Entropy and the
Second Law
Lecture 2
Getting to know Entropy
• Imagine a box containing
two different gases (for
example, He and Ne) on
either side of a
removable partition.
• What happens when you
remove the partition?
• Did the energy state of
the system change?
• Is the process reversible?
(All the kings horses and
all the kings men…)
• What changed?
The Second Law
• It is impossible to construct a machine that is able to
convey heat by a cyclical process from one
reservoir at a lower temperature to another at a
higher temperature unless work is done by some
outside agency (i.e., air conditioning is never free).
• Heat cannot be entirely extracted from a body and
turned into work (thus car engines always have
cooling systems).
• Every system left to itself will, change toward a
condition of maximum probability.
• The entropy of the universe always increases.
The Second Law
• My favorite:
o You can’t shovel manure into
the rear end of a horse and
expect to get hay out of its
mouth.
The Second Law states that there is a natural
direction in which reactions will tend to proceed.
Statistical Mechanics and
Entropy
The Microscopic Viewpoint
Back to our Box
•
Imagine there were just two atoms of
in each side of our box.
o
•
•
•
•
•
Only one possible arrangement
If we remove the partition, there are
24 = 16 possible arrangements.
Basic Postulate of Statistical
Mechanics: a system is equally likely
to be found in any of the states
accessible to it.
Odds of atoms being arranged the
same as the initial way: 1 in 16.
When we removed the partition, we
simply increased the number of
possible arrangements.
Suppose we had a mole of gas. What
are the odds of atoms being
arranged the same way?
Another Example
• Imagine two copper
blocks at different
temperatures separated
by an insulator.
• What happens if we
remove the insulator?
• Suppose we initially had 1
unit (quanta) of energy in
the left block and 5 in the
right (total of 6). How will
they be distributed after
we remove the insulator?
How many ways can we
arrange the energy?
• How many combinations are there that correspond
to the first block having 1 quantum?
• Each of these arrangements is equally possible.
o Although since we can’t tell the individual quanta apart, some of these
arrangements –combinations– are identical.
• How many when the left
block has 2?
• Is there a rule we can use to
figure out questions like this?
W(e)=
•
E!
e!(E-e)!
Where E is the total number
of energy units and e is the
number the left block has.
• Doing the math, we find
there are 15 ways.
How many all together?
•
•
•
•
The system is symmetrical, so
there are also 15 ways of
distributing the energy when
the left block has 4 and right
2.
There are 20 ways of
distributing our 6 units of
energy so that each block
has three, and only one way
each when all the energy is
on one side.
Dong the math, there are a
total of 64 ways.
A simpler way: 6 quanta
distributed between 2 blocks
can be distributed in 26 = 64.
o
The chances of any particular state
occurring are (½)6.
What’s the point?
• There were 20 ways of
distributing the energy
equally, 2x15 ways where
one side had 4 and the
other 2, 2x6 ways where one
side had one and 2 ways
where one side had none.
• The point is that there are
many ways to distribute
energy for some values of e
and only a few for other
values.
• The chances of an equal
distribution are 20x(1/2)6 =
20/64 = 0.3125. The chances
of the left block having all
the energy are only 1/64.
Calculating Probabilities
• Suppose there are 20 quanta of energy to distribute. Too
many to count individual combinations! We need an
equation.
• The equation will simply be the number of combinations
corresponding to a given state of the system, Ω(ƒ) (e.g., one
block having ƒ= 5 quanta) times the probability of any
particular combination occurring, in this case (½)20:
P(ƒ) = Ω(ƒ) × C
o
where C is the probability of any combination
occurring: (½)20 and Ω is
E!
W(e)=
• More generally,
o
o
P(e)=
e!(E-e)!
E! peqE-e
e!(E-e)!
where p is the probability of an energy unit being in the left block and q is the
probability of it being in the right. This equation is known as the binomial distribution
(can be computed with the this is the BINOMDIST function in Excel).
In this case, p and q are equal, so it reduces
E! to: E
P(e)=
e!(E-e)!
p
Energy Distribution
• We can use the
BINOMDIST function in
Excel to compute the
probability.
• We see an equal or
nearly equal distribution is
the most likely outcome.
o Imagine if we had 1020 quanta to
distribute – one block having a
few more or less would make little
difference.
• The point is that energy is
distribution equally
between the blocks
simply because that is the
most likely outcome.
Entropy and
Thermodynamics
• Unlike a simple physical system (a ball rolling down a hill),
in thermodynamics whether or not a chemical system is
at equilibrium depends not on its total energy, but on
how that energy is internally distributed.
• Clearly, it would be useful to have a function that could
predict the internal distribution of energy at equilibrium.
o The function that predicts how energy (or atoms) will be distributed at
equilibrium is entropy.
• We found that in this case (blocks of equal size and
composition), energy will be distributed equally between
them at equilibrium.
• How do we develop this mathematically?
Predicting the
Equilibrium Distribution
Probability!
• Consider again the
energy distribution.
• What is maximized
when the system is at
equilibrium?
• How do we know when
that function is
maximized?
• What is mathematical
characteristic does the
maximum of function
have?
Its first differential is 0.
Probability Function
• So the function we want is the probability, P of the
system (a copper block in this case) being found in a
particular state, E. As we have seen, this is proportional
to the number of combinations, Ω(E), corresponding to
that state:
P(E) = CΩ(E)
o So we can use the function Ω(E) in place of P(E)
• A difficulty here is that both P(E) and Ω(E) are
multiplicative. So the number of assessable states for the
two blocks is Ωleft xΩright. Our other thermodynamic
properties are additive.
• How do we convert a multiplicative property to an
additive one?
o Take the log: ln (Ω).
Defining Entropy
• Ludwig Boltzmann defined
entropy, S, as
S = k ln Ω
o Where k is Boltzmann’s constant.
o R = k × NA
• Entropy is a measure of the
randomness of a system.
• An increase in entropy of a
system corresponds to a
decrease in knowledge of it.
• We can decrease the
entropy of a “system”, but
only by increasing the
entropy of its surroundings.
Temperature & Statistical
Mechanics
• We noted that we can find the maximum of the
probability function by taking its derivative.
• For our blocks, equilibrium occurs where
d lnW(E)left d lnW(E)right
=
d Eleft
d Eright
• So this function ∂Ω/∂E, appears to be a useful one as
well; we’ll call this function β.
• The two blocks are in equilibrium when βleft = βright.
• We other variable is equal when the two blocks are in
equilibrium?
o Temperature!
• It turns out that β= 1/kT
The Second Law
•
•
•
Entropy also has the interesting •
property that in any
spontaneous reaction, the
total entropy of the system plus
its surroundings must never
decrease.
In our example, this is a simple
consequence of the
observation that the final
probability, P(E), and therefore •
also Ω, will be maximum and
hence never be less than the
original one.
Any decrease in entropy of
one of the blocks must be at
least compensated for by an
•
increase in entropy of the
other block.
Experience has shown that
in any spontaneous
transition, the increase in
entropy will always exceed
the ratio of heat
exchanged to
temperature.
Formally, the Second Law is
written as:
In the case of a (fictive)
reversible process, this becomes
an equality.
Integrating factors and
exact differentials
• Any inexact differential
that is a function of only
two variables can be
converted to an exact
differential.
• dW is an inexact
differential, and dV is an
exact differential. Since
dWrev = -PdV,
• dWrev can be converted
to a state function by
dividing by P since
dV =-dWrev/P
• Similarly for heat
• dQrev/T = dS
• Here we convert heat to
the state function
entropy by dividing by T.
Back to our box
• Entropy relates not just to
the distribution of energy,
but of atoms and molecules
as well.
• In our box with two gases,
the most likely state after we
removed the partition was
one in which the He and Ne
atoms would be randomly
mixed, with equal numbers
on both sides.
• What changed when we
removed the partition from
our box was that the
entropy of the system
increased.