HW%204%20Solutions

HW%204%20Solutions - HW 4 Solutions CH 353 Physical...

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HW 4 Solutions CH 353: Physical Chemistry I (54655) Spring 2008 Bersuker (10 points) I. An ideal gas containing 2.5 moles for which 2 / 3 , R C m V = undergoes the following reversible cyclical process from an initial state characterized by 450 = T K and 00 . 1 = p bar: (a) It is expanded reversibly and adiabatically until the volume doubles. You can calculate the final temperature of an adiabatic expansion via 630 . 1 = = γ i f i f V V T T , giving K. We can thus calculate the initial and final pressure and volume using the ideal gas equation m 3 , with 283 = f T V 2 10 25 . 9 / × = = i i i P nRT 185 . 0 2 = = i f V V m 3 , leading to 314 . 0 = i f f i T T = i f V V p p bar. Clearly the adiabatic process has q equal to zero. . 0 Thus . 0 since 0 equality. Clausius the and zero is heat reversible the since 0, J 10 68 . 8 ) 450 283 ( 2 K mol J 314 . 8 5 5 . 2 J 10 5.21 K) 450 K 283 ( 2 K 8.314Jmol 3 mol 5 . 2 3 1 - 1 - , 3 -1 -1 , = Δ = = Δ = Δ × = × × × = Δ = Δ × = × × × = Δ = = Δ total gs surroundin m p m V S q S S T nC H T nC w U (b) The pressure is increased in an isothermal reversible compression until
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HW%204%20Solutions - HW 4 Solutions CH 353 Physical...

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