27 February 2017
L17: The Carnot Cycle and Third Law of Thermodynamics
I. Second Law of Thermodynamics
A. Entropy
1. Derivation by thermocycler path using an isotherm, adiabat, and constant volume
Γ°ππππ£
process to find ππ = π was path independent
2. Derivation of Entropy as a State Function (Math Proof for Ideal gas)
a. ππ = Γ°π πππ£+ π€ πππ£
b. πΆ ππ = Γ°π β πππ (P = nRT/V)
π£ πππ£ ππ
π
c. Γ°π πππ£= πΆ π£π + ππ
π£
β’ Now check to see if the cross terms are equal, if so, it is a state function, if
not, it is path dependent
ππ
π
ππΆπ£ π( π )
β’ (ππ )π= ( ππ )π?
ππ
β’ 0 β π ο these are not equal, and thus Γ°π πππ£is path dependent
Γ°ππππ£ πΆπ£ππ ππ
π
d. Now, consider π = π + π£π ππ (divide through by T)
ππΆπ£ ππ
π
β’ ( π ) = ( π( π )) = 0 ο they are equal, so this function is a state function
ππ π ππ π
Γ°ππππ£
β’ π = π, thus entropy is a state function
B. Heat Engines and Refrigerators
1. Clausius
a. It is impossible for any system to operate in a cycle that takes heat from a cold
reservoir and transfers it to a hot reservoir without at the same time converting
some work into heat
b. The left violates the 2 law if |q1| = |2 |
Γ°π
β’ Consider ππ = πππ£
π
β’ On a number line: have βS coldq2/T2) which is negative, and βShot(q1/T1)
β’ βS = βS - βS < 0 (violates second law!)
uni cold hot
c. The right side is ok since work in will increase βSuni,or a reversible process,
βS uni= 0
β’ When work is added, βS coldis the same, but βShotincreases because work is
added for going into the hot reservoir
β’ Since S = klnΞ©, as you ad heat there is a larger distribution of energies and
microstates, adding the same amount of heat again spreads out energy levels a
little bit more β’ Can calculate βS as (S f S >i0) using the Boltzman equation
β’ Higher βS for starting at a colder temperature: smaller heat added gives a
larger difference in entropy at colder temperatures
d. If we were to couple an engine (giving work output) and a refrigerator:
β’ Left is the heat engine, right is the refrigerator
β’ Overall βS uni= 0 since both a heat engine and refrigerator are reversible, and
note that this machine wouldnβt function since neither process is
irreversible/spontaneous
β’ Overall positive heat since system accepts heat
β’ Always think about microstate expansion/contraction with heat changes using
S = klnΞ©
2. The Carnot Cycle (a typical heat engine, ideal gases, all steps reversible)
a. Cycle:
β’ 1ο 2 is an isothermal expansion at T (h1t), βU = q + w1= 0, 1 = -w 1 1
π
π ln( )2
1 π1
β’ 2 ο 3 is an adiabatic expansion (q = 0), βU = w ` 1 C (Tvβ T2) 1
π2 π2 ο§β1
i. Reversible adiabat:

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