Lecture 6. Entropy of an Ideal Gas (Ch. 3)

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Lecture 6. Entropy of an Ideal Gas (Ch. 3)
Today we will achieve an important goal well
derive the equation(s) of state for an ideal gas
from the principles of statistical mechanics. We
will follow the path outlined in the previous
lecture

• Find ? (U,V,N,...) the most challenging step
• S (U,V,N,...) kB ln ? (U,V,N,...)
• Solve for U f (T,V,N,...)

So far we have treated quantum systems whose
states in the configuration (phase) space may be
enumerated. When dealing with classical systems
with translational degrees of freedom, we need to
learn how to calculate the multiplicity.
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Multiplicity for a Single particle

• is more complicated than that for an Einstein
  solid, because it depends on three rather than
  two macro parameters (e.g., N, U, V).

Example particle in a one-dimensional box
-L
L
The total number of ways of filling up the cells
in phase space is the product of the number of
ways the space cells can be filled times the
number of ways the momentum cells can be filled
px
The total number of microstates
Quantum mechanics (the uncertainty principle)
helps us to numerate all different states in the
configuration (phase) space
-L
L
x
?px
-px
?x
Q.M.
The number of microstates
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Multiplicity of a Monatomic Ideal Gas (simplified)
For a molecule in a three-dimensional box the
state of the molecule is a point in the 6D space
- its position (x,y,z) and its momentum
(px,py,pz). The number of space microstates is
For N molecules
There is some momentum distribution of molecules
in an ideal gas (Maxwell), with a long tail
that goes all the way up to p (2mU)1/2 (U is
the total energy of the gas). However, the
momentum vector of an average molecule is
confined within a sphere of radius p (2mU/N)1/2
(U/N is the average energy per molecule). Thus,
for a single average molecule
The total number of microstates for N molecules
However, we have over-counted the multiplicity,
because we have assumed that the atoms are
distinguishable. For indistinguishable quantum
particles, the result should be divided by N!
(the number of ways of arranging N identical
atoms in a given set of boxes)
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More Accurate Calculation of ?p
Momentum constraints
1 particle -
2 particles -
The accessible momentum volume for N particles
the area of a 3N-dimensional hyper-sphere ??p
N 1
The reason why m matters for a given U, a
molecule with a larger mass has a larger
momentum, thus a larger volume accessible in
the momentum space.
Monatomic ideal gas (3N degrees of freedom)
f N- the total of quadratic degrees of
freedom
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Entropy of an Ideal Gas
the Sackur-Tetrode equation
Monatomic ideal gas
an average volume per molecule
an average energy per molecule
In general, for a gas of polyatomic molecules
f ? 3 (monatomic), 5 (diatomic), 6
(polyatomic)
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Problem
Two cylinders (V 1 liter each) are connected by
a valve. In one of the cylinders Hydrogen (H2)
at P 105 Pa, T 200C , in another one Helium
(He) at P 3105 Pa, T1000C. Find the entropy
change after mixing and equilibrating.
For each gas
The temperature after mixing
H2
He
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Entropy of Mixing
Consider two different ideal gases (N1, N2) kept
in two separate volumes (V1,V2) at the same
temperature. To calculate the increase of entropy
in the mixing process, we can treat each gas as a
separate system. In the mixing process, U/N
remains the same (T will be the same after
mixing). The parameter that changes is V/N
if N1N21/2N , V1V21/2V
The total entropy of the system is greater after
mixing thus, mixing is irreversible.
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Gibbs Paradox
- applies only if two gases are different !
If two mixing gases are of the same kind
(indistinguishable molecules)
?Stotal 0 because U/N and V/N available for
each molecule remain the same after mixing.
Quantum-mechanical indistinguishability is
important! (even though this equation applies
only in the low density limit, which is
classical in the sense that the distinction
between fermions and bosons disappear.
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Problem
Two identical perfect gases with the same
pressure P and the same number of particles N,
but with different temperatures T1 and T2, are
confined in two vessels, of volume V1 and V2 ,
which are then connected. find the change in
entropy after the system has reached equilibrium.
- prove it!
at T1T2, ?S0, as it should be (Gibbs paradox)
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An Ideal Gas from S(N,V,U) - to U(N,V,T)
Ideal gas (fN degrees of freedom)
?
- the energy equation of state
- in agreement with the equipartition theorem,
the total energy should be ½kBT times the number
of degrees of freedom.
The heat capacity for a monatomic ideal gas
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Partial Derivatives of the Entropy
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, e.g
for an ideal gas
When all macroscopic quantities S,V,N,U are
allowed to vary
We are familiar with the physical meaning 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 the other two partial
derivatives of the entropy
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Thermodynamic Identity for dU(S,V,N)
? if monotonic as a function of U
(quadratic degrees of freedom!), may be
inverted to give
pressure
chemical potential
compare with
? 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.
- the so-called thermodynamic identity for U
This holds for quasi-static processes (T, P, ?
are well-define throughout the system).
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The Exact Differential of S(U,V,N)
The coefficients may be identified as
Again, this holds for quasi-static processes (T
and P are well defined).
Type of interaction Exchanged quantity Governing variable Formula
thermal energy temperature
mechanical volume pressure
diffusive particles chemical potential
connection between thermodynamics and statistical
mechanics
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Mechanical Equilibrium and Pressure
Lets 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 Pressure Equation 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!
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Quasi-Static Processes
(all processes)
(quasi-static processes with fixed N)
Thus, for quasi-static processes
Comment on State Functions
P
- is an exact differential (S is a state
function). Thus, the factor 1/T converts ?Q into
an exact differential for quasi-static processes.
V
Quasistatic adiabatic (?Q 0) processes
? isentropic processes
The quasi-static adiabatic process with an ideal
gas
- weve derived these equations from the 1st Law
and PVRT
On the other hand, from the Sackur-Tetrode
equation for an isentropic process
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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.
You should be able to do this using (a)
Sackur-Tetrode eq. and (b)
Lets verify that we get the same result with
approaches a) and b) (e.g., for Tconst)
?
Since ?U 0,
(Pr. 2.34)
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Problem

• A body of mass M with heat capacity (per unit
  mass) C, initially at temperature T0?T, is
  brought into thermal contact with a heat bath at
  temperature T0..
• (a) Show that if ?TltltT0, the increase ?S in the
  entropy of the entire system (bodyheat bath)
  when equilibrium is reached is proportional to
  (?T)2.
• Find ?S if the body is a bacteria of mass
  10-15kg with C4 kJ/(kgK), T0300K, ?T0.03K.
• What is the probability of finding the bacteria
  at its initial T0?T for ?t 10-12s over the
  lifetime of the Universe (1018s).

(a)
(b)
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Problem (cont.)
? for the (non-equilibrium) state with Tbacteria
300.03K is greater than ? in the
equilibrium state with Tbacteria 300K by a
factor of
(b)
The number of 1ps trials over the lifetime of
the Universe
Thus, the probability of the event happening in
1030 trials
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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.
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P
S const along the isentropic line
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holds for all processes, energy conservation
1
quasistatic, T and P are well-defined for any
intermediate state
Vi
Vf
V
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 adiabat 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
3
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