problem20_40

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

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20.40: (a) The temperature at point c is , from since K 1000 nRT pV T c = = the maximum temperature occurs when the pressure and volume are both maximum. So ( 29 ( 29 ( 29 ( 29 mol. 16 . 2 K 1000 K mol J 3145 . 8 m 0300 . 0 Pa 10 00 . 6 3 5 = × = = c c c RT V p n (b) Heat enters the gas along paths ab and bc, so the heat input per cycle is = H Q . ac ac ac U W Q + = Path ab has constant volume and path bc has constant pressure, so J. 10 20 . 1 ) m 0100 . 0 m 0300 . 0 )( Pa 10 00 . 6 ( ) ( 0 4 3 3 5 × = - × = - + = + = b c c bc ab ac V V p W W W For an ideal gas, mol.K, J 46 . 28 , CO For . using , ) ( ) T ( 2 = = - = - = V a a c c V a c V ac C R pV nT R V p V p C T nC U so J. 10 48 . 5 )) m Pa)(0.0100 10 00 . 2 ( ) m Pa)(0.0300 10 00 . 6 (( K mol J 3145 . 8 K mol J 46 . 28 4 3 5 3 5 × = × - × = ac U Then J. 10 6.68 J 10 5.48 J 10 20 . 1 4 4 4 H × = × + × = Q (c) Heat is removed from the gas along paths cd and da , so the waste heat per cycle is . ca ca ca C U W Q Q
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Unformatted text preview: + = = Path cd has constant volume and path da has constant pressure, so J. 10 400 . ) m 0300 . m Pa)(0.0100 10 00 . 2 ( ) ( 4 3 3 5 ×-=-× =-+ = + = d a d da cd ca V V p W W W From (b), J. 10 5.88 J 10 5.48 J 10 0.400 so J, 10 48 . 5 4 4 4 C 4 ×-= ×-×-= ×-= ∆-= ∆ Q U U ac ca (d) The work is the area enclosed by the rectangular path abcd, J. 8000 J 10 5.86 J 10 68 . 6 or ), )( ( 4 4 C H = ×-× = + =--= Q Q W V V p p W a c a c (e) 0.120. J) 10 (6.68 J) 8000 ( 4 H = × = = Q W e...
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This document was uploaded on 02/06/2008.

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