Question

At $100°C$ the rms speed of nitrogen molecules is $576m/s$. Nitrogen at $100°C$ and a pressure of $2.0atm$ is held in a container with a $10cm\times 10cm$ square wall. Estimate the rate of molecular collisions (collisions/s) on this wall.

Short answer

The rate of molecular collisions on the wall is$3.78\times {10}^{25}collisions/sec$

Step-by-step solution

Given information and formula used

Given : At $100°C$the rms speed of nitrogen molecules : $576m/s$Nitrogen temperature : $100°C$

Nitrogen pressure : $2.0atm$

Container's square wall : $10cm\times 10cm$

Theory used :

The gas molecules collide with a wall in an elastic collision, rebounding with a change in velocity. The rate of molecular collisions if the molecules collide $N_{coll}$time $∆t$in time will be given by

$\frac{N_{coll}}{∆t}=\frac{1}{2}\frac{N}{V}A{v}_x$

Where $A$is the wall's area, $v_x$is the $x$velocity component, and $(N/V)$is the number density. The container's surface area is

$A=10cm\times 10cm=100c{m}^2=100\times {10}^{-4}{m}^2$

The following equation relates linear velocity to root mean square velocity :

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

As a result,

$\frac{N_{coll}}{∆t}=\frac{1}{6}\frac{N}{V}A{v}_{rms}$

The ideal gas law is :

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

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

Estimating the rate of molecular collisions 

Because the pressure is given in atm, we must convert it to Pascal.

$p=(2atm)\frac{1.013x{10}^{5}Pa}{1atm}=2.026x{10}^{5}Pa$

The Celsius scale is supplied by and the Kelvin scale is given by

$T_K=T_c+273=100°C+273=373K$

We use $N/V=\frac{p}{k_{B}T}$to calculate the number density number by plugging the values for $p,T\mathrm{and}{k}_B$:

$$\begin{gathered} \frac{N}{V}=\frac{2.026\times {10}^{5}Pa}{(1.38x{10}^{-23}J/K)(373K)} \\ =3.93\times {10}^{25}{m}^{-3} \end{gathered}$$

To determine the rate $\frac{N_{coll}}{∆t}$, we now insert the values for $(N/V),A,\mathrm{and}{v}_{rms}$into the equation

$$\begin{gathered} =\frac{1}{6}\frac{N}{V}{v}_{rms} \\ =\frac{1}{6}(3.9(3.93\times {10}^{25}{m}^{-3})(100\times {10}^{-4}{m}^2)(576m/s) \\ =\mathbf{3}\mathbf{.}\mathbf{78}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{25}}collisions/sec \end{gathered}$$