607_Physics ProblemsTechnical Physics

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

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Chapter 21 609 P21.25 (a) PV P V i i f f γγ = PP V V fi i f = F H G I K J = F H G I K J = γ 500 12 0 30 0 139 1.40 . . . . atm atm (b) T nR i ii == ×× = 5 00 1 013 10 12 0 10 200 365 53 .. . . Pa m mol 8.314 J mol K K 3 ej e j bg T nR f ff = 1 39 1 013 10 30 0 10 253 . . Pa m mol 8.314 J mol K K 3 e j (c) The process is adiabatic: Q = 0 === + 140 . C C RC C P V V V , CR V = 5 2 ∆∆ En C T WEQ 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 6 604 6 6 . af P21.26 V i = × F H G I K J π 250 10 0500 245 10 2 2 4 . m 2 m m 3 The quantity of air we find from nRT i = n RT n i 1013 10 8314 300 997 10 54 3 . . Pa m Jmo lK K mol 3 e j Adiabatic compression: P f =+ = 101 3 800 901 3 kPa kPa kPa (a) i i f f = VV P P V i f f = F H G I K J F H G I K J 1 4 57 5 101 3 901 3 515 10 . . . . m m 3 3 (b) f = TT T P P P P T P P T i f i i f i i f f F H G I K J = F H G I K J = F H G I K J =
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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