previous    next

Fluids

Kinetic energy and temperature

Let us consider a cubic box (dimensions V=L×L×L) filled with a quantity of N identical gas molecules such that the ideal-gas conditions hold. Let the mass of one molecule be represented by m.

As we have observed in a previous page, all molecules move with some velocity. As a consequence the molecules have kinetic energy. At higher temperatures the molecules move faster, hence have more kinetic energy. The question is: How does the kinetic energy of a molecule depend on temperature?

We also observed in the page which we have mentioned above, that the molecules all have different velocities which moreover vary a lot due to the collisions amongst each other. However, their average speed v is constant, because all collisions are elastic.

Here, we will simplify the picture by assuming that all molecules move with the same speed v and that no collisions occur. Furthermore, instead of dealing with the x, y and z components of the velocities, we will assume that one third (N/3) of the molecules moves along the x direction, one third along the y direction and one third along the z direction. That situation is shown in the figure below.



In that situation we may calculate how many molecules bounce off each of the walls of the box per unit of time. Let us concentrate on a molecule that moves from the left to the right, bounces elastically off the righthand-side wall, then moves from the right to the left, bounces elastically off the lefthand-side wall and returns towards the righthand-side wall, always with a speed v. Hence, one molecule bounces v/2L times per unit of time off the righthand-side wall. Moreover, since N/3 molecules move in that fashion, the righthand-side wall suffers (N/3)×(v/2L)=Nv/6L collisions from bouncing molecules per unit of time.

Each time that a molecule bounces off the righthand-side wall its linear momentum changes from +mv to -mv. Consequently the total change of momentum during one collision of one of the molecules is equal to -2mv. However, for a change of momentum equal to Δp during an interval of time equal to Δt a force is needed which is equal to

F = Δp/Δt

The wall suffers Nv/6L collisions per unit of time, whereas at each collision the momentum of a molecule is changed by an amount of -2mv. Therefor, the total momentum change at the righthand-side wall per unit of time is given by

Δp = (Nv/6L)×(-2mv) = -Nmv²/3L

For one unit of time Δt=1. Consequently, the righthand-side wall excerts a force F on the gas which is equal to

F = -Nmv²/3L

The reaction force -F of the gas is related to its pressure P by P=-F/L², since the area of the righthand-side wall equals L².
We thus end up with the relation

P = Nmv²/3L³ = Nmv²/3V

Furthermore, the kinetic energy of one molecule is given by mv²/2. So, we obtain for the kinetic energy of one molecule

E_kin(1 molecule) = mv²/2 = 3PV/2N

Finally, we may use the ideal-gas law PV=NkT, to find

E_kin(1 molecule) = 3kT/2

for the relation between the average kinetic energy of one molecule and the temperature of the gas.

Kinetic energy has units

units of kinetic energy = mass×velocity squared = J (Joule)

more on kinetic energy