PS1Solutions

# PS1Solutions - 1 APSC 132 PROBLEM SET 1 SOLUTIONS Note In...

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1 APSC 132 PROBLEM SET 1 SOLUTIONS Note : In this problem set, no chemical reactions occurred in the system. Problem 1 Isothermal reversible This means that D T = 0. Since D H=nc p D T and D E=nc v D T, D H = D E =0 Therefore, 0 = q + w or q = -w w system = -nRTln(P 1 /P 2 )= -(0.500)(8.31)(273)ln(5.000/0.500)J= -2.62 x10 3 J q system = -w system = 2.62 x10 3 J Isothermal irreversible D H = D E =0 (changes in state functions of system are identical for both reversible, irreversible systems) w = -P ext D V (assume final pressure of system equal to external pressure unless told otherwise) P 1 =(5.00 atm)(101 kPa/atm)=505 kPa P 2 =(0.500 atm)(101 kPa/atm)=50.5 kPa V =nRT/P; V 1 =(0.500)(8.31)(273)/505 L=2.25 L;V 2 =(0.500)(8.31)(273)/50.5L= 22.46L Therefore, w = -(50.5kPa)(22.46L – 2.25L) = -1021J ; q = -w = 1021 J Problem 2 These problems involve finding changes in entropy for the system only. Note: For system, D S rev = D S irr (state functions). Therefore, reversible equations can always be used to calculate D S for the system even if the changes in the system are irreversible.

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