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 55 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 4.0 m 6.1 10 J W P V       (b) 3 3.0 m V   , so   5 3 5 1.52 3.0 m 4.6 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 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 11 22 1 4.00 1.00 1.013 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 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 fi VV of the area under the curve on the PV diagram.
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Thus,     63 6.00 10 Pa 2.00 1.00 m 1 6.00 2.00 2.00 2 2.00 4.00 2.00 m if W     7 1.20 10 J 12.0 MJ if W   (b)
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This note was uploaded on 03/07/2011 for the course PHYS 102 taught by Professor Armenn.kocharian during the Winter '11 term at California State University Los Angeles .

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PHYS102hw3 - HOMEWORK Ch 12 CH 12:...

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