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problem_set_2_solutionsFall2008

# problem_set_2_solutionsFall2008 - Step 3 2-> Isothermal...

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∆g1847 = 0 g1836 g1875=−g1869=−g1866g1844g1846 g2870 ln g3023 g3119 g3023 g3118 g2880g2870g2874.g2870g3011 Step 3: 2 -> Isothermal compression to 1.5 L (State 3) ∆g1847 g3030g3052g3030g3039g3032 = 0 g1836 g1875 g3030g3052g3030g3039g3032 = −68.7−17.77+26.2+17.77=−42.5 g1869 g3030g3052g3030g3039g3032 g2880 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: g1858g1867g1870 g1853g1866 g1853g1856g1861g1853g1854g1853g1872g1861g1855 g1868g1870g1867g1855g1857g1871g1871: g1846 g2872 g1846 g2871 =( g1848 g2871 g1848 g2872 ) g3082g2879g2869 g1846 g2872 =473 g1837 ∆g1847=g1875=g1866g1829 g3049 (g1846 g2872 −g1846 g2871 )=61.26 g1836 g1869=0g1836 Cycle: g1875 g3030g3052g3030g3039g3032 =−68.7−17.77+26.2+61.26=1g1836 g1869 g3030g3052g3030g3039g3032 g2880 68.7 −26.2=42.5 ∆g1847 g3030g3052g3030g3039g3032 = 43.5g1836 The cycle is not closed!

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problem_set_2_solutionsFall2008 - Step 3 2-> Isothermal...

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