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`        System A and system B above are identical closed  systems that undergo a change of state from the  same initial conditions (P1, V1, T1) to the same  final conditions (P2, V2, T2), but by a different  process. What are DU and q for the change in  state for system B?`
2 months ago

Kapil Khare
80 Points
```							Work Done in system A = W = -300JSince, system A has gone through adiabatic process,Heat transfer in system A = q = 0Change in internal energy of A = W =  -300J Since, initial and final condition of A and B are same,Change in internal energy of A = change in internal energy of B$\implies$ Change in internal energy of B = -300 For system B,q + W = change in internal energy of B-200 + q = -300$\implies$ Heat transformmed = q = -100
```
2 months ago
Rajdeep
227 Points
```							HELLO THERE! The foremost concept that is going to be used in the question is that, Internal energy is a state function, that does not depend on the path (process) of the system, but on the initial and final state of the system. For A, the process is Adiabatic, so heat exchance Q = 0.U = Q + W (From First Law of Thermodynamics)or, dU = W = -300J. Now, for A and B, it is given that the Initial conditions and final conditions for each is the same, so the change in internal energy will also be the same. Hence, For B, dU = -300J Again for B,From First Law of Thermodynamics,dU = dQ + dW=> -300 = dQ – 200or, dQ = -100 J Thus, Option (B) is the correct answer!
```
2 months ago
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