# Lect13 - Le cture13 Heat Engines Thermodynamic processes...

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Lecture Lecture 13 Heat Engines Thermodynamic processes and entropy Thermodynamic cycles Extracting work from heat - How do we define engine efficiency? - Carnot cycle: the best possible efficiency Reading for Lecture 14: Elements Ch 10 Reading for this Lecture: Elements Ch 4D-F

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Lecture Review Entropy in Macroscopic Systems Traditional thermodynamic entropy: S = k ln = k σ We want to calculate S from macrostate information (p, V, T, U, N, etc .) Start with the definition of temperature in terms of entropy: The entropy changes when T changes : (We’re keeping V and N fixed.) If C V is constant: 2 1 2 1 2 1 ln T V V T T V V T C dT C dT dU dS S T T T T dT C C T T = = = = = , , 1 1 , or V N V N S kT U T U ∂σ
Lecture Entropy in Quasi-static Heat Flow When V is constant: dS dU/T = dQ/T W = 0, so dU = dQ In fact, dS = dQ/T during any reversible (quasi-static) process , even if V changes. The reason: In a reversible process, S tot (system plus environment) doesn’t change: 0 sys E E sys sys dS dS dU dS T dQ dS T = + = + = - for any reversible process , or final init dQ dQ dS S T T = ∆ = S tot = 0 if process is reversible. The reservoir is supplying (or absorbing) heat. The reservoir’s energy gain is the system’s heat loss. That’s how they interact.

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Lecture S in Isothermal Processes p V 1 2 T=constant Suppose V & p change but T doesn’t. Work is done (dW by = pdV). Heat must enter to keep T constant: dQ = dU+dW by . So: Special case, ideal gas: For an ideal gas, if dT = 0, then dU = 0. by dU dW dQ dU pdV dS T T T + + = = = 2 1 2 1 ln V V pdV NkTdV NkdV dS T VT V V NkdV S Nk V V = = =   ∆ = =     Remember: This holds for quasi-static processes, in which the system remains near thermal equilibrium at all times.
Lecture ACT 1 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? 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 Remove the barrier gas

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Lecture 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? 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 Solution vacuum Remove the barrier gas Energy (heat) leaves the environment, so its entropy decreases. In fact, since the environment and gas have the same T, the two entropy changes cancel: S tot = 0. This is a reversible process.
Lecture 1 ) We just saw that the entropy of a gas increases during a quasi-static isothermal expansion.

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## Lect13 - Le cture13 Heat Engines Thermodynamic processes...

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