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Ideal Gases

We have argued on an abstract basis that the state of highest entropy (and hence the most probable state) for any complicated system is the one whose macroscopic properties can be obtained in the largest possible number of different ways; if the model systems we have considered are any indication, a good rule of thumb for how to do this is to let each "degree of freedom" of the system contain (on average) an equal fraction of the total energy  U. We can justify this argument by treating that degree of freedom as a "system" in its own right (almost anything can be a "system") and applying Boltzmann's logic to show that the probability of that microsystem having an energy $\varepsilon$ while in thermal equilibrium at temperature $\tau$ decays exponentially as $\exp(-\varepsilon/\tau)$. This implies a mean $\varepsilon$ on the order of $\tau$, if we don't quibble over factors comparable to 1.

The Equipartition Theorem, which is more rigourously valid than the above hand-waving would suggest,15.25 specifies the factor to be exactly 1/2:

A system in thermal equilibrium with a heat reservoir at temperature $\tau$ will have a mean energy of   ${1 \over 2}\tau$   per degree of freedom.

In an ideal monatomic gas of  N  atoms at temperature  $\tau$  each atom has three degrees of freedom: left-right (x), back-forth (y) and up-down (z). Thus the average internal energy of our monatomic ideal gas is

 \begin{displaymath}U \; = \; {3 \over 2} \, N \, \tau
\end{displaymath} (15.18)

In spite of the simplicity of the above argument15.26 this is a profound and useful result. It tells us, for instance, that the energy  U  of an ideal gas does not depend upon its pressure15.27 p! This is not strictly true, of course; interactions between the atoms of a gas make its potential energy different when the atoms are (on average) close together or far apart. But for most gases at (human) room temperature and (Earth) atmospheric pressure, the ideal-gas approximation is extremely accurate!

It also means that if we change the temperature of a container of gas, the rate of change of the internal energy  U  with temperature, which is the definition of the HEAT CAPACITY

\begin{displaymath}C \; \equiv \; {\partial U \over \partial T} ,
\end{displaymath} (15.19)

is extremely simple: since   $\tau \equiv \kB T$  and   $U = {3 \over 2} N \tau$,   $U = {3 \over 2} N \kB T$  and so the heat capacity of an ideal gas is constant:

\begin{displaymath}C \hbox{\rm ~[ideal gas]} \; = \; {3 \over 2} \, N \, \kB
\end{displaymath} (15.20)

Now let's examine our gas from a more microscopic, "mechanical" point of view: picture one atom bouncing around inside a cubical container which is a length  L  on a side. In the "ideal" approximation, atoms never hit each other, but only bounce off the walls, so our consideration of a single atom should be independent of whether or not there are other atoms in there with it. Suppose the atom in question has a velocity  $\Vec{v}$  with components  vx, vy and vz along the three axes of the cube.

Thinking only of the wall at the  +x  end of the box, our atom will bounce off this wall at a rate  1/t  where  t  is the time taken to travel a distance  2L  (to the far wall and back again) at a speed  vx:   t = 2L/vx. We assume perfectly elastic collisions -- i.e. the magnitude of  vx  does not change when the particle bounces, it just changes sign.  Each time our atom bounces off the wall in question, it imparts an impulse of  2 m vx  to that wall. The average impulse per unit time (force) exerted on said wall by said atom is thus   F1 = 2 m vx/t  or   F1 = m vx2/L.  This force is (on average) spread out all over the wall, an area  A = L2,  so that the force per unit area (or pressure) due to that one particle is given by   p1 = F1/A = m vx2/L3.  Since  L3 = V,  the volume of the container, we can write   p1 = m vx2/V  or

\begin{displaymath}p_1 \, V \; = \; m \, v_x^2
\end{displaymath}

The average pressure  p  exerted by all  N  atoms together is just  N  times the mean value of  p1:   $p = N \langle p_1 \rangle$,  where the " $\langle \cdots \rangle$" notation means the average of the quantity within the angle brackets. Thus

 \begin{displaymath}p \, V \; = \; N \, m \, \langle v_x^2 \rangle
\end{displaymath} (15.21)

Now, the kinetic energy of our original atom is explicitly given by

\begin{displaymath}{1 \over 2}mv^2 = {1 \over 2}m(v_x^2 + v_y^2 + v_z^2)
\end{displaymath}

since  $\Vec{v}$  is the vector velocity. We expect each of the mean square velocity components   $\langle v_x^2 \rangle$,   $\langle v_y^2 \rangle$  and   $\langle v_z^2 \rangle$  to average about the same in a random gas, so each one has an average value of   ${1 \over 3}$  of their sum.15.28 Thus   $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle
= \onethird \langle v^2 \rangle$  and the mean kinetic energy of a single particle is   $U_1 = {3 \over 2} \, m \langle v_x^2 \rangle$.  The kinetic energy of all  N  atoms is just  U = N U1,  or

 \begin{displaymath}U \; = \; {3 \over 2} \, N \, m \, \langle v_x^2 \rangle
\end{displaymath} (15.22)

But according to Eq. (18) we have   $U = {3 \over 2} \, N \tau$;  so we may write15.29

 \begin{displaymath}m \, \langle v_x^2 \rangle \; = \; \tau
\end{displaymath} (15.23)

Combining Eqs. (21) and (23), we obtain the famous IDEAL GAS LAW:

\begin{displaymath}p \, V \; = \; N \, \tau
\end{displaymath} (15.24)

Despite the flimsiness of the foregoing arguments, the IDEAL GAS LAW is a quantum mechanically correct description of the interrelationship between the pressure  p,  the volume  V  and the temperature   $\tau \equiv \kB T$  of an ideal gas of  N  particles, as long as the only way to store energy in the gas is in the form of the kinetic energy of individual particles (usually atoms or molecules). Real gases can also store some energy in the form of rotation or vibration of larger molecules made of several atoms or in the form of potential energies of interaction (attraction or repulsion) between the particles themselves. It is the latter interaction that causes gases to spontaneously condense, below a certain boiling point  Tb,  into liquids and, at a still lower temperature  Tm  (called the melting point), into solids. However, in the gaseous phase even carbon [vaporized diamond] will behave very much like an ideal gas at sufficiently high temperature and low pressure. It is a pretty good Law!


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Next: Things I Left Out Up: Thermal Physics Previous: How Big are Atoms?
Jess H. Brewer - Last modified: Mon Nov 16 16:13:26 PST 2015