problem19_32

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

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19.32: a) See also Exercise 19.36; ( 29 Pa. 10 76 . 4 m 0400 . 0 m 0800 . 0 Pa 10 50 . 1 5 3 3 5 2 1 1 2 3 5 × = × = = γ V V p p b) This result may be substituted into Eq. (19.26), or, substituting the above form for 2 p , ( 29 ( 29 J. 10 60 . 1 0400 . 0 0800 . 0 1 m 0800 . 0 Pa 10 50 . 1 2 3 1 1 1 4 3 5 1 2 1 1 1 3 2 × - = - × = - - = - γ V V V p γ W c) From Eq. (19.22),
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Unformatted text preview: ( 29 ( 29 ( 29 , 59 . 1 0400 . 0800 . 3 2 1 1 2 1 2 = = =-γ V V T T and since the final temperature is higher than the initial temperature, the gas is heated (see the note in Section 19.8 regarding “heating” and “cooling.”)...
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• Heat, 0.0800 M, 0.0400 m, 0.0400 2 0.0800 m, 1 1 W

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