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?


> 
> 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?




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](https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution#Typical_speeds) 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. 


 6

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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