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Unformatted text preview: 19.1: a) b) . J 10 33 . 1 ) C 80 )( K mol J 3145 . 8 )( mol 00 . 2 ( 3 × = ° ⋅ = ∆ = ∆ T nR V p 19.2: a) b) If the pressure is reduced to 40.0% of its original value, the final volume is ) 2 5 ( of its original value. From Eq. (19.4), J. 10 15 . 9 2 5 ln ) K )(400.15 K mol J 3145 . 8 )( 3 ( ln 3 1 2 × = ⋅ = = V V nRT W 19.3: nRT pV = T constant, so when p increases, V decrease . : 194 At constant pressure, so , T nR V p W ∆ = ∆ = C. . 62 1 C . 35 1 C . 27 so , T and K 1 . 35 K) mol J ( . 8 3145 mol) ( 6 J 10 75 . 1 2 C K 3 ° = ° + ° = ∆ = ∆ = ⋅ × = = ∆ T T nR W T . : 195 a) b) At constant volume, . so and = = W dV . : 196 b) J. 10 50 . 4 ) m 0900 . m ( . 0 0600 Pa) 10 ( . 1 50 3 3 3 5 × = × = ∆ V p 19.7: a) b) In the first process, 1 = ∆ = V p W . In the second process, J. 10 00 . 4 ) m 0.080 ( Pa) 10 00 5 ( 4 3 5 2 × = × = ∆ = . V p W 19.8: a) . and ) ( , ), ( 41 2 1 2 24 32 1 2 1 13 = = = = W V V p W W V V p W The total work done by the system is ), )( ( 1 2 2 1 41 24 32 13 V V p p W W W = + + + W which is the area in the p V plane enclosed by the loop. b) For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part (a), so the total work is negative of the work found in part (a). 19.9: J 73 J, 254 = = W Q (work is done on the system), and so J. 327 = = ∆ W Q U 19.10: a) J. 10 78 . 3 ) m Pa)(0.210 10 80 . 1 ( 4 2 5 × = × = ∆ V p b) J. 10 7.72 J 10 3.78 J 10 15 . 1 4 4 5 × = × × = = ∆ W Q U c) The relations W Q U V p W = ∆ ∆ = and hold for any system. 19.11: The type of process is not specified. We can use W Q U = ∆ because this applies to all processes. Q is positive since heat goes into the gas; J 1200 + = Q W positive since gas expands; J 2100 + = W J 900 J 2100 J 1200 = = ∆ U We can also use ( 29 T R n U ∆ = ∆ 2 3 since this is true for any process for an ideal gas. ° = ⋅ = ∆ = ∆ C 4 . 14 K) mol J 5 mol)(8.314 3(5.00 J) 900 ( 2 3 2 nR U T C 113 C 14.4 C 127 1 2 ° = ° ° = ∆ + = T T T 19.12: At constant volume, the work done by the system is zero, so . Q W Q U = = ∆ Because heat flows into the system, Q is positive, so the internal energy of the system increases. 19.13: a) J 10 15 . 1 ) m Pa)(0.50 10 30 . 2 ( 5 3 5 × = × = ∆ V p . (b J 10 55 . 2 J) 10 15 . 1 ( J 10 40 . 1 5 5 5 × = × + × = + ∆ = W U Q (heat flows out of the gas). c) No; the first law of thermodynamics is valid for any system. 19.14: a) The greatest work is done along the path that bounds the largest area above the Vaxis in the p V plane (see Fig. (19.8)), which is path 1. The least work is done along path 3. W b) in all three cases; so , + ∆ = Q W U Q for all three, with the greatest Q for the greatest work, that along path 1. When , Q heat is absorbed....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.
 Fall '07
 Nearing
 Physics

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