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The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be?
1. 2
2. 1.58
3. 1.6
4. 1.31

I think the answer should be 1.6 km/h but the answer given in book is 1.31 km/h. Most probable speed should be the mode of the given speeds, which is equal to 1.6 km/h. Am I correct? If I am wrong, then what's the definition of most probable speed and how do I solve this problem?

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    $\begingroup$ I would give the textbook -1 on this question. Most probable is ambiguous here, leading me to an answer I had to delete. $\endgroup$ – Oscar Lanzi Mar 9 '19 at 23:30
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    $\begingroup$ (-1) The question does not make sense. I don't see any compelling reason why one should assume that the question is asking for the mode of the population (1.40 or 1.31, which depends on how you calculate(!)) rather than the mode of the sample (1.60). Furthermore, it is illogical to claim that the answer is 1.31 simply because there is no answer corresponding to 1.40. We are supposed to answer based on chemical principles, not by reading people's minds and reverse-engineering the questions they set. $\endgroup$ – orthocresol♦ Mar 10 '19 at 2:07
  • $\begingroup$ @orthocresol - I think we're having a violent agreement. I totally agree that this is a very lousy problem. It was more of a matter of trying to figure out the logic of the 1.31 answer rather than agreeing that 1.31 was "right." $\endgroup$ – MaxW Mar 10 '19 at 5:56
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    $\begingroup$ I don't understand why this was closed as a homework question. The OP did state his analysis. Look at all the debate that it took us to come to a consensus on the book's "answer." $\endgroup$ – MaxW Mar 10 '19 at 18:38
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    $\begingroup$ Mode. - The number which appears most often in a set of numbers. The OP was right the answer should be 1.6. The book answer is just wrong. $\endgroup$ – MaxW May 15 '20 at 16:22
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Following up on user andselisk's answer ...

The problem is weird for several different reasons.

  • First the solution depends on assuming that the velocity of the gas molecules follow the Maxwell-Boltzmann distribution which is weird for only 5 molecules. The problem also doesn't state that the 5 molecules are a sample from a much larger number of molecules.

  • Second the answer depends on how you calculate to get a result.

For the Maxwell-Boltzmann distribution there are relationships between the most probable speed $v_p$, the average speed $\langle v \rangle$, and the root-mean-square speed $v_{rms}$. Following the Wikipedia article:

\begin{align} \langle v \rangle &= \frac{2}{\sqrt{\pi}}v_p\\ v_\mathrm{rms} &= \sqrt{\frac{3}{2}}v_p \end{align}

Rearranging these two equations we get two different ways to calculate the most probable speed.

\begin{align} v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle\\ v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms} \end{align}

Thus using the mean (average) speed we get:

\begin{align} \langle v \rangle &= \frac{2 + 1.5 + 1.6 + 1.6 + 1.2}{5} = 1.58\\ v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle = 1.40 \end{align}

However using the rms speed we get:

\begin{align} v_\mathrm{rms} &= \sqrt{\frac{2^2 + 1.5^2 + 2\cdot 1.6^2 + 1.2^2}{5}} = 1.60\\ v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms} = 1.31 \end{align}

The difference is due to small sample statistics and not really a statistically significant result. However the difference is absolute proof that the sample doesn't follow the Maxwell-Boltzmann distribution.

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  • $\begingroup$ I don't necessarily agree with the implications of the final statement. Like anything else in statistical mechanics or thermodynamics, the results of the MB distribution are not guaranteed to hold true for small systems. In the (admittedly meaningless) limit of having one particle, it is literally impossible to have a different rms, mean, and most probable speed, no matter how accurately you measure the speed of that one particle. So... $\endgroup$ – orthocresol♦ Mar 10 '19 at 2:15
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    $\begingroup$ ... the problem is not that the sample is obeying some different non-MB distribution, nor is it about measurement accuracy or confidence intervals. The problem is that the sample is too small, so the predictions of the MB distribution are not manifest. $\endgroup$ – orthocresol♦ Mar 10 '19 at 2:15
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    $\begingroup$ @orthocresol - The point is that MB is a distribution which assumes a "large" number of particles. It is absolutely meaningless to discuss MB with only 5 particles. // If the 5 particles were from large number of particles which did obey MB, then the sample size just isn't large enough to get a decent measurement. $\endgroup$ – MaxW Mar 10 '19 at 5:53
  • $\begingroup$ Yup, I agree with that assessment. Consider it just me being rather nitpicky about the wording, then. $\endgroup$ – orthocresol♦ Mar 10 '19 at 22:27
  • $\begingroup$ @orthocresol The question is probably from a book for exam preparation, and as such wants the answers to be ~calculated~ according to a pre-mentioned formula. Yes, the problem doesn't make sense if we do about the definitions "strictly", but i don't think thats the answer OP wants. A lot of times on Chem SE such exam questions are shot down because they are based on outdated concepts, but simply stating "Well we don't believe that any longer" doesn't help anyone- not the other students having the same problem and wanting an explanation fitting the demands of the pattern, nor the asker, anyone $\endgroup$ – Micelle May 15 '20 at 14:04
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The book is absolutely correct. Average speed is not the same as rms speed which is defined as

$$c_\text{rms} = (3kT/m)^{1/2} = \left(\frac{\sum_{i = 1}^N v_i^2}{N}\right)^{1/2}, \tag{1}$$

where $v_i$ are the velocities of each particle with the total of $N$, and the most probable speed is even smaller:

$$c_\text{mp} = (2kT/m)^{1/2} \tag{2}$$

For the derivations of the formulas above see e.g. [1, p. 69]. Combining both equations

$$ \begin{align} c_\text{mp} &= \left(\frac{2\sum_{i = 1}^N v_i^2}{3N}\right)^{1/2} \\ &= \left(\frac{2 × \left[(\pu{2 km s-1})^2 + (\pu{1.5 km s-1})^2 + 2 × (\pu{1.6 km s-1})^2 + (\pu{1.2 km s-1})^2\right]}{3 × 5}\right)^{1/2} \\ &= \pu{1.31 km s-1} \end{align} $$

References

  1. Castellan, G. W. Physical Chemistry, 3rd ed.; Addison-Wesley, 1983. ISBN 978-0-201-10386-1.
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    $\begingroup$ Agree with the answer totally But when I started with the average velocty which comes out to be $1.58 Km/s$ now I must be able to multiply it by $\sqrt{\pi}/2$ (as $v_{av}=\frac{2}{\sqrt{\pi}} \sqrt{\frac{2RT}{M}}$) and should get the answer as 1.30 but the answer comes out to be 1.400 , I am not getting it please help $\endgroup$ – Advil Sell Mar 9 '19 at 19:54
  • $\begingroup$ @AdvilSell I'm afraid I don't understand what you are using average velocity for (and how you determined it) and where $\sqrt{π}/2$ term came from. $\endgroup$ – andselisk♦ Mar 9 '19 at 20:00
  • $\begingroup$ Desmos can you please check this out I have shown calculations and formula ... please check it once $\endgroup$ – Advil Sell Mar 9 '19 at 20:03
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    $\begingroup$ @andselisk - $\sqrt{\pi}/2$ factor came from relationship between average speed and most probable speed. See en.wikipedia.org/wiki/… $\endgroup$ – MaxW Mar 9 '19 at 20:31
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    $\begingroup$ The gist is that the problem ignores the confidence interval of the measurement. So if you compute average speed and then use that to get most probable speed, then you get 1.40. However if you calculate rms and use that to calculate most probable speed then you get 1.31. There are so few measurements that the two values are not statistically significantly different. $\endgroup$ – MaxW Mar 9 '19 at 20:34

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