Question

$5.0\times {10}^{23}$nitrogen molecules collide with a $10c{m}^2$ wall each second. Assume that the molecules all travel with a speed of $400m/s$ and strike the wall head-on. What is the pressure on the wall?

Short answer

The pressure on the wall is$\mathbf{1}\mathbf{.}\mathbf{86}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}\mathbf{}}\mathit{P}\mathit{a}$

Step-by-step solution

Given information and formula used 

Given :

Number of nitrogen molecules collide each second : $5.0\times {10}^{23}$

Area of wall :$10c{m}^2$

The molecules all travel with a speed of : $400m/s$

Theory used :

The nitrogen molecules collide with the wall and bounce back, changing their momentum in the process. The force exerted by the gas on the wall can be calculated by :

$\left(F_{onwall}\right)=2m{v}_x\frac{N_{coll}}{∆t}$ ...(1)

Calculating the pressure on the wall 

The rate of collision is :

$\begin{array}{rcl}\frac{N_{coll}}{∆t}& =& \frac{5.0\times {10}^{23}}{1s}\\ & =& 5.0\times {10}^{23}{s}^{-1}\end{array}$

Nitrogen has a molecular mass of $14u$. However, because nitrogen is a diatomic gas, one molecule has a molecular mass of $m=28u$. When we convert this to kilograms, we obtain :

$m=28u\times \frac{1.66x{10}^{-27}kg}{1u}=46.5x{10}^{-27}kg$

To get ($F_{onwall}$), we plug the values for localid="1649065101990" $m,v_x\mathrm{and}\frac{\mathit{}{\mathit{N}}_{\mathit{c}\mathit{o}\mathit{l}\mathit{l}}\mathit{}}{\mathit{∆}\mathit{t}}$, and into equation (1) :

$$\begin{gathered} F_{onwall}=2(46.5x{10}^{-27})(400m/s)(5.0x{10}^{23}{s}^{-1}) \\ =18.6N \end{gathered}$$

The pressure is the force per area, so we can get the pressure by :

$$\begin{gathered} p=\frac{{\left(F_{onwall}\right)}_x}{A} \\ =\frac{18.6N}{10\times {10}^{-4}m²} \\ =\boxed{\mathbf{1}\mathbf{.}\mathbf{86}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{4}}\mathit{P}\mathit{a}\mathbf{}} \end{gathered}$$