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607_Physics ProblemsTechnical Physics

607_Physics ProblemsTechnical Physics - Chapter 21 P21.25(a...

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Chapter 21 609 P21.25 (a) PV P V i i f f γ γ = P P V V f i i f = F H G I K J = F H G I K J = γ 5 00 12 0 30 0 1 39 1.40 . . . . atm atm (b) T PV nR i i i = = × × = 5 00 1 013 10 12 0 10 2 00 365 5 3 . . . . Pa m mol 8.314 J mol K K 3 e je j b g T P V nR f f f = = × × = 1 39 1 013 10 30 0 10 2 00 253 5 3 . . . . Pa m mol 8.314 J mol K K 3 e je j b g (c) The process is adiabatic: Q = 0 γ = = = + 1 40 . C C R C C P V V V , C R V = 5 2 E nC T W E Q V int int mol 5 2 J mol K K K kJ kJ kJ = = F H G I K J = = = − = 2 00 8 314 253 365 4 66 4 66 0 4 66 . . . . . b g a f P21.26 V i = × F H G I K J = × π 2 50 10 0 500 2 45 10 2 2 4 . . . m 2 m m 3 The quantity of air we find from PV nRT i i i = n PV RT n i i i = = × × = × 1 013 10 2 45 10 8 314 300 9 97 10 5 4 3 . . . . Pa m J mol K K mol 3 e je j b ga f Adiabatic compression: P f = + = 101 3 800 901 3 . . kPa kPa kPa (a) PV P V i i f f γ γ = V V P P V f i i f f = F H G I K J = × F H G I K J = × 1 4 5 7 5 2 45 10 101 3 901 3 5 15 10 γ
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