Let $Δ U_{1}$ and $Δ U_{2}$ be the change in internal energy in process A and B respectively, $Δ Q$ be the net heat given to the system in process (A+B) and $Δ W$ be the net work done by the system in the process (A+ B).
i) $Δ U_{1} + Δ U_{2} = 0$
ii) $Δ U_{1} - Δ U_{2} = 0$
iii) $Δ Q + Δ W = 0$
iv) $Δ Q + Δ W = 0$

(i) , (iii)

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(ii), (iii)

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(iii), (iv)

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(i), (iv)

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Solution

The correct option is A (i) , (iii)
The process that takes place through $A$ and returns back to the same state through $B$ is cyclic. Being a state function, the net change in internal energy, $Δ U$ will be zero, i.e.
$Δ U_{1}$ (change in internal energy in process A)= $Δ U_{2}$ ( Change in internal energy in process B)
$\Rightarrow Δ U = Δ U_{1} + Δ U_{2} = 0$
Here, $Δ U$ is the total change in internal energy in the cyclic process.
Using the first law of thermodynamics, we get
$Δ Q - Δ W = Δ U$

Here, $Δ Q$ is the net heat given to the system in process $A + B$ and $Δ W$ is the net work done by the system in the process $A + B$
Thus,
$Δ Q - Δ W = 0$

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Let ΔU1 and ΔU2 be the change in internal energy in process A and B respectively, ΔQ be the net heat given to the system in process (A+B) and ΔW be the net work done by the system in the process (A+ B).i) ΔU1+ΔU2=0ii) ΔU1−ΔU2=0iii) ΔQ+ΔW=0iv) ΔQ+ΔW=0