{X}^T] - E[\vec{X}](E[\vec{X}])^T$
• $Cov(\vec{X}|\vec{Y}) = E[(\vec{X} - E[\vec{X}|\vec{Y}])(\vec{X} - E[\vec{X}|\vec{Y}])^T|\vec{Y}]$
6. Mean Square Error: $MSE(\hat{\theta}) = E[(\hat{\theta} - \theta)^2] = E[(\hat{\theta}-E[\hat{\theta}])^2] + (E[\hat{\theta}]-\theta)^2 = Var(\hat{\theta}) + Bias(\hat{\theta})^2$, more detail explanation can be found on wikipedia.
7. Inequalities
• Boole’s inequality (Union bound): for a countable set of events $A_1, A_2,...$, $\mathbb{P}[\bigcup\limits_i A_i] \le \sum\limits_{i} \mathbb{P}[A_i]$
• Jensen: for convex $f$, $f(E[X]) \le E[f(X)]$; for concave $f$, $f(E[X]) \ge E[f(X)]$
• Markov: If $X \ge 0$ then for all $t >0$, $\mathbb[P][X\ge t] \le \frac{E[X]}{t}$
• Chebyshev-Cantelli: for $t>0$, $\mathbb{P}[|X-E[X]|\ge t] \le \frac{Var(X)}{Var(X) + t^2}$
• Chebyshev’s Association: let $f$ and $g$ be nondecreasing (nonincreasing) real-valued functions defined on the real line. If $X$ is a real-valued random variable then, $E[f(X)g(X)] \ge E[f(X)]E[g(X)]$
• $E[f(X)g(X)] \le E[f(X)]E[g(X)]$
• Harris’: extends Chebyshev’s Association to functions $f,g:\mathbb{R}^n\rightarrow \mathbb{R}$
• Chernoff bound: for any $t\in \mathbb{R}$, for any random variable $X$, $\mathbb{P}[X\ge t] = \mathbb{P}[e^{sX}\ge e^{st}] \le \frac{E[s^{sX}]}{e^{st}}$
• Cauchy-Schwarz: if $E[X^2]<\infty$ $E[Y^2]<\infty$, then $|E[XY]|\le \sqrt{E[X^2]E[Y^2]}$
• Hoeffding’s inequality: Let $X$ be a random variable with $E[X]=0, a \le X \le b$. Then for $s>0, E[\exp(sX)] \le \exp(s^2\frac{(b-a)^2}{8})$
• Hoeffding’s tail: Let $X_1,...,X_n$ be independent bounded random variable such that $X_i$ falls in the interval $[a_i,b_i]$ with probability one. Then for any $t>0$, $\mathbb{P}[S_n-E[S_n]\ge t] \le \exp\Big(\frac{-2t^2}{\sum\limits_{i=1}^n(b_i-a_i)^2}\Big)$ and $\mathbb{P}[S_n-E[S_n]\le -t] \le \exp\Big(\frac{-2t^2}{\sum\limits_{i=1}^n(b_i-a_i)^2}\Big)$
• Bernstein’s inequality: Let $X_1,...,X_n$ be independent real-valued random variables with $0$ mean, and assume that $|X_i|\le c$ with probability $1$. Let $\sigma^2 = \frac{1}{n}\sum\limits_{i=1}^n Var[X_i]$. Then, for any $\epsilon > 0$, $\mathbb{P}[\frac{1}{n}\sum\limits_{i=1}^n X_i > \epsilon] \le \exp(-\frac{n\epsilon^2}{2\sigma^2 + 2c\epsilon/3})$.
• when $\sigma^2 < \epsilon$, the upper bound behaves like $\exp(-n\epsilon)$ instead of the $\exp(-n\epsilon^2)$ guaranteed by Hoeffding’s tail.
• McDiarmid’s inequality: let $X_1, X_2,...,X_n$ be i.i.d. random variables, and let $f: X^n \mapsto \mathbb{R}$ be s.t. $\forall i, a_1, a_2, ..., a_n, a'_i, |f(a_1, ..., a_i, ..., a_n) - f(a_1, ..., a'_i, ..., a_n)| \le c_i$. Then $\mathbb{P}[f(x_1, x_2, ..., x_n) - E[f(x_1, x_2,...,x_n)] \ge \epsilon] \le \exp\Big(-\frac{2\epsilon^2}{\sum\limits_{i=1}^n c_i^2}\Big)$.
• Azuma-Hoeffding’s inequality: let $x_0, x_1, ..., x_n$ be a martingale (see definition of martingale below) s.t. $\forall k, |x_k - x_{k-1}| < c_k$ for some constant $c_k$, then $\mathbb{P}[x_n - x_0 \ge \epsilon] \le \exp\Big(- \frac{\epsilon^2}{2\sum\limits_{k=1}^n c_k^2}\Big)$.
8. Variables
• Rademacher random variables: $\sigma_1,...,\sigma_n$ s.t. $\mathbb{P}[\sigma_i = 1] = \mathbb{P}[\sigma_i = -1] = 1/2, \forall i \in [n]$
• Rademacher averages: for a set of vectors $S \subseteq \mathbb{R}^n$, the Rademacher average is defined as $\mbox{Rad}(S) = E_{\sigma}[\sup\limits_{v\in S}\langle v,\sigma \rangle_n]$, where $\langle v, \sigma \rangle_n = \frac{1}{n}\sum\limits_{i=1}^n v_i\sigma_i$
• Martingale is a special sequence of random variables
• A discrete-time martingale is a discrete-time stochastic process $X_1, X_2, ... , X_m$ that satisfies
1. $E[|X_i|] < \infty$
2. Let filtration $\mathcal{F}_i = \sigma(X_1, X_2, ..., X_i)$ then $E[X_{i+1}|\mathcal{F}_i] = X_i$. e.g. $\mathcal{F}_i = \{X_1,X_2,...,X_i\}$
9.  Other
• If $\mathbb{P}[X > t]\le F(t)$, then with probability at least $1 - \delta$, $X\le F^{-1}(\delta)$.
• Moment generating function of a random variable $X$ is $M_X(t) = E[e^{tX}]$ where $t\in\mathbb{R}$