problem20_39

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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20.39: The initial volume is . m 10 62 . 8 3 3 1 1 1 - × = = p nRT V a) At point 1, the pressure is given as atmospheric, and Pa, 10 01 . 1 5 1 × = p with the volume found above, = 1 V Pa 10 03 . 2 2 and , m 10 62 . 8 . m 10 62 . 8 5 1 1 2 3 3 1 2 3 3 1 2 × = = = × = = × - - p p p V V T T ( 29 ( 29 ( 29 ( 29 %. 50 500 . 0 1 is efficiency thermal the K, 600 and K 300 between operating engine cycle - Carnot a For %. 4 . 10 104 . 0 e) J. 227 J 1956 J 2183 is cycle one for engine the into flow heat the 2, - 1 process the for ns calculatio in the figures extra Keeping d) J. 227 J 559 J 786 is done net work The c) . J 1956 ) K 192 )( K mol J 3145 . 8 )( 2 7 )( mol 350 . 0 ( and J 1397 ) K 192 )( K mol J 3145 . 8 )( 2 5 ( J, 559 ) K 192 )( K mol J 3145 . 8 )( mol 350 . 0 ( isobaric; is 1 - 3 process The 0). ( J 786 ) K 108 )( K mol J 3145 . 8 )( 2 5 )( mol 350 . 0 ( and , 0 adiabatic, is 3 - 2 process The
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Unformatted text preview: 300 K J/mol 3145 . 8 2 5 mol 350 . . so isochoric, is 2-1 Process b) . m 10 41 . 1 and Pa 10 01 . 1 Pa). 10 013 . 1 using ( 600 300 J 2183 J 227 3 3 2 1 3 5 1 3 5 1 3 = =-= = = =-=-+ ∆ =-=-⋅ = ∆ =-=-⋅ = ∆ = ∆-=-⋅ = ∆ = ∆ =-=-⋅ = ∆ =-= ∆ = × = ⋅ × = ∆ = = ∆ = = ∆ × = = × = = × =-e W U T nC Q n T nC U T nR V p W W T nC W U Q T nC Q U W V V V p p p p V V V T T a...
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