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Lecture 17

Chem 402 Lecture 17: L17 2:27:17
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by OneClass1401264 , Spring 2017
3 Pages
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Spring 2017

Department
University College - Chemistry
Course Code
University College - Chemistry Chem 402
Professor
Barnes Alexander
Lecture
17

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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
process to find   
was path independent
2. Derivation of Entropy as a State Function (Math Proof for Ideal gas)
a.     
b.      (P = nRT/V)
c.    

Now check to see if the cross terms are equal, if so, it is a state function, if
not, it is path dependent

  
 ?
  
these are not equal, and thus  is path dependent
d. Now, consider


  (divide through by T)

  
   they are equal, so this function is a state function

 , 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 2nd law if |q1| = |q2|
Consider  
On a number line: have ∆Scold (q2/T2) which is negative, and ∆Shot (q1/T1)
Suni = ∆Scold - ∆Shot < 0 (violates second law!)
c. The right side is ok since work in will increase ∆Suni, for a reversible process,
Suni = 0
When work is added, ∆Scold is the same, but ∆Shot increases because work is
added for going into the hot reservoir
Since S = kln, 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
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Description
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 ð𝑞𝑟𝑒𝑣 process to find 𝑑𝑆 = 𝑇 was path independent 2. Derivation of Entropy as a State Function (Math Proof for Ideal gas) a. 𝑑𝑈 = ð𝑞 𝑟𝑒𝑣+ 𝑤 𝑟𝑒𝑣 b. 𝐶 𝑑𝑇 = ð𝑞 − 𝑝𝑑𝑉 (P = nRT/V) 𝑣 𝑟𝑒𝑣 𝑛𝑅𝑇 c. ð𝑞 𝑟𝑒𝑣= 𝐶 𝑣𝑇 + 𝑑𝑉 𝑣 • Now check to see if the cross terms are equal, if so, it is a state function, if not, it is path dependent 𝑛𝑅𝑇 𝜕𝐶𝑣 𝜕( 𝑉 ) • (𝜕𝑉 )𝑇= ( 𝜕𝑇 )𝑉? 𝑛𝑅 • 0 ≠ 𝑉  these are not equal, and thus ð𝑞 𝑟𝑒𝑣is path dependent ð𝑞𝑟𝑒𝑣 𝐶𝑣𝑑𝑇 𝑛𝑅𝑇 d. Now, consider 𝑇 = 𝑇 + 𝑣𝑇 𝑑𝑉 (divide through by T) 𝜕𝐶𝑣 𝑛𝑅𝑇 • ( 𝑇 ) = ( 𝜕( 𝑉 )) = 0  they are equal, so this function is a state function 𝜕𝑉 𝑇 𝜕𝑇 𝑉 ð𝑞𝑟𝑒𝑣 • 𝑇 = 𝑆, 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 | ð𝑞 • Consider 𝑑𝑆 = 𝑟𝑒𝑣 𝑇 • On a number line: have ∆S coldq2/T2) which is negative, and ∆Shot(q1/T1) • ∆S = ∆S - ∆S < 0 (violates second law!) uni cold hot c. The right side is ok since work in will increase ∆Suni,or a reversible process, ∆S uni= 0 • When work is added, ∆S coldis the same, but ∆Shotincreases because work is added for going into the hot reservoir • Since S = klnΩ, 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 • Can calculate ∆S as (S f S >i0) using the Boltzman equation • Higher ∆S 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: • Left is the heat engine, right is the refrigerator • Overall ∆S uni= 0 since both a heat engine and refrigerator are reversible, and note that this machine wouldn’t function since neither process is irreversible/spontaneous • Overall positive heat since system accepts heat • Always think about microstate expansion/contraction with heat changes using S = klnΩ 2. The Carnot Cycle (a typical heat engine, ideal gases, all steps reversible) a. Cycle: • 1 2 is an isothermal expansion at T (h1t), ∆U = q + w1= 0, 1 = -w 1 1 𝑅𝑇 ln( )2 1 𝑉1 • 2  3 is an adiabatic expansion (q = 0), ∆U = w ` 1 C (Tv– T2) 1 𝑇2 𝑉2 −1 i. Reversible adiabat:
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