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Principle of entropy increase
Consider a closed system in adiabatic enclosure undergoing
. b –Sa?
• Return the system to state a by a reversible process, withany heat exchange happening with a single reservoir at Tr .Clearly Qb→a ≠ 0
• or e cyc e, b→a = b→a + b→a• Can Qb→a be positive? No, but why not?
• Because Kelvin-Planck statement will be violated
• Therefore Qb→a < 0, ie., Sa – Sb < 0
• ., a.enclosure cannot decrease. Isolated system a particular case
• (∆S)isolated ≥ 0
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Entropy creation and transfer
• Let a closed system undergo an infinitesimal process
ur ng w c ere s a ea rans er w a reservo r aTr.
• dSsystem + dSreservoir ≥0. But dSr = - δQ/Tr
• dSs + dSr = δSgen ≥ 0, defining the entropy generation
• s r gen
• Entropy transfer accompanies heat transfer
Process Entropy
changeEntropy
transferEntropycreation
Work Lost
work
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1 R ln 2 R ln 2 0 RT ln 2 0
2 R ln 2 0 R ln 2 0 ?
3 R ln 2 0.5 R 0.19 R 0.5 RT ?
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Clausius inequality
s r gen, gen ,
• dSs ≥ δQ/Tr , the Clausius inequality (meaning?)
• For a cycle ∫ δQ/Tr ≤ 0
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What is entropy?
Reversible work versus reversible heat transfer
Consider an ideal gas (non-interacting point particles)
What is the effect of a reversible adiabatic decrease in the
volume?
What is the effect of a reversible transfer of heat at constant
volume?
What happens in the course of the free expansion process
that we have discussed in some detail?
What is entropy?
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Spontaneity and Equilibrium
Isolated system spontaneously evolves to states of greater
entropy
What therefore is the criterion that characterizes the
equilibrium state of an isolated system?
• Maximum Entropy
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Irreversibility and entropy
• Due to nature of matter at microscopic level
• Large scale natural processes eventually some kind of mixing –
of particles or over available space
• In some cases the mixing or sharing is of energy, as in the
equa za on o e empera ure o oc s
• In friction, kinetic energy of body as a whole into random energy
of component molecules
• Chemical reaction – at e uilibrium the available ener is
distributed over the available quantum states in the most random
possible manner
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Entropy and Mixing - Spreading
Two distinct kinds
• Spreading of particles over space
• Spreading of available energy of system among available
articles and their states
The two effects may oppose one another – under adiabatic
isolation, equilibrium determined by maximum in the overall‘randomness’ or s readin – confi urational + thermal
Illustrate with discussion of diffusion
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Natural processes and Microstates
• Thermodynamic state or macrostate
• Quantum state or microstate
• A macrostate comprises an astronomical number of
microstates in a t ical macrosco ic s stem
• Natural process is one in which there is an increase in Ω,
the number of microstates
• ,
microstate the system is in
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Entropy andΩ
• Adiabatic mixing of A and B
• Increase in Ω, which in turn is a measure of the decrease
in information about the actual state
• Increase in entro S
• Ω = Ω(U,V,N) and S = S(U,V,N), both are functions of
state. How are they related to one another?
• = –
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Entropy andΩ
• S = k ln Ω. In previous example, assuming no energetic
distinctions, ∆S = k ln(70)
• S as a measure of thermodynamic probability, also
measured by Ω
• Mixed-up-ness, order and disorder – case of adiabatic
change of super cooled liquid water to ice – greater thermaldisorder, greater configurational order
•Increase of S – system spread over larger number of
possible quantum states – loss of information
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Ideal monatomic gas
• Non-interacting structure-less point particles
• Translational states, given by particle in box model, very
closely spaced, virtually a continuum
• Translational state ener ies de end on the size ie.
volume of the box – work versus heat
• Accessible states for a particle, z = (2πmkT/h2
)3/2
V
• = N ,
of thermally accessible quantum states or microstates
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Ideal monatomic gas
S = Nk ln V + 3/2 Nk ln T + constant
• If T1 = T2, ∆S = Nk ln(V2/V1)
Comparing to the result we had earlier, for 1 mole, can find k.
Entro increase due to more accessible levels at lower
energy
• If V1 = V2, ∆S = 3/2 Nk ln(T2/T1)
• If ∆S = 0, TV2/3 = constant
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