*©Fernando Caracena *2014

**Introduction**

**Introduction**

This blog looks into the meaning of temperature and relates the phenomenon of pressure in gases to the dynamics of molecules. It completes the previous discussion of Grokking "The Kinetic Theory of Gases Part I", which builds bottom-up from observations; but this time, the discussion begins from the atomic-molecular hypothesis and Newtonian to show that the gas laws follow theoretically.

Even before direct evidence was discovered for the existence of atoms and molecules, some scientists such as John Dalton proposed that such structures were the basic constituents of matter to make sense of the gas laws and those of chemistry. Today, thermodynamics is an important component of physics, which is critical in understanding several other areas of physics.

*The Ideal Gas Law*

*The Ideal Gas Law*

The following form of the ideal gas laws was found empirically in Part I of this series:

P V = N k_{0}T, (I.2)

where N represents the number of gas molecules within the volume in question (V), the temperature is in Kelvins (T) and the constant per molecule (k_{0} ) is left to be determined experimentally (or theoretically, from other measurements or theory). Further, an important conclusion of the previous blog was that the terms on both sides of (I.2) represented energy, so that the combination, k_{0}T, represented some form of energy per molecules.

This blog continues this theme top-down from an analysis of the motions of individual molecules to show that a bunch of particles, such as molecules, exert pressure over a given volume in proportion to the average, translational kinetic energy per particle and the number of particles in the sample.

The analysis sheds light on just what temperature measures and how molecular motions cause a gas to exert pressure.

Pressure is the average force per unit area exerted at right angles to the walls of the enclosing container of a gas or liquid. Consider the pressure of a gas on an infinitesimal area (ΔA) of the wall of a container (shaded in Fig.1), which is oriented in a way that the outward normal to the wall is in the positive x-direction. How is pressure force exerted by molecules? The molecules acquire kinetic energy by banging together so that that this energy is distributed among all the molecules of a gas in some kind of distribution. Eventually some of these molecules fall on a portion of the wall of the container by bouncing off it elastically. The molecular impacts on the walls, although spiky, happen at such a high enough frequency that the spikes average out on the time scale of our awareness to be a constant force, which over a small area of wall, manifests as a constant pressure. We assume that the pressure is exerted statically, which results from having the enclosed gas at rest within the containing vessel. The present discussion could be extended to include streaming effects, by including a streaming velocity in the gas motion. But why complicate the discussion when we want to reduce the discussion to the simplest, essential details?

## Stepping gingerly through the mine-field of statistics

Consider the collision of a single molecule with a small area (ΔA) of the walls of a container full of gas (as depicted in Fig. 1). Let this area be big enough to allow it to experience many molecular impacts per unit time, but small enough to be considered infinitesimal. The area is situated on a plane that is perpendicular to the x-axis. Consider that each molecular collision with this area element is elastic, which means that both the energy and momentum of the system are conserved in the process. Only the component of momentum of the molecule normal to the wall (along the x-axis) is changed by the collision and the wall remains stationary because it is held rigidly in place by other forces exerted through the material of the container. The conditions for this elastic impact model are expressed by the following equations:

the change in the x-component of momentum of each molecule striking ΔA is along the x-direction only,

Δ(mv_{x}) = mv_{x} - (-mv_{x }),

or

Δ(mv_{x}) =2 mv_{x} , (1a)

where we restrict the impacts on ΔA to those arriving from the left, or moving along the positive x-axis

v_{x} > 0;

and no change, in the other two components of momentum

Δ(mv_{y}) = 0 , (1b)

Δ(mv_{z}) = 0 . (1c)

The force exerted on the wall element (ΔA) by the impact of molecules is given by Newton's second law of motion,

ΔF_{x} = Δ(mv_{x}) / Δt . (2)

We can estimate the momentum flux on the RHS of (2) by using the frequency of arrival of molecules on the wall element having the momentum mv_{x} (which here is restricted to the positive range in x-component of velocity).

First, note that if ρ is the density of the gas, and m is the average mass of each molecule, then the number density of molecules (Δn/ΔV) approaching the wall from the left is given by the expression,

Δn/ΔV = ρ/m , (3a)

or,

Δn= ρ/m ΔV.

But, the volume element, ΔV, is what is swept out by by a cylindrical column of base area, ΔA, and height, 1/2 v_{x} Δt that includes Δn molecules, which is given by

Δn^{+} = ρ/m (1/2) v_{x} Δt dA. (3b)

Note that a factor of 1/2 appears in (3b) because we have restricted the population of particles to those impacting the surface ΔA only from the left, because this surface is a reflecting boundary. On an interior surface, not a boundary, a flux of particles of equal magnitude crosses the surface parallel to the x-axis from both the positive and negative x-directions, the net flux being zero, viz.,

<v_{x} >=<v_{y} > = <v_{z} > =0.

We can solve for the collision frequency by dividing both sides of (3b) by Δt,

Δn^{+}/Δt = (1/2) ρ/m v_{x} dA . (3c)

At this point, somebody in the audience would criticize the above argument, by pointing out that I had neglected motions at right angles to the x-axis, which would sweep molecules in and out of the impact cylinder, such as are depicted in Fig. 2. I would answer that since these motions are just as likely to bring in new molecules as well as take others out of the impact cylinder, their effects would tend to cancel out.

To calculate the average force on the wall, multiply the x-momentum change of a molecule on the bounding area by the frequency of arrival of molecules having this component of momentum to get the momentum flux, which equals the force that the rebounding molecules exert on the wall,

ΔF_{x} = Δ(mv_{x}) Δn^{+}/Δt . (4a)

ΔF_{x} = 2 (mv_{x}) (1/2) ρ/m v_{x} ΔA . (4b)

To get the pressure exerted on the differential area (ΔA), divide both sides of (4b) by ΔA and average over the entire spectrum of v_{x}^{2} (we spare the reader the statistical details on how to do this) to get,

<ΔF_{x} /ΔA >= 2 ρ/m (1/2) <mv_{x}^{2}> ,

or,

P = 2 ρ/m <(1/2) mv_{x}^{2} > . (4c)

The number of molecules per unit volume is given by the expression,

N/V = ρ/m ,

which can be used to write (4c) as

P = 2 N/V <(1/2) mv_{x}^{2} > . (5a)

We recognize the bracketed expression on the RHS of (5a) as the contribution to the total kinetic energy by translational motion along the x-direction.

P = 2 N/V <KE_{x} > . (5b)

The total average kinetic energy of translation is related to that of motion along one of the axes. To slow this, expand the square of the velocity into components and take the average,

<**v**^{2} > = <v_{x}^{2} >+<v_{y}^{2} >+<v_{z}^{2} > . (5c)

But, assuming that there are no external forces such as gravity or electric and magnetic ones acting on the molecules, the averages of the squares of the velocity components are equal,

<v_{x}^{2} > = <v_{y}^{2} > = <v_{z}^{2} >,

or

<v_{x}^{2} > =(1/3)<**v**^{2} >.

The result is that the kinetic energy of thermal motion is shared equally in all directions, or

<KE_{x} >=(1/3)<KE >. (5d)

This result (5d) substituted in (5b) produces what looks like the ideal gas law (I.2) except that instead of the temperature term, k_{0}T, we have the average kinetic energy of the molecules of the gas multiplied by 2/3,

P V= (2/3) N <KE >, (5e)

where N is the number of molecules contained in the sample volume, V. Comparing (5d) with (I.2), we see that we can identify the average translation kinetic of a typical molecule with the Kelvin temperature of an ideal gas,

<KE > = (3/2) k_{b} T. (5f)

This substitution allows us to rewrite (5d) in terms of a standard volume measure and number of molecules contained therein,

P V= N k_{b} T, (6)

which is also an expression that followed theoretically from the work of

Ludwig Eduard Boltzmann (1844-1906). In honor of this pioneer, k_{b} is called the Boltzmann constant, the latest value of which has been measured at the UK National Physical Laboratory as

k_{b} =1.380 651 56 (98) × 10^{−23} J K^{−1}.

What Boltzmann showed was that the Newtonian dynamics of gas molecules could be used to explain the empirical properties of ideal gases. This closed the loop between bottom-up and top-down thinking in the physics of dilute gases.

Poor Bolzmann was too far ahead of his time, before more direct evidence was discovered for the existence of atoms and molecules. In his days, atoms and molecules were more like a simplifying picture that made it easier to understand the ideal gas laws and the laws of chemistry. Although many physicists recognized them as simplifying conceptual devices, they thought of them just as convenient fictions. But Bolzmann was quite convinced that atoms and molecules were real and that his rigorous statistical analysis of the dynamical effects of these entities as Newtonian particles proved the point. Perhaps many people became lost in statistical arguments, which is easy to do in the thick, dark forests of statistics. That is why we have stepped gingerly through and around this tangle, and I have led you past the scariest spots by basically diagramming what you do. Once students become familiar with the argument, it becomes possible to step them through the deep dark tangle of statistics rigorously without getting lost.

Alas, poor Bolzmann, the Austrian physicist after suffering a heavy hail of criticism from those who did not believe in atoms and molecules, suffered a fit of severe depression and committed suicide. It probably would not have happened to an Italian who would have balanced his statistics with wine, cheese, bread and hikes in the clear Alpine atmosphere.

**Moles and Molar Volume**

**Moles and Molar Volume**

The ideas that have evolved to become the basis for modern units of measure regarding gases have come out of the early work of chemists and physicists.

Even before scientists verified the existence of atoms and molecules, chemists found that thinking in terms of atoms and molecules simplified ideas of physical chemistry. The complicated empirical rules of combinations of substances from elements to chemical compounds became easier to understand

by picturing a pure elemental substance as consisting of a large number of identical atoms, each having the same mass and chemical properties. The elements came together to form chemical compounds, which at the atomic molecular level appeared as identical molecules composed of clusters atoms of the constituent elements bound together in identical structures (See Fig.4). Chemists quickly realized that volumetric measurements of various elemental gases (at the same temperature and pressure) combined together in the ratio of whole numbers to form gaseous compounds. This happened despite the fact that the masses associated with the various volumetric measurements were quite diverse. The simplifying assumption that explained this mystery was that equal volumes of different gases, at the same temperature and pressure, contained the same number of molecules. Amedeo Carlo Avogadro (1776-1856) this hypothesis in** **an article called, "*Essai d'une manière de déterminer les masses relatives des molécules élémentaires des corps, et les proportions selon lesquelles elles entrent dans ces combinaisons"* in the *Journal de Physique, de Chimie et d'Histoire naturelle*, which contained his law:

Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

At first, chemists thought that atoms should have masses that were integral multiples of the mass of a hydrogen atom, because by weighing equal volumetric gas samples (at the same temperature and pressure) they found that this assumption seemed to hold, except for measurement errors that could have come from lack of precision. As the precision of measurements improved, however, they found the assumption held only approximately but with subtle differences that were later explained by physics. Nevertheless, today the discovery of chemists that that equal volumes of different gases, at the same temperature and pressure, contained the same number of molecules has been verified by numerous experiments and observations. To standardize their observations, chemists adopted a convention of measuring sample volumes of gases at a standard temperature (0 °C or 273.16 °K) and pressure (1 atmosphere, ), abbreviated as STP. They assumed that the sample volume was equal in mass to the relative molecular mass of the gaseous compound in grams, which chemists called moles. The volume of one mole at STP would always be the same for any gaseous compound ( molar volume , 2nd Ref.), which chemists have found to be,

22.414 L/mol at 0 °C,

where L stands for litre (1,000 cm^{ 3} of volume).

All molar volumes of various gaseous compounds at STP have the same number of molecules, which should be a universal constant, defined as Avogadro's Number,

N_{A} = 6.022 141 79(30)×10^{23} mol^{−1}.

Note that molar volumes and Avagadro's number are not fundamental, but result from adopting the convention of measuring molar masses in grams. If we communicated with aliens of different planets, we would have to know their various units of measure to translate our measurements from their units of ours and *visa versa*.

We have concentrated on the meaning of temperature in this discussion. An important omission here is the meaning of entropy, which requires much more discussion, and so is left for another time.

Finally, I have received an intelligent comment that shows that the reader connected with my blog. The reader is correct in his interpretation, and perhaps I did not express that point well enough.

About the factor half in Eq (3 b): can we think in the following manner: the number of molecules in any volume can be divided into two parts, one's with positive vx and the ones with negative vx. Here, for colliding with the right wall from left only the first half matters. Or, I am wrong? Or is there a better way to think?