# Calculating force and pressure of an ideal gas on a wall [duplicate]

The text below is from a discussion on ideal gases and the kinetic theory of gases:

After the collision the molecule must travel a distance $$l$$ to the opposite wall, and then back across this same distance before colliding again with the wall in question. This determines the time between successive collisions with a given wall; the number of collisions per second will be $$v / 2 l$$. The force $$F$$ exerted on the wall is the rate of change of the momentum, given by the product of the momentum change per collision and the collision frequency:

$$F=\frac{d\left(m v_{x}\right)}{d t}=\left(2 m v_{x}\right) \times\left(\frac{v_{x}}{2 l}\right)=\frac{mv_{x}^{2}}{l}$$

Pressure is force per unit area, so the pressure $$P$$ exerted by the molecule on the wall of cross-section $$l^{2}$$ becomes

$$P=\frac{m v^{2}}{l^{3}}=\frac{m v^{2}}{V}$$

in which $$V$$ is the volume of the box.

To find pressure, the text considers the area of the wall instead of the area of contact between the molecule and the wall. Why is that the case?

According to the text, force on the wall exerted by the molecule is the change in momentum times frequency of collisions. Isn’t force supposed to be momentum over the time in which the molecule is in contact with the wall?

• @Mathew Mahindaratne I’m just not getting one thing -why is the time that’s considered to find the rate of change of momentum not the time for which the gas particle is in contact with the plane? Jun 11, 2020 at 14:01
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• @Richie Harvy: I thought I explained it clearly. Newton's secondlaw says $F = ma$ where $a$ is acceleration, which is rate of change of velocity, $v_x$. Each collision change velocity. Time take to change this velocity change is time between two collisions, meaning time take to travel $2\ell$. That equals to $2\ell v_x$. Jun 11, 2020 at 15:37