639_Physics ProblemsTechnical Physics

639_Physics ProblemsTechnical Physics - Chapter 22 P22.22...

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Chapter 22 641 P22.22 (a) First, consider the adiabatic process DA : PV D D A A γγ = so PP V V A D = F H G I K J = F H G I K J = γ 1400 712 53 kPa 10.0 L 15.0 L kPa . Also nRT V V nRT V V D D D A A A F H G I K J = F H G I K J or TT V V A D = F H G I K J = F H G I K J = 1 23 720 549 K 10.0 15.0 K . Now, consider the isothermal process CD : == 549 K . V V P V V V V VV D C A A D D C A A C D = F H G I K J = F H G I K J L N M M O Q P P F H G I K J = 1 P C 100 24 0 15 0 445 kPa L L L kPa . .. af Next, consider the adiabatic process BC : B B C C = . But, P C A A C D = 1 from above. Also considering the isothermal process, V V BA A B = F H G I K J . Hence, P V V V V A A B B A A C D C F H G I K J = F H G I K J 1 which reduces to V V B AC D = 10 0 15 0 16 0 . . . L 24.0 L L L . Finally, V V A B = F H G I K J = F H G I K J = 875 kPa 10.0 L 16.0 L kPa . State P (kPa) V (L) T (K) A B C D 1 400 875 445 712 10.0 16.0 24.0 15.0 720 720 549 549 (b) For the isothermal process AB : ∆∆ En C T V int 0 so QW n R T V V B A =− = F H G I K J =⋅ F H G
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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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