Hw12 - SOLUTION HOMEWORK 12 1. (a) With T = 283 K, we...

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Unformatted text preview: SOLUTION HOMEWORK 12 1. (a) With T = 283 K, we obtain ( )( ) ( )( ) 3 3 100 10 Pa 2.50m 106mol. 8.31J/mol K 283K pV n RT p = = = ) (b) We can use the answer to part (a) with the new values of pressure and temperature, and solve the ideal gas law for the new volume, or we could set up the gas law in ratio form as in Sample Problem 19-1 (where n i = n f and thus cancels out): ( 29 3 100kPa 303K 2.50m 300kPa 283K f f f f i i i p V T V pV T = ⇒ = which yields a final volume of V f = 0.892 m 3 . 2. (a) At point a , we know enough information to compute n : ( 29 ( 29 ( 29 ( 29 3 2500Pa 1.0m 1.5mol. 8.31 J/mol K 200K pV n RT = = = (b) We can use the answer to part (a) with the new values of pressure and volume, and solve the ideal gas law for the new temperature, or we could set up the gas law as in Sample Problem 19-1 in terms of ratios (note: n a = n b and cancels out): ( 29 3 3 7.5kPa 3.0m 200K 2.5kPa 1.0m b b b b a a a p V T T p V T = ⇒ = which yields an absolute temperature at b of T b = 1.8×10 3 K. (c) As in the previous part, we choose to approach this using the gas law in ratio form (see Sample Problem 19-1): ( 29 3 3 2.5kPa 3.0m 200K 2.5kPa 1.0m c c c c a a a p V T T p V T = ⇒ = which yields an absolute temperature at c of T c = 6.0×10 2 K. (d) The net energy added to the gas (as heat) is equal to the net work that is done as it progresses through the cycle (represented as a right triangle in the pV diagram shown in Fig. 19-19). This, in turn, is related to ± “area” inside that triangle (with 1 2 area = (base)(height)), where we choose the plus sign because the volume change at...
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This note was uploaded on 04/07/2008 for the course PETE 2001 taught by Professor Thurber during the Spring '07 term at LSU.

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Hw12 - SOLUTION HOMEWORK 12 1. (a) With T = 283 K, we...

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