Chem 402 Lecture 17: L17 2:27:17
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Department
University College - Chemistry
Course Code
University College - Chemistry Chem 402
Professor
Barnes Alexander

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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 Γ°π‘žπ‘Ÿπ‘’π‘£ process to find 𝑑𝑆 = 𝑇 was path independent 2. Derivation of Entropy as a State Function (Math Proof for Ideal gas) a. π‘‘π‘ˆ = Γ°π‘ž π‘Ÿπ‘’π‘£+ 𝑀 π‘Ÿπ‘’π‘£ b. 𝐢 𝑑𝑇 = Γ°π‘ž βˆ’ 𝑝𝑑𝑉 (P = nRT/V) 𝑣 π‘Ÿπ‘’π‘£ 𝑛𝑅𝑇 c. Γ°π‘ž π‘Ÿπ‘’π‘£= 𝐢 𝑣𝑇 + 𝑑𝑉 𝑣 β€’ Now check to see if the cross terms are equal, if so, it is a state function, if not, it is path dependent 𝑛𝑅𝑇 πœ•πΆπ‘£ πœ•( 𝑉 ) β€’ (πœ•π‘‰ )𝑇= ( πœ•π‘‡ )𝑉? 𝑛𝑅 β€’ 0 β‰  𝑉 οƒ  these are not equal, and thus Γ°π‘ž π‘Ÿπ‘’π‘£is path dependent Γ°π‘žπ‘Ÿπ‘’π‘£ 𝐢𝑣𝑑𝑇 𝑛𝑅𝑇 d. Now, consider 𝑇 = 𝑇 + 𝑣𝑇 𝑑𝑉 (divide through by T) πœ•πΆπ‘£ 𝑛𝑅𝑇 β€’ ( 𝑇 ) = ( πœ•( 𝑉 )) = 0 οƒ  they are equal, so this function is a state function πœ•π‘‰ 𝑇 πœ•π‘‡ 𝑉 Γ°π‘žπ‘Ÿπ‘’π‘£ β€’ 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 | Γ°π‘ž β€’ Consider 𝑑𝑆 = π‘Ÿπ‘’π‘£ 𝑇 β€’ On a number line: have βˆ†S coldq2/T2) which is negative, and βˆ†Shot(q1/T1) β€’ βˆ†S = βˆ†S - βˆ†S < 0 (violates second law!) uni cold hot c. The right side is ok since work in will increase βˆ†Suni,or a reversible process, βˆ†S uni= 0 β€’ When work is added, βˆ†S coldis the same, but βˆ†Shotincreases 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 βˆ†S as (S f S >i0) using the Boltzman equation β€’ Higher βˆ†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: β€’ Left is the heat engine, right is the refrigerator β€’ Overall βˆ†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 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€’ 1οƒ  2 is an isothermal expansion at T (h1t), βˆ†U = q + w1= 0, 1 = -w 1 1 𝑅𝑇 ln( )2 1 𝑉1 β€’ 2 οƒ  3 is an adiabatic expansion (q = 0), βˆ†U = w ` 1 C (Tv– T2) 1 𝑇2 𝑉2 ο§βˆ’1 i. Reversible adiabat:
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