problem19_27

University Physics with Modern Physics with Mastering Physics (11th Edition)

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19.27: a) For an isothermal process, 4) 1 K)ln( K)(350.15 mol J 5 mol)(8.314 150 . 0 ( ) ( ln 1 2 = = V V nRT W J. 605 - = b) For an isothermal process for an ideal gas,
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Unformatted text preview: . and = ∆ = ∆ U T c) For a process with , = ∆ U J 605-= = W Q ; 605 J are liberated....
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  • Isothermal process, nRT ln

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