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.
Figure
(a) ∆U1 + ∆U2 = 0.
(b) ∆U1 − ∆U2 = 0.
(c) ∆Q − ∆W = 0.
(d) ∆Q + ∆W = 0

Answer 1

option (A) and (C) are correct

Explanation:

The cycle that occurs through A and by B returns to the same state is cyclic. Being a state function, net internal energy transition, some ∆U is going to be zero, i.e.

$\\ \Delta U_{1}=\Delta U_{2} \\ \Delta U_{1}= In Process A change in internal energy\\ \Delta U_{2}= In Phase B shift in internal energy\\ \Delta U=\Delta U_{1}+\Delta U_{2}=0$

Here, though ΔU is the total change in the cyclic cycle in internal energy.

Using the first thermodynamic rule, we obtain

$\Delta \mathrm{Q}-\Delta \mathrm{W}=\Delta \mathrm{U}$

Here, in process A + B, ΔQ is the net heat provided to the system and ΔW is the net work done by the system in process A+B.

$\Delta Q-\Delta W=0$