Lecture 8. Thermodynamic Identities (Ch. 3)

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Lecture 8. Thermodynamic Identities (Ch. 3)
We have been considering the entropy changes in
the processes where two interacting systems
exchanged the thermal energy but the volume and
the number of particles in these systems were
fixed. In general, however, we need more than
just one parameter to specify a macrostate
We are familiar with only one partial derivative
of entropy
Today we will explore what happens if we let the
other two parameters (V and N) vary, and analyze
the physical meaning of two partial derivatives
of the entropy
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General Approach
? if monotonic as a function of U
(quadratic degrees of freedom!), may be
inverted to give
When all macroscopic quantities S,V,N,U are
allowed to vary
So far, we have abbreviated one of these partial
derivatives
The other partial derivatives are
pressure
chemical potential
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Thermodynamic Identities
- the so-called thermodynamic identity
With these abbreviations
? shows how much the systems energy changes
when one particle is added to the system at fixed
S and V. The chemical potential units J.
? is an intensive variable, independent of the
size of the system (like P, T, density).
Extensive variables (U, N, S,V...) have a
magnitude proportional to the size of the system.
If two identical systems are combined into one,
each extensive variable is doubled in value.
The thermodynamic identity holds for the
quasi-static processes (T, P, ? are well-define
throughout the system)
The 1st Law for quasi-static processes (N
const)
This identity holds for small changes ?S provided
T and P are well defined.
The coefficients may be identified as
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Quasi-static Adiabatic Processes
Lets compare two forms of the 1st Law
(quasi-static processes, N is fixed, and P is
uniform throughout the system)
(holds for all processes)
Thus, for quasi-static processes
? isentropic processes
Quasistatic adiabatic (Q 0) processes
In this case,

reduces to
- the change in internal energy is due to a
purely mechanical compression (or expansion)
that involves no change in entropy. (No entropy
change no heat flow).
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An example of a non-quasistatic adiabatic process
Caution for non-quasistatic adiabatic processes,
?S might be non-zero!!!
Pr. 3.32. A non-quasistatic compression. A
cylinder with air (V 10-3 m3, T 300K, P 105
Pa) is compressed (very fast, non-quasistatic) by
a piston (S 0.01 m2, F 2000N, ?x 10-3m).
Calculate ?W, ?Q, ?U, and ?S.
holds for all processes, energy conservation
quasistatic, T and P are well-defined for any
intermediate state
quasistatic adiabatic ? isentropic
non-quasistatic adiabatic
?Q 0 for both
The non-quasistatic process results in a higher T
and a greater entropy of the final state.
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Direct approach
adiabatic quasistatic ? isentropic
adiabatic non-quasistatic
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2
To calculate ?S, we can consider any quasistatic
process that would bring the gas into the final
state (S is a state function). For example, along
the red line that coincides with the adiabata and
then shoots straight up. Lets neglect small
variations of T along this path (? U ltlt U, so it
wont be a big mistake to assume T ? const)
P
?U Q 1J
1
Vi
Vf
V
The entropy is created because it is an
irreversible, non-quasistatic compression.
2
P
For any quasi-static path from 1 to 2, we must
have the same ?S. Lets take another path along
the isotherm and then straight up
?U Q 2J
isotherm
1
Vi
Vf
V
straight up
Total gain of entropy
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The inverse process, sudden expansion of an
ideal gas (2 3) also generates entropy
(adiabatic but not quasistatic). Neither heat nor
work is transferred W Q 0 (we assume the
whole process occurs rapidly enough so that no
heat flows in through the walls).
2
P
Because U is unchanged, T of the ideal gas is
unchanged. The final state is identical with the
state that results from a reversible isothermal
expansion with the gas in thermal equilibrium
with a reservoir. The work done on the gas in the
reversible expansion from volume Vf to Vi
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1
Vi
Vf
V
The work done on the gas is negative, the gas
does positive work on the piston in an amount
equal to the heat transfer into the system
Thus, by going 1 ? 2 ? 3 , we will increase the
gas entropy by
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Comment on State Functions
is an exact differential (S is a state function).
Thus, the factor 1/T converts ?Q into an exact
differential.
U, S and V are the state functions, Q and W are
not. Specifying an initial and final states of a
system does not fix the values of ?Q and ?W, we
need to know the whole process (the intermediate
states). In math terms, ?Q and ?W are not exact
differentials and we need to use ? instead of d
for their increments. Analogy in classical
mechanics, if a force is not conservative (e.g.,
friction), the initial and final positions do
not determine the work, the entire path must be
specified.
z(x1,y1)
y
- it is an exact differential if it is
the difference between the values of some (state)
function z(x,y) at these points
z(x2,y2)
x
A necessary and sufficient condition for this
If this condition holds
- cross derivatives are not equal
e.g., for an ideal gas
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Entropy Change for Different Processes
The partial derivatives of S play very important
roles because they determine how much the entropy
is affected when U, V and N change
Type of interaction Exchanged quantity Governing variable Formula
thermal energy temperature
mechanical volume pressure
diffusive particles chemical potential
The last column provides the connection between
statistical physics and thermodynamics.
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Temperature
To discuss the physical meaning of these partial
derivatives of S, lets consider an ideal gas
isolated from the environment, and consider two
sub-systems, A and B, separated by a membrane,
which, initially, is insulating, impermeable for
gas molecules, and fixed at a certain position.
Thus, the sub-systems are isolated from each
other. (We have already considered such a system
when we discussed T, but we want to proceed with
P and ?)
UA, VA, NA
UB, VB, NB
Assume that the membrane becomes thermally
conducting (allow exchange U while V and N remain
fixed). The system will evolve to the state of
equilibrium that is characterized by a maximum
entropy. One of the sub-systems can increase its
energy only at the expense of the other
sub-system, and its crucial that the sub-system
receiving energy increases its entropy more than
the donor loses. Thus, the
sub-system with a larger ?S/?U should receive
energy, and this process will continue until
?SA/?UA and ?SB/?UB become the same
This was our logic when we defined the stat.
phys. temperature as
- the energy should flow from higher T to lower
T in thermal equilibrium, TA and TB should be
the same.
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Mechanical Equilibrium and Pressure
Lets now fix UA,NA and UB,NB , but allow V to
vary (the membrane is insulating, impermeable for
gas molecules, but its position is not fixed).
Following the same logic, spontaneous exchange
of volume between sub-systems will drive the
system towards mechanical equilibrium (the
membrane at rest). The equilibrium macropartition
should have the largest (by far) multiplicity ?
(U, V) and entropy S (U, V).
- the sub-system with a smaller
volume-per-molecule (larger P at the same T) will
have a larger ?S/?V, it will expand at the
expense of the other sub-system.
In mechanical equilibrium
- the volume-per-molecule should be the same for
both sub-systems, or, if T is the same, P must
be the same on both sides of the membrane.
The stat. phys. definition of pressure
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The Equation(s) of State for an Ideal Gas
Ideal gas (fN degrees of freedom)
The energy equation of state (U ? T)
The pressure equation of state (P ? T)
- we have finally derived the equation of state
of an ideal gas from first principles! In other
words, we can calculate the thermodynamic
information for an isolated system by counting
all the accessible microstates as a function of
N, V, and U.
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Problem
(all the processes are quasi-static)
(a) Calculate the entropy increase of an ideal
gas in an isothermal process. (b) Calculate the
entropy increase of an ideal gas in an isochoric
process. (c) What is the process in which the
entropy remains constant?
You should be able to do this using (a)
Sackur-Tetrode eq. and (b)
(a)
(b)
(c)
a quasistatic adiabatic process an isentropic
process
an adiabatic process
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Diffusive Equilibrium and Chemical Potential
Lets fix VA and VB (the membranes position is
fixed), but assume that the membrane becomes
permeable for gas molecules (exchange of both U
and N between the sub-systems, the molecules in A
and B are the same ).
For sub-systems in diffusive equilibrium
UA, VA, SA
UB, VB, SB
In equilibrium,
- the chemical potential
Sign - out of equilibrium, the system with the
larger ?S/?N will get more particles. In other
words, particles will flow from from a high ?/T
to a low ?/T.
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Chemical Potential examples
Einstein solid consider a small one, with N
3 and q 3.
lets add one more oscillator
To keep dS 0, we need to decrease the energy,
by subtracting one energy quantum.
?
Thus, for this system
Monatomic ideal gas
At normal T and P, ln(...) gt 0, and ? lt 0 (e.g.,
for He, ? - 510-20 J - 0.3 eV.
Sign - usually, by adding particles to the
system, we increase its entropy. To keep dS 0,
we need to subtract some energy, thus ?U is
negative.
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The Quantum Concentration
?
nN/V the concentration of molecules
0
when n increases
The chemical potential increases with the density
of the gas or with its pressure. Thus, the
molecules will flow from regions of high density
to regions of lower density or from regions of
high pressure to those of low pressure .
?
when n ? nQ, ? ? 0
- the so-called quantum concentration (one
particle per cube of side equal to the thermal de
Broglie wavelength). When nQ gtgt n, the gas is in
the classical regime.
At T300K, P105 Pa , n ltlt nQ. When n ? nQ, the
quantum statistics comes into play.
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Ideal Gas in a Gravitational Field
Pr. 3.37. Consider a monatomic ideal gas at a
height z above sea level, so each molecule has
potential energy mgz in addition to its kinetic
energy. Assuming that the atmosphere is
isothermal (not quite right), find ? and
re-derive the barometric equation.
note that the U that appears in the
Sackur-Tetrode equation represents only the
kinetic energy
In equilibrium, the two chemical potentials must
be equal
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S and CP
- this is a useful result that removes the
restriction of constant volume.
For P const processes, Q CPdT ?
Pr. 3.29. Sketch a graph of the entropy of H20
as a function of T at P const.
S
?
At T ?0, the graph goes to 0 with zero slope.
After initial fast increase of the slope, the
rate of the S increase slows down (CP almost
const). When solid melts, there is a large ?S at
T const, another jump at liquidgas phase
transformation. Curving down in both liquid and
gas phases because CP ? const.
water
ice
vapor
T
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Future Directions
Although the microcanonical ensemble is
conceptually simple, it is not the most practical
ensemble. The major problem is that we must
specify U - isolated systems are very difficult
to realize experimentally, and T rather than U
is a more natural independent variable.
Ensemble Macrostate Probability Thermodynamics
micro-canonical U, V, N (T fluctuates)
canonical T, V, N (U fluctuates)
grand canonical U, V, ? (N fluctuates)