Pressure:

Consul Engel and Reid (Ch. 33.1) for a discussion of the derivation for the pressure of a rarefied collection of particles of mass $m$. In the following, we provide a connection from the one-dimensional version to the full scalar pressure. The connection is not quite direct from the discussion of Engel and Reid.

In 3-D, for a collection of many particles (on the order of Avogadro number), using average values of velocity and velocity components (in Cartesian coordinates); these are not generalized coordinates (as physicists would consider), the total kinetic energy is:

$$KE_{total} \ = \ \frac{N}{2} \ m \ \left ( \vec{v} \cdot \vec{v} \right )$$

$$= \ \frac{N}{2} \ m \ \left ( v_x^2 + v_y^2 + v_z^2 \right )$$

$$= \frac{N}{2} \ m \ v_x^2 \ + \ \frac{N}{2} \ m \ v_y^2 \ + \ \frac{N}{2} \ m \ v_z^2$$

$$= (KE)_x + (KE)_y + (KE)_z$$

The last equality is really just a notational trick; there really does not exist a thermodynamic or kinetic property $(KE)_x$ or $(KE)_y$ or $(KE)_z$! The lowercase $m$ is mass.

The velocity is a vector so the $v^2$ we treat casually is really a dot product.

The pressure components for the x , y , and z directions as we determined in class are:

$$p_x \ = \ \left ( \frac{N}{V} \right ) m \ v_x^2 = \frac{1}{V} 2 \ (KE)_x$$

$$p_y \ = \ \left ( \frac{N}{V} \right ) m \ v_y^2 \ = \frac{1}{V} \ 2 \ (KE)_y \$$

$$p_z \ = \ \left ( \frac{N}{V} \right ) m \ v_z^2 \ = \frac{1}{V} \ 2 \ (KE)_z \$$

$N$ is the number of particles. $V$ is the volume of space we are considering.

From statistical mechanics (which you will learn more about in the future) we have the relation for the special case of a fluid or state of matter with extremely weak interactions (or no interactions):

$$(KE)_x \ = \ \frac{N}{2 \ N_{Avogadro}} \ R T$$

$$(KE)_y \ = \ \frac{N}{2 \ N_{Avogadro}} \ R T$$

$$(KE)_z \ = \ \frac{N}{2 \ N_{Avogadro}} \ R T$$

Thus,

$$2 \ (KE)_x \ = \ \frac{N}{ N_{Avogadro}} \ R T$$

$$2 \ (KE)_y \ = \ \frac{N}{ N_{Avogadro}} \ R T$$

$$2 \ (KE)_z \ = \ \frac{N}{ N_{Avogadro}} \ R T$$

Substituting the above relations for $2 \ (KE_x)$, etc. into the pressure equations yields:

$$p_x \ = \ \left (\frac{N}{V} \right ) \ \frac{1}{ \ N_{Avogadro}} \ R T$$

$$p_y \ = \ \left (\frac{N}{V} \right ) \ \frac{1}{ \ N_{Avogadro}} \ R T$$

$$p_z \ = \ \left (\frac{N}{V} \right ) \ \frac{1}{ \ N_{Avogadro}} \ R T$$

Recall that $\frac{N}{N_{Avogadro}}$ is the number of moles $N_{moles}$. Thus, the equations for pressure become:

$$p_x \ = \ \left (\frac{N_{moles}}{V} \right ) \ R T$$

$$p_y \ = \ \left (\frac{N_{moles}}{V} \right ) \ R T$$

$$p_z \ = \ \left (\frac{N_{moles}}{V} \right ) \ R T$$

Here, we stop and realize that we have a x-component of pressure, a y-component and a z-component. This is not an artificial result, as rigorously, pressure is a tensorial property (it is a 3x3 matrix). The diagonal elements (which we have computed) have special meaning in that they can be used to determine the pressure as we know it. Rigorously, the scalar pressure (that we normally measure and talk about) is determined from the trace of the pressure tensor (or matrix). This is:

$$p_{total}^{scalar} \ = \ \frac{1}{3} \ ( p_x + p_y + p_z)$$

$$= \frac{1}{3} \ \left (\frac{N_{moles}}{V} \right ) ( RT + RT + RT )$$

$$= \ \left (\frac{N_{moles}}{V} \right ) \ R T$$

$$pressure = \ \left (\frac{N_{moles}}{V} \right ) \ R T$$

This should be a more convincing argument for the equality of kinetic theory description of fluids to that of an ideal gas.