According to the kinetic theory of gases, the pressure exerted by a gas on the walls equals.

Answer Details

In the kinetic theory, gas molecules move randomly and collide elastically with the container walls. Each collision reverses the component of a molecule's momentum normal to the wall, so one molecule of mass \(m\) striking a wall at speed \(u\) and rebounding changes its momentum by \(2mu\), and the wall receives that momentum. Newton's second law in its general form says force is the rate at which momentum is transferred: \[F = \frac{\Delta p}{\Delta t}.\] Pressure is force spread over area, \(P = F/A\), so combining the two gives \[P = \frac{1}{A}\,\frac{\Delta p}{\Delta t}.\] Pressure is therefore the rate of change of momentum imparted to the walls, per unit area, which is the description the question requires.

The time factor is the part most often dropped. Momentum imparted per unit area alone is an impulse per unit area, with units \(\text{N s m}^{-2}\), not \(\text{N m}^{-2}\); it would grow without limit the longer you waited, whereas the pressure of a gas in a sealed vessel is steady. Only by dividing the momentum transfer by the time over which it happens do you obtain a constant force and hence a constant pressure. Dividing momentum change by volume instead of area is wrong in the same way, and also produces the units of momentum density rather than pressure.

This reasoning is what leads to the kinetic-theory result \[P = \frac{1}{3}\rho \overline{c^{2}} = \frac{1}{3}\frac{Nm}{V}\overline{c^{2}},\] where \(\overline{c^{2}}\) is the mean square speed. When a question offers several verbal definitions, test each one by its units: the correct statement for pressure must reduce to \(\text{N m}^{-2}\), and only "momentum per second per unit area" does so.