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Carnot Cycle, PV Diagram

Carnot Cycle, PV Diagram. The work done by the engine is shown by the area enclosed by the curve, W eng The net work is equal to | Q h | – | Q c | D E int = 0 for the entire cycle. Carnot Cycle, Entropy D S = 0. Heat is exchanged in the isothermal portions:

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Carnot Cycle, PV Diagram

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  1. Carnot Cycle, PV Diagram • The work done by the engine is shown by the area enclosed by the curve, Weng • The net work is equal to |Qh| – |Qc| • DEint = 0 for the entire cycle

  2. Carnot Cycle, Entropy DS = 0 • Heat is exchanged in the isothermal portions: Qh = DEint-WAB = nRThln(VB/VA) Qc = nRTcln(VC/VD) • So entropy is also changed: DS = +Qh/Th - Qc/Tc = Rln{(VBDD)/(VAVC)} • For the adiabatic portions ThVBg-1 = TcVCg-1 ThVAg-1 = TcVDg-1 Therefore, ln{(VBVD)/(VAVC)} = 0 • So DS = 0 as expected for a reversible cycle.  VB/VA = VC/VD

  3. Heat Conduction, Entropy DS > 0 • Heat is exchanged inward at the high temperature: DSh = +DQh/Th • Heat is exchanged outward at the low temperature: DSc = -DQc/Tc • There is now work done (DW=0) and no change in internal energy (DEint=0) so DQh=DQc=DQ • So entropy change is: DS = D Q(1/Th -1/Tc) > 0 • DS>0 as expected, since change is IRREVERSIBLE Th DQh DQc Tc

  4. DS in a Free Expansion • Consider an adiabatic free expansion • DQ =0 and DW =0 so DEint=0 (ie T = constant) • DQ =0 but that is for an irreversible process and entropy change is dS = dQr/T Make change reversibly at constant T Change (removing membrane) is irreversible.

  5. DS in Free Expansion, cont • For an isothermal process with an ideal gas, • Since Vf > Vi , DS is positive • This indicates that both the entropy and the disorder of the gas increase as a result of the irreversible adiabatic expansion

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