Review-part II

Review-part II - Not required Chapter 5 p162-164 p188-190...

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Helpful Practice Problems Chapter 5: 29, 33, 34, 38 Chapter 6: 28, 32, 41, 67 Chapter 7: 12, 23, 32, 48 Not required: Chapter 5: p162-164; p188-190 Chapter 6: p217-219; p225 (Thermodynamic diagrams); p238-239; Chapter 7: p257-263;

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The second law No apparatus can operate in such a way that its only effect is to convert heat absorbed by a system completely into work done by the system. No process is possible which consists solely into the transfer of heat from one temperature level to a higher one. From the first law: C H Q Q W - = The thermal efficiency of the engine: H C H C H H Q Q Q Q Q Q W - = - = 1 η | Q H |: Heat absorbed from a hot reservoir |Q C |: heat discarded to a cold reservoir Chapter 5,
T H T C b c a d P V |Q H | |Q C | For the isothermal steps b → c and d → a b c H H V V RT Q ln = and a d C C V V RT Q ln = For adiabatic processes a → b and c → d b a T T V V V T dT R C H C ln = and c d T T V V V T dT R C H C ln = H C b c a d H C H C T T V V V V T T Q Q = = ) / ln( ) / ln( H C H C H T T Q Q Q W - = - = = 1 1 η 5 . 0 600 300 1 = - = Actual heat engines are irreversible and η rarely exceed 0.35. Because left side of the two equations are the same, so c d b a V V V V ln ln = or a d b c V V V V ln ln = For ideal gases If T C =300k and T H = 600 K, Carnot engine

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Carnot’s Theorem For two given heat reservoirs no engine can have a thermal efficiency higher than that of a Carnot engine The thermal efficiency of a Carnot engine depends only on the temperature levels and not upon the working substances of the engine. 0 2 2 3 3 = + T dQ T dQ C C H H T Q T Q = 0 = + C C H H T Q T Q A Carnot engine: From Entropy Because Q H is positive and Q C is negative Any reversible cycles can be divided into many Carnot cycles. If the isomer steps are infinitesimal 0 1 1 2 2 = + T dQ T dQ 0 = T dQ rev T dQ dS rev t = or t rev TdS dQ = 0 3 3 4 4 = + T dQ T dQ ………. define C H a d b c C H C H T T V V V V T T Q Q = = ) / ln( ) / ln( so Where S t is the total (rather than molar) entropy of the system
For two reversible paths ACB and ADB T dQ S rev ACB t ACB = T dQ S rev BDA t BDA = Because 0 = T dQ rev It can be concluded that S = S A – S B is independent of path A B D C V P 0 = + t BDA t ACB S S 0 = - ADB t ACB t S S t ADB t ACB S S = S is a state function

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= T dQ S irrev t 0 Q 0 t S Assume a fluid adiabatically and reversibly changing from A to B as shown in the P-V diagram. dQ rev = 0, dS t = 0 Now, assume that the fluid irreversibly changes from A to B’ then to B So Is this correct? No. 0 Q may be correct, but 0 t S is incorrect because = T dQ S rev t but not P V A B B’ Reversible adiabatic Any irreversible Any irreversible Entropy cannot be calculated using heat change in an irreversible process For an irreversible process The entropy change of the system ΔS is evaluated by an arbitrarily chosen reversible process. Since entropy is a state function, the entropy changes of the irreversible and reversible processes are identical for the same starting and final states
Entropy changes of an ideal gas as a function of P and T

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This note was uploaded on 05/01/2011 for the course CHBE 2110 taught by Professor Gallivan during the Spring '08 term at Georgia Tech.

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Review-part II - Not required Chapter 5 p162-164 p188-190...

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