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HW1 - Homework 1 Jan 25 2007 1 a At constant pressure qp =...

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Homework 1 Jan. 25, 2007
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1. a. At constant pressure, q p = Δ H and Δ H does not depend on paths. q p = Δ H (20 C ) = Δ(100 C ) + C p ( )(100 - 20) + C p ( g )(20 - 100) = Δ(100 C ) + ( C p ( ) - C p ( g ))(100 - 20) = 2257 + (4 . 18 - 1 . 874) × 80 = 2441 . 48 kJ b. Δ T hiker = q p 60 × 4 . 18 = 9 . 73 K c The reaction is sucrose(s) + 12O 2 ( g ) 12CO 2 + 11H 2 O . From TABLE A.5 and 7, Δ H = 11 × ( - 285 . 83) + 12 × ( - 393 . 51) - 0 - ( - 2222 . 1) = - 5644 kJ / mol Molecular weight of sucrose is 342 g/mol. Thus, the amount of sucrose which replace the heat of evaporating 1.00 L of water is 342 × 2441 . 48 5644 = 148 g . Note that the Δ ¯ H 0 of sucrose given in TABLE A.7 should be for solid . 3. a. w = 10 × g × 10 = 10 × 9 . 8 × 10 = 980 J b. w = 6 × 5 . 5 × 7200 = 237600 J 1
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c. w = 1 2 k Δ x 2 = 1 2 × 5 (10 . 5 - 10 . 0) × 10 - 2 × (10 - 2 ) 2 = 0 . 05 J d. 1 atm = 101325 N / m 2 w = - 2 × 10 - 3 × 101325 = - 2 . 03 × 10 2 J e. w = - 2 × 10 - 3 × 101325 × 10 - 6 = - 2 . 03 × 10 - 4 J f. Since T is constant, P = f ( V ) = nRT V n = PV RT = 1 × 1 0 . 082 × 293 (unit of R here is L · atm · K - 1 · mol - 1 ) = 0 . 041 w = - V 2 V 1 nRT V dV = - nRT ln V 2 V 1 = - 0 . 041 × 8 . 31 × 298 × 1 . 10 (unit of R here is J · K - 1 · mol - 1 ) = - 111 . 68 J 2
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11.
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