Question

A beam of hydrogen molecules $\left(H_2\right)$ is directed towards a wall, at an angle of ${55}^o$ with the normal to the wall. Each molecule in the beam has a speed of $1.0km/s$ and a mass of $3.3\times {10}^{-24}g$. The beam strikes the wall over an area of $2.0c{m}^2$, at the rate of ${10}^{23}$ molecules per second. What is the beam's pressure on the wall?

Answer 1

In the reflection process, only the normal component of the momentum changes, so for one molecule the change in momentum is $2mv\cos \theta$, where $m$ is the mass of the molecule $v$ is its speed and $\theta$ is the angle between its velocity and the normal to the wall. If $N$ molecules collide with the wall. then the change in their total momentum is $2Nmv\cos \theta$, and if the total time taken for the collision is $\Delta t$, then the average rate fo change of the total momentum is $2\left(N/\Delta t\right)mv\cos \theta$. This is the average force exerted by the $N$ molecules on the wall, and the pressure is the average force per unit area:
$p=\frac{2}{A}\left(\frac{N}{\Delta t}\right)mv\cos \theta$
$=\left(\frac{2}{2.0\times {10}^{-4}{m}^2}\right)(1.0\times {10}^{23}{s}^{-1})(3.3\times {10}^{-27}kg)(1.0\times {10}^{3}m/s)\cos {55}^o$
$=1.9\times {10}^{3}Pa$
We note that the value given for the mass was converted to $kg$ and the value given for the area was converted to $m^2$.