Tutorial 2 Notes - CHM2132 Tutorial notes for Sept. 18th,...

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1 C A B D p V CHM2132 Tutorial notes for Sept. 18 th , 22 nd . Question 1) When a system is taken from state A to state B along the path ACB (shown below), 80J of heat flows into the system and the system does 30 J of work. a) How much heat flows into the system along path ADB if the work done is 10 J? b) When the system is returned from state B to A along the curved path the work done on the system is 20 J. Does the system absorb or liberate heat, and how much? c) If U D - U A =40J, find the heat absorbed in the processes AD and DB. Solution: Refering to the diagram above, we are told that following path ACB gives 80 J to the system ( q ACB = 80 J) and that the system does 30 J of work ( w ACB = -30 J). a) How much heat flows into this system along path ADB if 10 J of work is done? From the information we have been given about path ACB, we can first calculate the change in internal energy to go from state A to B: ! U AB = q ACB + w ACB = 80 J " 30 J =50 J This change in internal energy is the same regardless of the path we take to get there since U is a state function. Now since we have been given the work done for path ADB ( w ADB = -10 J since the system did the work) we can calculate the heat from the first law: ! U AB = q ADB + w ADB q ADB = ! U AB " w ADB = 50 J " " 10 J ( ) = 60 J Therefore 60 J of work was done for process ADB. b) When the system is returned from B to A along the curved path, the work done on the system is 20 J. Does the system absorb or liberate heat and how much? This time, work is done on the system (positive work) and so w BA = 20 J. However, the change in internal energy is now opposite from that in part a) since we are going from B to A instead of A to B. Therefore the heat of going from B to A is:
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2 q BA = ! U BA " w BA = " 50 J " 20 J = " 70 J Since q AB is negative, heat is being liberated. c) If U D - U A = 40 J then what is the heat of going from A to D and from D to B? First we must recognize that the energy term that we were given was Δ U AD = 40 J. As we already know the energy of going from A to B we can calculate the energy for the remaining step in the cycle: ! U AB = ! U AD + ! U DB ! U DB = ! U AB " ! U AD = 50 J " 40 J =10 J Now in order to use this information to find the heat of steps AD and DB, we must
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Tutorial 2 Notes - CHM2132 Tutorial notes for Sept. 18th,...

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