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

Answer 1

(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