# Kinetic theory of gases and its derivation

In the kinetic theory of gases, the time taken for a change in momentum is $$2l/v(x)$$, where $$l$$ is the length of a cube and $$v(x)$$ is the velocity in the $$x$$ direction.

Shouldn't the time correspond to the momentum change of the molecule, such that the force can be calculated as $$F=\Delta p/\Delta t$$, where $$\Delta t$$ is the time taken by molecule just to strike and rebound from the wall?

• Mar 20, 2023 at 19:14

If I understand your question correctly, you are asking why the time interval $$\Delta t$$ in $$F=\frac{\Delta p}{\Delta t}$$ is $$\frac{2 \ell}{v_x}$$ rather than $$\frac{\ell}{v_x}$$.

Recall that in the classical, kinetic theory-based derivation of the ideal gas law, we are interested in the force imparted against one particular wall of the (say cubical) container, so that we can compute the pressure as $$P=\frac{F}{A}.$$ Thus, we only care about collisions that a gas molecule has with that wall. This immediately restores the factor of two, since after each molecule collides with a wall, it has to travel to the opposite wall, and bounce back to collide again.