NWedSep24

# NWedSep24 - Thermodynamics 1 2 Overall U U = U1 U 2 U 3 U 4...

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1 Thermodynamics

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3 Overall U 4 3 2 1 U U U U U + + + = 0 ) ( 0 ) ( 0 = - + + - + = c h V h c V T T C T T C U 0 = U

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4 Overall q 4 3 2 1 q q q q q + + + = 0 ln 0 ln 3 4 1 2 + + + = V V RT V V RT q c h ( 29 - = 1 2 ln V V T T R q c h since …………….
5 Since …. 0 ln 0 ln 3 4 1 2 + + + = V V RT V V RT q c h ( 29 1 - γ TV For an adiabatic process …. . is a constant, or ( 29 ( 29 1 2 3 1 1 4 - - = = V V V V T T c h therefore = 2 1 3 4 V V V V and ( 29 ( 29 ( 29 ( 29 1 3 1 2 1 4 1 1 and - - - - = = V T V T V T V T c h c h = 2 3 1 4 V V V V

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6 and ………… 0 ln 0 ln 3 4 1 2 + + + = V V RT V V RT q c h - = 1 2 1 2 ln ln V V RT V V RT q c h ( 29 - = 1 2 ln V V T T R q c h = 2 1 3 4 V V V V
7 Overall w If U = 0 then q = - w, so ( 29 - = 2 1 ln V V T T R w c h ( 29 - = 1 2 ln V V T T R q c h

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8 Interpretation The expression of the heat is positive since V 2 /V 1 is greater than 1 and therefore ln(V 2 /V 1 ) is greater than 0 ( 29 - = 1 2 ln V V T T R q c h ( 29 - = 2 1 ln V V T T R w c h The expression of the work is negative since V 1 /V 2 is less than 1 and therefore ln(V 1 /V 2 ) is less than 0
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NWedSep24 - Thermodynamics 1 2 Overall U U = U1 U 2 U 3 U 4...

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