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586_Physics ProblemsTechnical Physics

# 586_Physics ProblemsTechnical Physics - 588 P20.37 Heat and...

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588 Heat and the First Law of Thermodynamics P20.37 WP VP V V PnRT P P sw =− + L N M M O Q P P bg af ej e j 18 0 10 6 . g 1.00 g cm cm m 33 3 W Qm L EQ W v + × F H G I K J == × = =+ = 100 8314 373 1013 10 18 0 310 00180 407 37 6 5 .. . . . . mol J K mol K N m g 10 g m kJ kg 2.26 10 J kg kJ kJ 2 63 6 int P20.38 (a) The work done during each step of the cycle equals the negative of the area under that segment of the PV curve. WW W W W VV P V V P V DA AB BC CD ii i i i i i i =+++ + − + = − 30 3 3 04 (b) The initial and final values of T for the system are equal. Therefore, E int = 0 and QW P V = 4 . (c) Vn R T i = − 4 4 4 1 00 8 314 273 9 08 . a fa fa f kJ FIG. P20.38 P20.39 (a) PV P V nRT f f = = × 200 300 499 10 3 mol 8.314 J K mol K J a f V nRT P V nRT P V i i f f i × × 1 3 00410 3 3 . . . J 0.400 atm J 1.20 atm m 3 (b) d V n R T V V f i F H G I K J × F H G I K J =+ z ln . ln . 1 3 548 3 kJ (c) W int == + 0 Q . k J P20.40 ∆∆ EE ABC AC int, int, = (conservation of energy) (a) W ABC ABC ABC int, (First Law) Q ABC =+= 800 500 1 300 J J J (b) V CD C CD =− ∆ , AB CD
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