# PHYS102hw3 - HOMEWORK Ch 12 CH 12...

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HOMEWORK Ch 12 CH 12: 3,5,10,19,21,24,28,34,48,58 12.3 The constant pressure is 5 5 1.5 atm 1.013 10 Pa atm 1.52 10 Pa P and the work done on the gas is W P V   (a) 3 4.0 m V   and   5 3 5 1.52 10 Pa 4.0 m 6.1 10 J W P V     (b) 3 3.0 m V   , so   5 3 5 1.52 10 Pa 3.0 m 4.6 10 J W P V     12.5 In each case, the work done on the gas is given by the negative of the area under the path on the PV diagram. Along those parts of the path where volume is constant, no work is done. Note that 5 3 3 1 atm 1.013 10 Pa and 1 Liter 10 m . (a) 5 3 3 0 4.00 1.013 10 Pa 4.00 2.00 10 m 810 J IAF IA AF I A I W W W P V V       (b)     5 3 3 triangular area rectangular area 1 1 2 2 1 4.00 1.00 1.013 10 Pa 4.00 2.00 10 m 2 507 J IF I B F B B F B I B F B W P P V V P V V P P V V         (c) 5 3 3 0 1.00 1.013 10 Pa 4.00 2.00 10 m 203 J IBF IB BF B F I W W W P V V     12.10 (a) The work done on the fluid is the negative since f i V V of the area under the curve on the PV diagram.

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Thus, 6 3 6 3 6 3 6.00 10 Pa 2.00 1.00 m 1 6.00 2.00 10 Pa 2.00 1.00 m 2 2.00 10 Pa 4.00 2.00 m if W
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