Lect09 - Lecture 9 Heat Engines q q q Thermodynamic...

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Physics 213: Lecture 9, Pg 1 Lecture 9 Lecture 9 Heat Engines Heat Engines Thermodynamic processes and entropy Thermodynamic processes and entropy Thermodynamic cycles Thermodynamic cycles Extracting work from heat Extracting work from heat How do we define engine efficiency? How do we define engine efficiency? Carnot cycle Carnot cycle --- best possible --- best possible Reference for Lect. 10: Elements Ch 10 References for this Lecture: Elements Ch 4D-F
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Physics 213: Lecture 9, Pg 2 Here’s the entropy change when T changes from T 1 to T 2 , keeping V and N fixed: Review: Entropy in Macroscopic Systems Review: Entropy in Macroscopic Systems Traditional thermodynamic entropy, S: σ k k S ) ln( k = Boltzmann constant How can we find out about S from macrostate information (p, V, T, U, N, etc.) ? o start with expression defining temperature 2 1 T V V T C dT C dT dU dS S T T T = = ⇒ ∆ = = = 1 2 2 1 T T C T dT C S V T T V ln o Special case: if C v is constant V,N V,N 1 1 S so kT U T U ∂σ ÷ ÷
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Physics 213: Lecture 9, Pg 3 Example: Example: S in Quasi-static S in Quasi-static Constant-Temperature Process Constant-Temperature Process Work (dW = -pdV) is done o heat must enter to keep T constant p V 1 2 T=T 1 o is S now zero?? T pdV dU T dW dU T dQ dS + = - = = 2 1 2 1 ln V V pdV NkTdV NkdV dS T VT V V NkdV S Nk V V = = =   ∆ = =  ÷   Specialize to IDEAL GAS: Now if dT = 0, then dU = 0
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Physics 213: Lecture 9, Pg 4 ACT 1: Total entropy change in isothermal processes ACT 1: Total entropy change in isothermal processes 1. We just saw that the entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment during this process? a. S env < 0 b. S env = 0 c. S env > 0 2. Consider instead the ‘free’ expansion (i.e., not quasi-static) of a gas. What happens to the total entropy during this process? a. S tot < 0 b. S tot = 0 c. S tot > 0 vacuum T Pull out barrier
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Physics 213: Lecture 9, Pg 5 Example: ADIABATIC process, i.e., no Q o V changes as applied p changes In equilibrium, the system must pick V to maximize its own S: dS/dV = 0 (no Q) WHY? o because no other S is changing o if, for example, it expands, GAIN in S from bigger V is exactly cancelled by LOSS in S due to reduced T. quasi-static adiabatic process Let’s find S for “quasi-static ” processes o everything stays very close to being in equilibrium, i.e., V, T, p, etc. not changing rapidly Quasi-static Processes Quasi-static Processes s ys t e m Insulated Cylinder p V 1 2 T H T C S = 0
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Physics 213: Lecture 9, Pg 6 Quasi-static Heat Flow and Entropy Quasi-static Heat Flow and Entropy In quasi-static reversible process, total S (of system plus environment) doesn’t change
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Lect09 - Lecture 9 Heat Engines q q q Thermodynamic...

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