An ideal gas is made to undergo the cyclic process shown in the figure below. Let $\Delta W$ depict the work done, $\Delta U$ be the change in internal energy of the gas and $Q$ be the heat added to the gas. Sign of each of these three quantities for the whole cycle will be (0 refers to no change)
KVPY 2018, Medium
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$(a)$ Given cyclic process is
Area under compression process $C A$ is more than area under expansion process AB. So, net work done is negative.
i.e. $\Delta W < 0$
Also, in a cyclic process, change in internal energy is zero.
i.e. $\Delta U=0$
Now, by using first law of thermodynamics, we have
$\Delta Q=0+\Delta W$
we see that, $\quad \Delta Q < 0$
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