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596_PartUniversity Physics Solution

# 596_PartUniversity Physics Solution - The Second Law of...

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The Second Law of Thermodynamics 20-11 E VALUATE : For a cycle 0, U Δ = so by U Q W Δ = it must be that net net Q W = for a cycle. We can also check that net 0: U Δ = net 1 2 2 3 3 1 2180 J 1050 J 1130 J 0 U U U U Δ = Δ + Δ + Δ = = (Carnot), e e < as it must. 20.41. I DENTIFY : , pV nRT = so pV is constant when T is constant. Use the appropriate expression to calculate Q and W for each process in the cycle. H . W e Q = S ET U P : For an ideal diatomic gas, 5 2 V C R = and 7 2 . p C R = E XECUTE : (a) 3 2.0 10 J. a a p V = × 3 2.0 10 J. b b p V = × pV nRT = so a a b b p V p V = says . a b T T = (b) For an isothermal process, 2 1 ln( / ). Q W nRT V V = = ab is a compression, with , b a V V < so 0 Q < and heat is rejected. bc is at constant pressure, so . p p C Q nC T p V R = Δ = Δ V Δ is positive, so 0 Q > and heat is absorbed. cd is at constant volume, so . V V C Q nC T V p R = Δ = Δ p Δ is negative, so 0 Q < and heat is rejected. (c) 3 2.0 10 J 241 K. (1.00)(8.314 J/mol K) a a a p V T nR × = = = 241 K. b b b a p V T T nR = = = 3 4.0 10 J 481 K. (1.00)(8.314 J/mol K) c c c p V T nR × = = = (d) 3 3 3 0.0050 m ln (1.00 mol)(8.314 J/mol K)(241 K)ln 1.39 10 J. 0.010 m b ab a V Q nRT V = = = − × 3 7 2 (1.00)( )(8.314 J/mol K)(241 K) 7.01 10 J. bc p Q nC T = Δ = = × 3 5 2 (1.00)( )(8.314 J/mol K)( 241 K) 5.01 10 J. ca V Q nC T = Δ = = − × net 610 J. ab bc ca Q Q Q Q = + + = net net 610 J. W Q = = (e) 3 H 610 J 0.087 8.7% 7.01 10 J W e Q = = = = × E VALUATE : We can calculate W for each process in the cycle. 3 1.39 10 J. ab ab W Q = = − × 5 3 3 (4.0 10 Pa)(0.0050 m ) 2.00 10 J. bc W p V = Δ = × = × 0. ca W = net 610 J, ab bc ca W W W W = + + = which does equal net . Q 20.42. (a) I DENTIFY and S ET U P : Combine Eqs.(20.13) and (20.2) to eliminate C Q and obtain an expression for H Q in terms of W , C , T and H . T 1.00 J, W = C 268.15 K, T = H 290.15 K T = For the heat pump C 0 Q > and H 0 Q < E XECUTE : C H ; W Q Q = + combining this with C C H H Q T Q T = − gives H C H 1.00 J 13.2 J 1 / 1 (268.15/290.15) W Q T T = = = (b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required. (c) E VALUATE : From part (a), H C H . 1 / W Q T T = H Q decreases as C T decreases. The heat pump is less efficient as the temperature difference through which the heat has to be °pumped± increases. In an engine, heat flows from H T to C T and work is extracted. The engine is more efficient the larger the temperature difference through which the heat flows. 20.43. I DENTIFY : b c T T = and is equal to the maximum temperature. Use the ideal gas law to calculate .

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596_PartUniversity Physics Solution - The Second Law of...

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