Question

A $10cm\times 10cm\times 10cm$ box contains $0.010mol$ of nitrogen at $20°C$. What is the rate of collisions (collisions/s) on one wall of the box?

Short answer

The rate of collisions on one wall of the box is$8.9x{10}^{24}$

Step-by-step solution

Given information and formula used  

Given :

Volume of the box : $10cm\times 10cm\times 10cm$

The box contains nitrogen : $0.010mol$

Temperature : $20°C$.

Theory used :

The gas molecules collide with a wall in an elastic collision, rebounding with a change in velocity. The rate of molecular collisions will be given by equation .$\frac{N_{coll}}{∆t}=\frac{1}{2}\frac{N}{V}A{v}_x$ ...(1)

where $A$is the wall's area, is the velocity component, the molecules collide $N$times in time $∆t$and (N/V) is the number density.

The container's surface area is role="math" localid="1649066181954" $$\begin{gathered} A=10cmx10cm=10c{m}^2 \\ =100x{10}^{-4}{m}^2 \end{gathered}$$

The linear velocity to the root mean square velocity is :

$v_x=\frac{1}{\sqrt{3}}{v}_{rms}$

As a result, equation (1) will be

$\frac{N_{coll}}{∆t}=\frac{1}{2\sqrt{3}}\frac{N}{V}A{v}_{rms}$

The ideal gas law is :

$$\begin{gathered} pV=N{k}_{B}T \\ \Rightarrow \frac{N}{V}=\frac{p}{k_{B}T} \end{gathered}$$

Where $k_B$is Boltzmann's constant and in SI unit its value is $k_B=1.38x{10}^{-23}J/K$

The conversion between the Celsius scale and the Kelvin scale is given by $T_K=T_c+273=20°C+273=293K$

Calculating the rate of collisions on one wall of the box 

The ideal gas law in the form might be used to compute the pressure :

$$\begin{gathered} p=\frac{nRT}{V} \\ =\frac{(0.01mol)(8.314J/molK)(293K)}{10\times {10}^{-4}{m}^3} \\ =2.43x{10}^{4}Pa \end{gathered}$$

Plugging in the values of $p,T\mathrm{and}{k}_{\mathit{B}}$:

$$\begin{gathered} N/V=\frac{p}{k_{B}T} \\ =\frac{2.43x{10}^{4}Pa}{(1.38x10-23J/K)(293K)} \\ =6x{10}^{24}{m}^{-3} \end{gathered}$$

Nitrogen has a molecular mass of $14u$. However, because nitrogen is a diatomic gas, one molecule has a molecular mass of$m=28u$. Converting to kg :

$$\begin{gathered} m=28u\times \frac{1.66x10-27kg}{1u} \\ =46.48x{10}^{-27}kg \end{gathered}$$

The values are used to get $v_{rms}$:

$$\begin{gathered} v_{rms}=\sqrt{\frac{3k_{B}T}{m}} \\ =\sqrt{\frac{3(1.38x{10}^{-23}J/K)(293K)}{46.48x{10}^{27}kg}} \\ =511m/s \end{gathered}$$

To derive the rate$N_{coll}$, we plug the values for $(N/V),A,\mathrm{and}{v}_{rms}$to get :

$$\begin{gathered} N_{coll}=\frac{1}{2\sqrt{3}}\frac{N}{V}A{v}_{rms} \\ =\frac{1}{2\sqrt{3}}(6x{10}^{24}{m}^{-3})(100x{10}^{-4}{m}^2)(511m/s) \\ =\boxed{\mathbf{8}\mathbf{.}\mathbf{9}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{24}}}collisions/s \end{gathered}$$