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<p><font face="serif">This is a classic example of something that
initially looks hard, looks doable if you understand the concept
behind the problem, and then is almost trivial if you have a
program like Mathematica (or a calculator ) that will do
symbolic calculus.</font></p>
<p><font face="serif">So you are looking at the Boltzmann
distribution function which you should remember from CHM 332.</font></p>
<p><font face="serif"><img style="vertical-align: middle"
src="cid:part1.CF756387.46884ECE@mercer.edu" alt="$f(v)= 4 \pi
\left(\frac{m}{2 \pi kT}\right)^{3/2} v^2 e^{-mv^2/2kT}$"></font><br>
</p>
The most probable speed occurs at the top of the Boltzmann
distribution curve (its maximum), so we want to take the derivative
of the function above with repect to V, set it equal to zero, and
determine the value of v.<br>
<br>
When I do this in Mathematica I get three answers.<br>
<br>
<img style="vertical-align: middle"
src="cid:part2.C90A8B7B.465A6695@mercer.edu" alt="$m=0$">, <img
style="vertical-align: middle"
src="cid:part3.19E410FE.578D54D0@mercer.edu" alt="$v=0$">, and <img
style="vertical-align: middle"
src="cid:part4.92444530.B19F68D6@mercer.edu"
alt="$m=\frac{2kT}{v^2}$">.<br>
<br>
only one of these is physically correct -- and it has to be the
third one. Rearrange it for <img style="vertical-align: middle"
src="cid:part5.01D7C88E.46563813@mercer.edu" alt="$v$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part6.642249D4.978AFDFC@mercer.edu" alt="$v =
\sqrt{\frac{2kT}{m}}$"><br>
<br>
and that is the answer to the problem. <br>
<br>
I have attached my Mathematica notebook so you can see what I did.<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
</pre>
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