Question

The mass of the ${\mathrm{H}}_2$ molecule is $3.3\times {10}^{-24}\mathrm{g}$. If $1.6\times {10}^{23}$ hydrogen molecules per second strike $2.0{\mathrm{cm}}^2$ of wall at an angle of ${55}^{\circ }$ with the normal when moving with a speed of $1.0\times {10}^5\mathrm{cm}/\mathrm{s}$, what pressure do they exert on the wall?

Short answer

The pressure exerted on the wall by the hydrogen molecules is given by $(3.3\times {10}^{-24}g\cdot 1.0\times {10}^{5}cm/s\cdot \cos ({55}^{\circ })\cdot 1.6\times {10}^{23})/2.0c{m}^2$

Step-by-step solution

Find Momentum of H2 Molecule

The momentum $p$ of one ${\mathrm{H}}_2$ molecule can be found by the equation $p=mv$, where $m$ is the mass and $v$ is the speed. However, since only a component of the molecule's momentum is transferred to the wall, we multiply by the cosine of the angle: $p=m\cdot v\cdot \cos (\theta )=3.3\times {10}^{-24}g\cdot 1.0\times {10}^{5}cm/s\cdot \cos ({55}^{\circ })$

Find Force Exerted by One Molecule per Second

Force $F$ can be found using the definition of momentum and the fact that momentum per unit time is force. Therefore $F=dp/dt$ , and since the change of momentum is equal to the momentum for one molecule per second (as each molecule only hits the wall once), we find $F=p=3.3\times {10}^{-24}g\cdot 1.0\times {10}^{5}cm/s\cdot \cos ({55}^{\circ })$

Find Total Force Exerted per Second

The total force $F_{\mathrm{total}}$ is found by multiplying the force exerted by one molecule per second by the number of molecules hitting the wall per second: $F_{\mathrm{total}}=F\cdot N=3.3\times {10}^{-24}g\cdot 1.0\times {10}^{5}cm/s\cdot \cos ({55}^{\circ })\cdot 1.6\times {10}^{23}$

Calculate Pressure

Pressure $P$ is defined as force per area. So to find the pressure, divide the total force exerted per second by the area of the wall: $P=F_{\mathrm{total}}/A=(3.3\times {10}^{-24}g\cdot 1.0\times {10}^{5}cm/s\cdot \cos ({55}^{\circ })\cdot 1.6\times {10}^{23})/2.0c{m}^2$

Key concepts

Momentum

Momentum, a fundamental concept in physics, is the product of an object's mass and velocity. It's given by the equation
$p=m\times v$,
where $m$ is mass and $v$ is velocity. This quantity is crucial in understanding how objects interact through collisions and forces. For example, when a gas molecule like ${\text{H}}_2$ strikes a wall, the momentum transferred to the wall will determine the force exerted by the collision.
In our problem, due to the angle of impact, only a component of the molecule's momentum is transferred to the wall. Therefore, we actually use a modified equation to calculate the effective momentum:
$p=m\cdot v\cdot \cos (\theta )$. This modification takes into account the direction of the molecule's velocity relative to the surface normal.

Force Exerted by Molecules

The force exerted by molecules, particularly in gases, stems from their continuous motion and collisions against surfaces. According to Newton's second law of motion, force is the change in momentum over time ($F=\frac{dp}{dt}$).
In a situation where numerous molecules collide with a surface, the overall force is the sum of all individual forces exerted by each molecule. Each time a molecule with momentum $p$ hits the wall, it imparts a change in momentum, thus exerting a force. Given that our ${\text{H}}_2$ molecules strike the wall and then ideally bounce off, the force exerted by one molecule in one second is equivalent to its momentum.

Pressure Exerted on Surface

Pressure is defined as the force exerted per unit area of a surface. The formula is given by
$P=\frac{F}{A}$,
where $P$ is pressure, $F$ is force, and $A$ is the area over which the force is distributed. The pressure exerted by a gas on a surface is a result of the collective force of its molecules incessantly colliding with that surface.
For instance, in our problem, the ${\text{H}}_2$ molecules create pressure on the wall not just by their speed but also by their sheer number. By dividing the total force due to all molecules hitting the wall per second by the area of the wall, the pressure exerted on the wall is obtained.

Cosine Angle Effect in Momentum

The cosine angle effect is important in physics when calculating the component of a vector, like momentum, in a particular direction. In a collision where a particle strikes a surface at an angle, only the component of its momentum that is perpendicular to the surface results in force exerted on that surface.
This is why we use the cosine of the angle between the molecule's velocity and the normal to the surface. The modified momentum equation becomes
$p=m\cdot v\cdot \cos (\theta )$,
thus, ensuring we only account for the momentum that's effective in producing pressure on the wall. The cosine effectively scales down the momentum based on the angle, representing a smaller amount of force when the molecule hits the surface at a shallower angle.