problem19_69

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

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19.69: a) Solving for p as a function of V and T and integrating with respect to V, . 1 1 ln 1 2 2 1 2 2 2 2 1 - + - - = = - - = V V an nb V nb V nRT pdV W V an nb V nRT p V V When ( 29 , ln , 0 1 2 V V nRT W b a = = = as expected. b) Using the expression found in part (a), ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 J. 10 11 . 3 ln(2) ii) J. 10 80 . 2 m 10 00 . 2 1 m 10 00 . 4 1 mol 80 . 1 mol m J 554 . 0 mol / m 10 38 . 6 mol 80 . 1 m 10 00 . 2 mol / m 10 38 . 6 mol 80 . 1 m 10 00 . 4 ln K 300 K mol J 3145 . 8 mol 80 . 1 i) 3 3 3 3 3 3 2 2 3 2 5 3 3 2 5 3 3 × = × = × - × + × - × × - × ×
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Unformatted text preview: ⋅ =------nRT W c) 300 J to two figures, larger for the ideal gas. For this case, the difference due to nonzero a is more than that due to nonzero b . The presence of a nonzero a indicates that the molecules are attracted to each other and so do not do as much work in the expansion....
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