Elementary Principles 252 - 7.16. (a) E k = 0 u1 = u2 = 0 E...

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7- 7 7.16 . (a) Δ Δ Δ Δ Eu u E P WP V k p s == = = = = 00 0 0 1 2 no elevation change (the pressure is constant since restraining force is constant, and area is constrant) the only work done is expansion work bg . ± . . . ±± HT T =+ = × = ⇒= 34980 355 125 10 1 8 314 0 0295 480 3 2 (J / mol), V = 785 cm , T = 400 K, P =125 kPa, Q = 83.8 J n= PV RT Pa 785 cm m m Pa / mol K 400 K 10 cm mol Q = H = n(H - H ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol) 83.8 J = 0.0295 35.5T - 35.5(400) K 1 3 1 33 3 6 3 21 2 2 Δ iV nRT P ii W P V iii Q U P V U Q PV ) .. ). . . mol m Pa cm K Pa mol K m cm N (941- 785)cm m m cm J J J 3 3 3 3 3 3 2 3 ×⋅ = × = = = 00295 8 314 10 480 125 10 1 941 125 10 1 10 19 5 838 19 5 64 3 6 5 5 6 Δ ΔΔ Δ Δ (b) Δ E p = 0 7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could occur, the temperature would drop during these periods.) (b) Δ Δ Δ UE EQ t W t pR ++= ²² EE W U t Q pk = = = = × = 0 0 0 090 1 126 ,, ² , ± () ² . . 1.4 W J s 1 W Js Ut () . = 126 Moles in tank: 1 atm 2.10 L 1 mol K K L a t m mol nP VR T
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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