ch20-p022 - 22. (a) The final pressure is p f = ( 5.00 kPa...

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22. (a) The final pressure is () ( ) 33 3 1.00 m 2.00 m 1.00 m = 5.00 kPa = 5.00 kPa 1.84 kPa . i f f VV a pe e = (b) We use the ratio form of the gas law (see Sample Problem 19-1) to find the final temperature of the gas: 3 3 (1.84 kPa)(2.00 m ) 600 K 441 K . (5.00 kPa)(1.00 m ) ff fi ii pV TT §· == = ¨¸ ©¹ For later purposes, we note that this result can be written “exactly” as T f = T i (2 e –1 ). In our solution, we are avoiding using the “one mole” datum since it is not clear how precise it is. (c) The work done by the gas is ( ) / / 1.00 3 1.00 2.00 (5.00 kPa) 5.00 kPa 5.00 kPa 1.00 m 3.16 kJ . f f i i i i fV V
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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