20.3.1 Force Due to Collisions

FIGURE 20.5 shows a molecule with an x-component of velocity $v_{x}$ having a perfectly elastic collision with a wall and rebounding with its velocity changed to $- v_{x}$ . This one molecule has a momentum change

Δ p_{x} = m (- v_{x}) - m v_{x} = - 2 m v_{x}

(20.4)

The figure illustrates the velocity values of a gas molecule before and after collision within a box on the x y coordinate plane.

FIGURE 20.5 Full Alternative Text

Suppose there are $N_{coll}$ collisions with the wall during a small interval of time $Δ t$ . Further, suppose that all the molecules have the same speed. (This latter assumption isn’t really necessary, and we’ll soon remove this constraint, but it helps us focus on the physics without getting lost in the math.) Then the total momentum change of the gas during $Δ t$ is

Δ P_{x} = N_{coll} Δ p_{x} = - 2 N_{coll} m v_{x}

(20.5)

You learned in Section 11.1 that the momentum principle can be written as

Δ P_{x} = (F_{avg})_{x} Δ t

(20.6)

Thus the average force of the wall on the gas is

(F_{on gas})_{x} = \frac{Δ P_{x}}{Δ t} = - \frac{2 N_{coll} m v_{x}}{Δ t}

(20.7)

Equation 20.7 has a negative sign because, as we’ve set it up, the collision force of the wall on the gas molecules is to the left. But Newton’s third law is $(F_{on gas})_{x} = - (F_{on wall})_{x,}$ so the force on the wall due to these collisions is

(F_{on wall})_{x} = \frac{2 N_{coll} m v_{x}}{Δ t}

(20.8)

We need to determine how many collisions occur during $Δ t$ . Assume that $Δ t$ is smaller than the mean time between collisions, so no collisions alter the molecular speeds during this interval. FIGURE 20.6 has shaded a volume of the gas of length $Δ x = v_{x} Δ t$ . Every one of the molecules in this shaded region that is moving to the right will reach and collide with the wall during $Δ t$ . Molecules outside this region will not reach the wall and will not collide.

The figure illustrates determining the rate of collision using a container filled with gas molecules.

FIGURE 20.6 Full Alternative Text

The shaded region has volume $A Δ x$ , where A is the area of the wall. If the gas has number density $N / V$ , the number of molecules in the shaded region is $(N / V) A Δ x = (N / V) A v_{x} Δ t$ . But only half these molecules are moving to the right, so the number of collisions during $Δ t$ is

N_{coll} = \frac{1}{2} \frac{N}{V} A v_{x} Δ t

(20.9)

Substituting Equation 20.9 into Equation 20.8, we see that $Δ t$ cancels and that the force of the molecules on the wall is

F_{on wall} = \frac{N}{V} m v_{x}^{2} A

(20.10)

Notice that this expression for $F_{on wall}$ does not depend on any details of the collisions.

A key aspect of the micro/macro connection is recognizing that state variables, such as pressure, are related to appropriate averages of molecular properties. We can relax the assumption that all molecules have the same speed by replacing the squared velocity $v_{x}^{2}$ in Equation 20.10 with its average value. That is,

F_{on wall} = \frac{N}{V} m {(v_{x}^{2})}_{avg} A

(20.11)

where ${(v_{x}^{2})}_{avg}$ is the quantity $v_{x}^{2}$ averaged over all the molecules in the container.

20.3.1 Force Due to Collisions

FIGURE 20.5 A molecule colliding with the wall exerts an impulse on it.

FIGURE 20.6 Determining the rate of collisions.