657_Physics ProblemsTechnical Physics

657_Physics ProblemsTechnical Physics - Chapter 22 The work...

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Chapter 22 659 The work output in the isothermal expansion is W PdV nRT V dV nRT V V nRT nRT i C i i C i C i ii == = F H G I K J F H G I K J zz 1 1 3 1 3 ln ln ln γγ γ bg ej This is also the input heat, so the entropy change is S Q T nR F H G I K J 1 3 ln Since CC C R PV V + we have −= 1 CR V , C R V = 1 and C R P = 1 Then the result is Sn C P = ln3 (b) The pair of processes considered here carry the gas from the initial state in Problem 66 to the final state there. Entropy is a function of state. Entropy change does not depend on path. Therefore the entropy change in Problem 66 equals ∆∆ SS isothermal adiabatic + in this problem. Since S adiabatic = 0, the answers to Problems 66 and 67 (a) must be the same. P22.68 Simply evaluate the maximum (Carnot) efficiency.
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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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