One mole of monoatomic gas was taken through a cyclic process as shown in figure. Calculate. \underset{\text{ABCA}\textrm{ }}{\sum} \frac{\left(dq\right)

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Question

One mole of monoatomic gas was taken through a cyclic process as shown in figure. Calculate. $\sum_{ABCA} \frac{{dq}_{rev}}{T}$

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$dS = \frac{{dq}_{rev}}{T} \Rightarrow \sum \frac{{dq}_{rev}}{T} = ΔS$

As 'S' being a state function, ΔS = 0 in a cyclic process.

$∴ \sum \frac{{dq}_{rev}}{T} = 0 .$