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Question
Refer to figure. Let ∆U1 and ∆U2 be the change in internal energy in processes A and B respectively, ∆Q be the net heat given to the system in process A + B and ∆W be the net work done by the system in the process A + B.
(a) ∆U1 + ∆U2 = 0
(b) ∆U1 − ∆U2 = 0
(c) ∆Q − ∆W = 0
(d) ∆Q + ∆W = 0
Solution
(a) ∆U1 + ∆U2 = 0
(c) ∆Q − ∆W = 0
The process that takes place through A and returns back to the same state through B is cyclic. Being a state function, net change in internal energy, ∆U will be zero, i.e.
∆U1 (Change in internal energy in process A) = ∆U2 (Change in internal energy in process B)
\[\Rightarrow \Delta U = \Delta U_1 + \Delta U_2 = 0\]
Here, ∆U is the total change in internal energy in the cyclic process.
Using the first law of thermodynamics, we get
\[\Delta Q - \Delta W = \Delta U\]
Here, ∆Q is the net heat given to the system in process A + B and ∆W is the net work done by the system in the process A + B.
Thus,
∆Q - ∆W = 0
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