problem_set_2_solutionsFall2008

problem_set_2_solutionsFall2008 - ∆g = 0 G ¡ = −¢ =...

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Unformatted text preview: ∆g = 0 G ¡ = −¢ = −£¤¥ ¦ ln § ¨ § © ª¦«.¦¬ Step 3: 2 -> Isothermal compression to 1.5 L (State 3) ∆g ­®­¯° = 0 G ¡ ­®­¯° = −68.7 − 17.77 + 26.2 + 17.77 = −42.5 ¢ ­®­¯° ª 68.7 − 26.2 = 42.5 Cycle: NOTE: This problem was found not to be a closed cycle. If you use P 3 V 3 γ = P 4 V 4 γ , to the specified volume you will get P 4 = 1.587 bar and T 4 =473.04 K which would not close the cycle. Credit will be given to both people who assumed the final temperature was 298K and got a closed cycle (like the solutions above), and people who used P 3 V 3 γ = P 4 V 4 γ to get the final temperature and used this to solve for the energy. Alternate Solution: Step 4: 3 -> "0", Adiabatic compression to 0.5 L: gG¡ ¢£ ¢¤¥¢¦¢§¥¨ ©¡G¨ª««: ¬ ­ ¬ ® = ( ¯ ® ¯ ­ ) °±² ¬ ­ = 473 ³ ∆´ = µ = £¶ · (¬ ­ − ¬ ® ) = 61.26 ¸ ¹ = 0¸ Cycle: µ º»º¼½ = −68.7 − 17.77 + 26.2 + 61.26 = 1¸ ¹ º»º¼½ ¾ 68.7 − 26.2 = 42.5 ∆´ º»º¼½ = 43.5¸ The cycle is not closed! 3.) 3....
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This document was uploaded on 10/08/2008.

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problem_set_2_solutionsFall2008 - ∆g = 0 G ¡ = −¢ =...

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