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**Unformatted text preview: **19.39: See Exercise 19.32. a) p2 p1 (V1 V2 ) (1.10 10 5 Pa) ((5.00 10 3 m 3 1.100 10 2 m 3 ))1.29 4.50 10 4 Pa. b) Using Equation (19.26), ( p1V1 p2V2 ) W 1 [(1.1 105 N m 3 )(5.0 10 3 m 3 ) (4.5 104 N m 3 )(1.0 10 (1.29 1) 2 m 3 )] , and thus W 345 J c) (T2 T1 ) (V2 V1 ) 1 ((5.00 10 3 m 3 ) (1.00 10 2 m 3 )) 0.29 0.818 . The final temperature is lower than the initial temperature, and the gas is cooled. ...

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