We’ve looked at adiabatic expansion and compression.
These isentropic processes macroscopically reverse each other without requiring microscopic
reversal.
The gas begins in, and returns to, the macroscopic state of N atoms at temperature T in a volume
V.
But the individual atoms do not return to their original positions and velocities.
Now let’s consider another adiabatic processes in which gas increases its volume.
Instead of moving the center wall, we can simply remove it, allowing the gas to freely
expand into the greater volume.
As it does so, it never pushes on a moving surface.
Therefore, it has no opportunity to do work.
And, its internal energy remains constant.
So, in the free expansion of an ideal gas.
The gas does no work, and its internal energy, and hence its temperature, do not change,
unlike the case of normal adiabatic expansion.
Since the temperature remains constant, the ratio of temperatures in our entropy expression
can be taken to be one, and we are left with entropy as a function of volume alone.
During the free expansion the volume increased, so the entropy increased, indicating an irreversible
process.
In particular, the gas will not spontaneously reduce its volume thereby lowering its entropy.
Of course, we could mechanically compress the gas, and then remove heat equal to the
work done.
This would return the gas to its original state, but it would change the state of the
rest of the universe.
To be reversible, remember, we have to be able to return the entire universe to its
original state.
To gain more insight into our simplified entropy expression, let’s drop the S-0 term, at
least for now.
Note that V over V-0, a ratio of volumes, is a dimensionless quantity, a pure number.
Let’s denote this as M, and think of it as a number of elementary cells.
That is, we think of our entire volume V as divided into M cells, each with volume V-0.
We can then specify an atom’s position, to within the dimensions of an elementary
cell, by stating which cell it is located in.
The smaller V-0, the more accurate is the specification of the atom’s position.
With this definition, we can write the entropy as, S equals k N natural log M. Using the
power property of logarithms, this becomes k times natural log M to the Nth power.
Let’s denote M to the Nth power by capital gamma.
The significance of this quantity is that it is the number of ways to distribute N atoms
among the M cells.
For N equal 1, there are M ways to distribute one atom.
It can go in any of the M cells.
For N equal 2, there are M-squared arrangements.
For any of the M cells the first atom can be in, the second atom can be in any of the
M cells.
And M times M is M-squared.
Extending this idea to N atoms, the number of arrangements is M to the Nth power.
So, at least in this special case, entropy S equals Boltzmann’s constant k times the
natural log of gamma.
Which leads us to wonder if, maybe in general, the entropy of a gas is Boltzmann’s constant
times the logarithm of the number of ways the molecules can be arranged within the volume.
This is known as Boltzmann’s principle.
It was one of the great insights of Ludwig Boltzmann.
So much so that, as shown here, it’s etched on his grave marker, just above his rather
imposing bust.
Boltzmann used W for the number of arrangements.
We’ve been using W for work, so we’ll use the alternate symbol gamma.
This gives us a new way to think about the entropy change in free expansion.
And why a process like this doesn’t spontaneously reverse, even though that is possible in theory.
Let’s set M to the N equal to gamma-1.
This is the number of ways to arrange the N atoms among the M cells in our box.
Then we allow the gas to expand into twice the volume.
With twice the volume there will be twice the number of cells, 2 M. So the number of
ways to arrange the N atoms in this doubled volume is gamma-2 equals 2 M to Nth power,
which is 2 to the N times gamma-1.
Now we come to the new way of thinking.
In video 2 of this series, we saw that a dynamic system of elastic balls is chaotic.
Any uncertainty in its initial state grows exponentially with the number of collisions.
So it is practically impossible to predict the microscopic physics of such a system.
Let’s turn that problem to our advantage by treating the system as so unpredictable
as to effectively be a random process.
That is, we think of the dynamics as essentially randomly shuffling the atoms’ positions
after each short interval of time.
If we assume each of the possible gamma-2 atomic arrangements is equally likely, then
each has a probability of occurrence equal to one over gamma-2.
If at any time we look at the atomic arrangement in the large volume, what is the probability
that all atoms, or more generally molecules, will be in the left-half of the box?
This is simply the fraction of the gamma-2 arrangements that have all atoms in the left-half.
But the number of such arrangements is just gamma-1.
The number of arrangements in the original smaller box.
So the probability is gamma-1 over gamma-2, which equals one-half to the Nth power.
For large N, this is extremely small.
So the question, why can’t a gas sample spontaneously compress into half its original
volume, has the answer, it can.
But, it’s extremely unlikely.
So unlikely as to effectively be impossible.
Therefore, it is possible that if we keep watching our system we will eventually see
all the particles end up on the left side of the box.
The gas will have spontaneously compressed.
But, with 64 particles, the probability of this occurring at a given time is one over
2 to the 64th power.
This is about one over 18 times ten to the 18th power.
One over 18 million, million, million.
The time since the Big Bang, the age of the universe, is about 0.44 times ten to the 18th
seconds.
So, even if we had been looking at our system once a second since the beginning of the universe,
it is unlikely we would have seen this occur.
Now imagine the time scale for spontaneous compression in the case of a gas with trillions
of atoms.
From this point of view, the second law of thermodynamics does not tell us what is impossible,
only what is very, very improbable.
The physics of free expansion brings up an important point.
We start at a state 1, with temperature T, pressure P-1, and volume V-1.
We end up at a state 2, with the same temperature, lower pressure P-2, and greater volume V-2.
In isothermal expansion, we move between these states continuously along the isothermal curve
in a manner such that at any instant the system is effectively in equilibrium.
By contrast, in free expansion, we start at state 1.
Then, with the removal of the center wall, the system enters a non-equilibrium condition,
for which pressure and volume do not have well-defined values.
So, the system does not correspond to any point on our P-V diagram.
A rigorous description of the gas evolution would require the techniques of non-equilibrium
thermodynamics.
Eventually, after re-establishing equilibrium in the larger volume, the gas state can once
again be represented by a point on our P-V diagram with temperature T, pressure P-2,
and volume V-2.
Now let’s compare what happens in moving between these two states by these two processes,
isothermal, and free expansion.
We start with volume V-1, and end with volume V-2.
So the volume change in both cases is simply V-2 minus V-1.
We only need to know the initial and final states to determine this.
We don’t need to know anything about the process that moved the system between the
states.
What about added heat?
In isothermal expansion, as we’ve seen previously, the gas absorbs heat N k T log x-2 over x-1.
Where the x’s are the piston positions, and x-2 over x-1 equals V-2 over V-1.
But, in free expansion, no heat is absorbed by the gas.
There is no way to determine the added heat by simply knowing the initial and final states.
We need to know the process that moved the system between the states.
This is why we call quantities like volume, pressure, temperature and entropy, state variables,
or, alternately, state functions.
Quantities like heat and work are called process functions.
For a thermodynamic cycle, like the Carnot cycle.
If we sum all the changes in a state variable over the cycle, we get zero.
This is obvious, since starting at, say, state 1, going through the complete cycle returns
us to state 1.
Since the initial and final states are the same, all state variables return to the same
value, and the total change is zero.
The elongated S symbol with a superimposed circle represents this summation over a cycle.
But the total change over the cycle of a process function, such as added heat, is not necessarily
zero.
For the Carnot cycle it’s N k log x-2 over x-1 times T-hot minus T-cold.
Which is also the net work performed by the cycle.
It’s common in thermodynamics to use the “d” notation, such as d-T or d-S, for
an infinitesimal change in a state variable.
These are said to be exact differentials, which is another way of saying that summing
them up over a closed cycle is guaranteed to be zero.
The notation delta-Q represents an infinitesimal change in a process function.
These are not guaranteed to sum to zero over a closed cycle.
A variation on this applies when we consider changes over different paths between distinct
initial and final states.
For example, suppose we go from state 1 to state 3 of the Carnot cycle.
Path A takes us along the forward cycle through state 2.
Path B takes us along the reversed cycle through state 4.
The sum of changes to a state variable has to be the same for both paths.
Because the total change depends only on the initial and final states, which are the same
for both paths.
This is not true for a process function.
In the next video we will use the methods of statistical mechanics to flesh out the
idea that entropy is a measure of the number of ways to arrange the molecules of a substance.
And that the chaotic microdynamics of a gas effectively reshuffle that arrangement continuously.
One of our results will be an expression for the absolute entropy of an ideal monatomic
gas.
Curiously, although the analysis is completely in the realm of classical physics, we will
see that quantum mechanics sneaks into the final result.