Question

Derive an expression for the most probable translational energy for an ideal gas. Compare

your results to the mean translational energy for the same gas.

Answer 1

Let n moles of an ideal gas be confined to a cubical box of volume V. The molecules in the box move in all directions with varying speeds, colliding with each other and with the walls of the box. Figure 18.1 shows a molecule moving in the box. The molecule will collide with the right wall. The result of the collision is a reversal of the direction of the x-component of the momentum of the molecule:

Figure 18.1. Molecule moving in box.

The y and z components of the momentum of the molecule are left unchanged. The change in the momentum of the particle is therefore

After the molecule is scattered of the right wall, it will collide with the left wall, and finally return to the right wall. The time required to complete this path is given by

Each time the molecule collides with the right wall, it will change the momentum of the wall by ∆p. The force exerted on the wall by this molecule can be calculated easily

For n moles of gas, the corresponding force is equal to

The pressure exerted by the gas is equal to the force per unit area, and therefore

The term in parenthesis can be rewritten in terms of the average square velocity:

Thus, we conclude that

where M is the molecular weight of the gas. For every molecule the total velocity can be calculated easily

Since there are many molecules and since there is no preferred direction, the average square of the velocities in the x, y and z-direction are equal

and thus

Using this relation, the expression for the pressure p can be rewritten as

where vrms is called the root-mean-square speed of the molecule. The ideal gas law tells us that

Combining the last two equations we conclude that

and

For H at 300 K vrms = 1920 m/s; for 14N vrms = 517 m/s. The speed of sound in these two gases is 1350 m/s and 350 m/s, respectively.  The speed of sound in a gas will always be less than vrms since the sound propagates through the gas by disturbing the motion of the molecules. The disturbance is passed on from molecule to molecule by means of collisions; a sound wave can therefore never travel faster than the average speed of the molecules.

            The average translational kinetic energy of the molecule discussed in the previous section is given by

Using the previously derived expression for vrms, we obtain

The constant k is called the Boltzmann constant and is equal to the ratio of the gas constant R and the Avogadro constant NA

The calculation shows that for a given temperature, all gas molecules - no matter what their mass - have the same average translational kinetic energy, namely (3/2)kT. When we measure the temperature of a gas, we are measuring the average translational kinetic energy of its molecules.