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?

Question

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?

Answer 1

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:

$\langle v \rangle = \frac{2}{\sqrt{\pi}}v_p$

and

$v_{rms} = \sqrt{\frac{3}{2}}v_p$

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

$ v_p = \frac{\sqrt{\pi}}{2}\langle v \rangle$

and

$v_p = \sqrt{\frac{2}{3}}v_{rms}$

Thus using the mean (average) speed we get:

$\langle v \rangle = \frac{2 + 1.5 + 1.6 + 1.6 + 1.2}{5} = 1.58$

$ v_p = \frac{\sqrt{\pi}}{2}\langle v \rangle = 1.40$

However using the rms speed we get:

$v_{rms} = \sqrt{\frac{2^2 + 1.5^2 + 2\cdot 1.6^2 + 1.2^2}{5}} = 1.60$

$v_p = \sqrt{\frac{2}{3}}v_{rms} = 1.31$

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.

Answer 2

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

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

References

1. Castellan, G. W. Physical Chemistry, 3rd ed.; Addison-Wesley, 1983. ISBN 978-0-201-10386-1.