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f5ch21 - CHAPTER 21 ENTROPY AND THE SECOND LAW 589 CHAPTER...

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CHAPTER 21 ENTROPY AND THE SECOND LAW . .. 589 C HAPTER 21 Answer to Checkpoint Questions . a, b, c . smaller . c, b, a . a, d, c, b . b Answer to Questions . increase . unchanged . ( a ) increase; ( b ) same . b, a, c, d . equal . c, a, b . lower the temperature of the low temperature reservoir . same . ( a ) same; ( b ) increase; ( c ) decrease . increase . ( a ) same; ( b ) increase; ( c ) decrease . greater . more than the age of the universe . the chance of the air clumping in one corner of the room would be greater

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T S i f S E int i f S V i f 590 CHAPTER 21 ENTROPY AND THE SECOND LAW . .. E The temperature of the gas in Kelvin is T = + = K. The heat absorbed is ten Q = T S = ( K)( : J/K) = : J : E The heat absorbed by the gas in the isothermal process is Q = W = nRT ln( V f =V i ) = nRT ln . Thus the entropy change is S = Q T = nRT ln T = nR ln = ( : mol)( : J/mol K)( K)(ln ) = : J/K : E The entropy change is S = Q T = W T = nRT T ln V f V i : Thus the number of moles is n = S R ln V f V i = : J/K ( : J/mol K) ln : L : L = : mol : E
CHAPTER 21 ENTROPY AND THE SECOND LAW . .. 591 The graphs are shown in the last page. While the rst two are obvious, the last one, the S vs V curve, satis es S / ln V + const. (see the result of E). E The change in entropy for the ideal gas is given by S = Q T = W T = nRT T ln V f V i = nR ln ; which is independed of the temperature T of the reservoir. Thus the change in entropy of the reservoir, S = S = nR ln , is also independent of T . Here we noticed thhat the net change in entropy for the entire system (the ideal gas plus the reservoir) is S total = S + S = for a reversible process so S = S . E ( a ) Since the gas is ideal, its pressure p is given in terms of the number of moles n , the volume V and the temperature T by p = nRT=V . The work done by the gas during the isothermal expansion is W = Z V V p dV = nRT Z V V dV V = nRT ln V V : Substitute V = V to obtain W = nRT ln = ( mol)( : J/mol K)( K) ln = : J : ( b ) Since the expansion is isothermal, the change in entropy is given by S = R ( =T ) dQ = Q=T , where Q is the heat absorbed. According to the rst law of thermodynamics E int = Q W . Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume. Since the expansion is isothermal E int = and Q = W . Thus S = W T = : J K = : J/K : ( c ) S = for all reversible adiabatic processes. E ( a ) The heat absorbed is given by Q = mc T = ( J/kg K)( : kg)( C C) = : J : ( b ) According to Sample Problem - S = mc ln T f T i = ( : kg)( J/kg K)( : kg) ln + + = : J/K :

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592 CHAPTER 21 ENTROPY AND THE SECOND LAW . .. E ( a ) Obviously, process AE is isothermal. ( b
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f5ch21 - CHAPTER 21 ENTROPY AND THE SECOND LAW 589 CHAPTER...

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