Download THERMODYNAMICS

THERMODYNAMICS
Thermodynamics is the phenomenological science which describes the behavior of
macroscopic objects in terms of a small number of macroscopic parameters. As an example, to
describe a gas in terms of volume pressure temperature, number of particles, and their type. The
interesting aspect of this is that is possible at all. After all, a macroscopic object contains ~1025
particles or more. Each would appear to need at least 6 numbers to describe it—3 position and 3
momentum or velocity. So why are only a few (~ 10) needed in practice?
The subject is approached in two very different ways. The first is the phenomenological
method stated above. The second is from the microscopic laws of Newton or quantum
mechanics (if needed). This second approach is called statistical mechanics. We will use both,
alternating between them.
Temperature
The concepts of volume and pressure are already familiar, but temperature (in the
scientific sense) is not. We begin with an objective definition of temperature—we show how to
measure it.
Suppose we have 2 macroscopic objects, A and B.
Now suppose we place them in thermal contact so that it is possible for energy to flow between
them. In practice, it is nearly impossible to prevent such flow—by radiation if no other way. If
energy tends to flow from A to B we say that the temperature of A is higher than that of B.
If energy goes from A → B then
TA  TB
If there is no tendency for energy to flow we say that
TA  TB
Now add a third object C. Then the 0th Law of Thermodynamics states that if
TA  TB
and
TB  TC
Then
TA  TC
This seems obvious, but it contains a subtle assumption—that only temperature affects
the direction of heat flow. This turns out to be true experimentally but it did not have to be that
way. Thus it is generally stated as an independent law.
Temperature Scale
Having thus defined temperature it is a simple matter to obtain a scale and a method of
assigning a numerical temperature to an object.
We observe that a very large number of physical properties depend on a temperature
(length, volume, color, resistance, etc.). We need merely pick a property and two “fixed points”
and we have a scale and a thermometer.
For example, a very convenient choice is an ideal gas—which we will discuss in a
moment. First we consider the familiar mercury in a tube thermometer. Here the property is the
height of the column of mercury. The two final points are the melting and boiling points of
water (under certain conditions). We place the tube of mercury in the ice water mixture and
mark its height. We then place it in the boiling water and mark the new height. We then divide
the space between the marks into equal intervals. We now have a thermometer. If we take
melting ice to be 0 and boiling water to be 100, we have the familiar centigrade scale. If we take
them to be 32 ad 212, we have the even more familiar Fahrenheit scale. If we take them to be
273 and 373, we have the Kelvin scale. Note that neither Centigrade nor Fahrenheit is
intrinsically better than the other. Kelvin on the other hand is fundamentally superior as we will
see.
Ideal Gas
To get an idea of the fundamental microscopic meaning of temperature we consider an
ideal gas, defined to be a gas in which there is no interaction between the molecules except at the
instant of collision. This is never completely accurate. It becomes arbitrarily close to reality as
the density of the gas → 0. (Infinitely far apart, they can’t see each other.) Consider a cubical
box of side L containing N molecules. By the isotropy of space we will have equal numbers
moving in each direction. More accurately, the components of velocity in each direction will be
the same. Furthermore, the molecules will have some average magnitude of velocity v. Then in
one second each molecule moving in the x direction will make
v
2L
collisions with the wall at x = 0. When it does so its momentum will change by 2mv (mv in –mv
out). Thus the total change in momentum in the x-direction/sec is
N v
2mv
3 2L
However, by Newton’s 2nd law this must be the force exerted on the gas by the wall. Hence
F Nmv 2 2
2 N


 KEav 
3
A
2 3L
3 V
Where V = L3 is the volume of the gas. By Newton’s law however, this is also the F/A exerted
on the wall by the gas—or the pressure. Thus
PV 
2
N  KE av 
3
We now define the ideal gas temperature to be
2
KEav
T 3
k
where k = Boltzman’s constant = 1.38  10-23 J/°K. Note that the lowest possible temperature is
the state of lowest possible KE. In classical physics this would be KE = 0. Clearly this works
since if KE = 0 so does dvM/dt and hence, P. The Kelvin scale simply starts with 0 at the lowest
possible temperature—which happens to be -273°C. This is why the Kelvin scale is intrinsically
superior to the other two.
Thus we have an idea about the fundamental meaning of temperature at the microscopic
level—it is a measure of the random KE of the molecules making up the material.
Next we will turn to a statistical mechanics description of an ideal gas—where we do not
assume an average velocity at the outset.
Statistical Mechanics
We need a way to characterize the microstate of a system. Begin by considering a
monatomic ideal gas. Then each molecule can be located by giving its position and momentum.
This will require 6 numbers—3 position and 3 momentum. We thus imagine a 6-dimenstional
space. We split this space into infinitesimal boxes of “volume”
dxdydzdp x dp ydpz
We now define a distribution function, f, by


# of molecules in box  x, y, z, p x , p y , p z dx  dp z
A macroscopic state is then given by specifying the occupation of each box.
For the case of a monatomic ideal gas—or any other for that matter—we must recognize
three distinct possibilities.
Case 1 – Classical
In classical mechanics each particle is distinguishable from all the others—we can keep
track of which is which.
Case 2 – QM Fermi Dirac
In quantum mechanics identical particles are not distinguishable—we can’t keep track of
which is which. There turn out to be two different classes of particles and hence two different
statistics. One is the Fermi-Dirac particles and the other Bose-Einstein.
Case 3 – QM Bose-Einstein
We will, for now, only consider the classical case. We now make the fundamental
assumption of statistical mechanics that all possible microstates are equally likely and therefore
that the most probable macrostate will be that one which arises from the largest number of
microstates.
Hence we need to know how many microstates correspond to the same macrostate. A
macrostate is specified by giving the number of particles in each infinitesimal box. Let ni be the
number of particles in the ith box, and N be the total number of particles. How many ways can
we have ni particles in the 1st box? We care which particles they are, but not he order in which
they were chosen. The 1st particle can be chosen in N way, the second in N-1 ways, etc. Hence
there are
N   N  1  ... N  n i  1
ways to do this. However, there are n1! orders in which they can have been chosen. Hence the
number of distinct microstates is
N   N  1 ... N  n1  i 
n1 !
Now take the next box. The number of ways is
 N  n1  ... N  n1  n 2   1
n2 !
Hence the total number of ways is
N   N  1    1
n1 !n 2 ! n i !

N!
 ni !
i
Thus the probability of the macrostate specified by the ni is proportional to
N!
 ni !
i
or
Pq
N!
 ni !
i
We now assume that all the ni and N >> 1 (simply choose large enough boxes). This will not
work for quantum statistics, but is fine here. Then it is easiest to use ℓnP instead of P. Thus
nP  nq  nN!   nn i !
i
We now use Stiring’s Approximsation
n N!  N n N  N
Thus
nP  nq  N n N  N    n i n n i  n1   nq  N n N   n i n n i
i
i
Next we must impose 2 constraints on the system—the total number of particles is fixed
and the total energy fixed. In other words, we are dealing with an isolated system. There are
other possibilities, but this will do for now. Here
N   ni
i
E   n i i
i
where i is the energy of particles in the ith box. We then find the macrostate as the one for
which ℓnP is greatest consistent with the 2 constraints.


n
 n P  0  i  n i n n i  i n i     n n i  1 n i   n n i n i
ni
i
i


since  n i  0 and
i
 ni   ,i ni 0
i
i
We use the method of Lagrange Multipliers
 n i  0
i
 ,i n i  0
i
 n n i n i  0
i
Add these equations together to get
    ,i n n i n i  0
i
Since this must be true for any choice of ni we must have
n n i    ,i  n i  ee,i
subject to the conditions
e  e,i  N
i
e   ,i e ,i  E
i
In classical mechanics we replace the sum by an integral since the boxes can be made
infinitesimally small (this is not true in QM systems). Thus
N  e  dx  dpz e
, x1pz 
or
e  
N


 


dx  dpz e
, xpz 
Thus far we have not used the fact that we have a monatomic ideal gas. For that case the only
energy is kinetic
,
px 2  py2  pz 2
2m
Thus
N
e 

 dxdydz




dp x
p 2
 x
e  2m








dp y e

p y2 
2m




dp z
p 2
 z
e 2m


N


p2
 

2m
V   e
dp 

 

3
The Gaussian integral in [ ] can be easily evaluated as follows:
1/2



2
x
x 2
 y


 e
 e



dx
dx
dy
 e







 

2
1/2
    x 2  y2


    e
dxdy 



  


1/2
1/2
 2
 

2
2
r
r





   2  r e
dr 
    r drd e
 0 0

 0


r 2
e
  2 
2


Hence
1/2

0




1/2

 


e  
N
3/2
 2m 
V

  
To determine β we calculate the pressure this system would exert on a wall of the container. We
do it as before, except now the particles in box i have the velocity pi/m rather than the average
velocity. Thus the pressure exerted by the particles in box i is
2pi
ni
A
Now we must add this up for all the boxes. py and pz can be anything, but px must be > 0 (else
the particles won’t hit the wall. Further, the box must be within px/m of the wall or the particles
can’t reach the wall in a second. Hence
P




0









y z x  px px  0
m


NA2
 2m 
V

  
N2
3/2
A
1
1/2 m
 2m 
V

  

 2m 


  


p 2
 x

0




py

e
m



pz







dp x p x
px  0



2
2
2
2p x   2 p x  p y  pz
e e
dx  dp z
A




x 
px 2
2m dp 
x
e

px 2
2m dx
px
m



px 2




2N
d
2m dp 
e


1/2

 2m   d    


0
mV 

 2m 

   
1/2


 

   
2N
d
2N
1
N m N
1/2







 
1/2 
1/2
2
3/2
V  V
2   
 2m   2d         
 2m 
mV 



 2m    2m    mV   
 2m 
   


But we know that the ideal gas temperature is given by
PV  NkT
Hence
NkT N
1

 
V
V
kT
Thus we arrive at the famous “Boltzmann Distribution.”

kT
f e

with

 
  e kT

dx  dpz  N
Degrees of Freedom
Next we consider what happens when we do not have a monatomic ideal gas.