When a gas follows path 123 on the PV diagram in the figure below, 445.0 J of he
ID: 2253117 • Letter: W
Question
When a gas follows path 123 on the PV diagram in the figure below, 445.0 J of heat flows into the system and 159.0 J of work is done by the system. What is the change in the internal energy of the system? How much heat flows into the system if the process follows path 143? The work done by the gas along this path is 61.2 J. What net work would be done by the system if the system followed path 12341? What net work would be done by the system if the system followed path 14321? What is the change in internal energy of the system in the processes described in parts c and d?
Explanation / Answer
a) Q = U + PV So, 407 = U + 169 U = 238 J
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b) Again, reaching the point 3 means the energy would be conserved. So, Q = U + PV Q = 238 + 61.2 J Q = 299.2 J
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c) Q = U + PV So, as 12341 is followed, we get back to point 1. So change in internal energy is zero. Thus, Q1 - Q2 = PV Q1 - Q2 = heat flowing in - heat flowing out Net work done = PV = 407 - 299.2 J W = 107.8 J
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d) If the system followed 14321, again the change in internal energy would be zero. But this time, the signs of the heats will be different. So, Q2 - Q1 = PV So, net work done = -107.8 J (negative work done in opposite cycle)
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e) Change in internal energy in parts C & D is zero. This is because internal energy is a function of positions only. So. if you follow a path & come back to the same position, there is no change in internal energy.
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