Question

The speeds of 22particles are as follows (N₁ represents the number of particles that have speed $v_1$):

1. What is${\mathit{v}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$?
2. What is${\mathit{v}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$?
3. What is${\mathit{v}}_{\mathbf{p}}$?

Short answer

1. The average speed of molecules is $3.2\text{cm}/\text{s}$.
2. The RMS speed of molecules is $3.4\text{cm}/\text{s}$.
3. The most probable speed is $4cm/s$.

Step-by-step solution

Given data

The table for number of particles and their corresponding speeds are given in question.

Understanding the concept

Average speed is calculated by adding the speed of each molecules and then by dividing the total number of molecules.

The RMS speed of molecules is given by,

${\text{v}}_{rms}=\sqrt{\frac{{\displaystyle {\sum }_{}^{}{\text{v}}^2}}{\text{N}}}$

(a) Calculate vavg

The average speed of molecules is,

$\begin{array}{c}\text{v}=\frac{{\displaystyle \sum {\text{n}}_i{\text{v}}_i}}{N}\\ =\frac{2\left(1\right)+4\left(2\right)+6\left(3\right)+8\left(4\right)+2\left(5\right)}{2+4+6+8+2}\\ =3.2\text{cm}/\text{s}\end{array}$

Therefore the average speed of molecules is$3.2\text{cm}/\text{s}$.

(b) Calculate vrms

The RMS speed of molecules is

$$\begin{gathered} {\text{v}}_{rms}=\sqrt{\frac{{\displaystyle \sum n_i{v}_i^2}}{N}} \\ \begin{array}{c}\sum n_i{\text{v}}^2=\left(2\right)1^2+\left(4\right)2^2+6{\left(3\right)}^2+\left(8\right)4^2+\left(2\right)5^2\\ =250{\text{cm}}^2/{\text{s}}^2\end{array} \end{gathered}$$

Substitute$250{\text{cm}}^2/{\text{s}}^2$ for $\sum n_i{\text{v}}^2$into the above equation,

$\begin{array}{c}{\text{v}}_{rms}=\sqrt{\frac{250}{2+4+6+8+2}}\\ =3.37\\ \approx 3.4\text{cm}/\text{s}\end{array}$

Therefore the RMS speed of molecules is .$3.4\text{cm}/\text{s}$

(c) Calculate vp

Since the number of particles having speed 4 cm/s is maximum, the most probable speed is$4cm/s$.