solutions 2

# solutions 2 - PHYS 253 Assignment 2 Solutions Oct 2 2010 1...

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PHYS 253 Assignment 2 Solutions Oct 2 2010 1) – 2 marks a PV = nRT dW =− Fdx =− PAdx =− PdV V = nRT P dV dP = nRT P 2 W = P i P f dW =− P i P f PnRT P 2 dP W = nRT ln P f ∣− ln P i ∣= nRT ln P f P i b R=9.31 J/(mol*K) T=20C o =293K P i =20 atm P f =1 atm W = 1 mol 8.31 J mol K 293K ln 1 20 =− 7.29kJ 2) – 2 marks Because the system is adiabatic no thermal energy can be exchanged between the system and its surroundings. Additionally because of the rigid walls no work can be done on/by the system. Thus no net work can be done by this system. Because the expansion occurring in the system is not quasi-static , thermodynamic parameters cannot be defined during this process (one requires thermodynamic equilibrium to do so). Thus one cannot state that the dW = -P*dV as neither p nor V are well defined.

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## This note was uploaded on 12/15/2010 for the course PHYS 253 taught by Professor Petergrutter during the Fall '10 term at McGill.

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solutions 2 - PHYS 253 Assignment 2 Solutions Oct 2 2010 1...

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