The figure shows $P-V$ diagram of a thermodynamic cycle. Which corresponding curve is correct?
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$\begin{array}{lll}\text { For } A-B: V=\text { constant } & \text { (ideal qas equation } \Rightarrow) \\ (P \propto T) & (\text { straight line) } & (P V=n R T \\ A B & & (P \alpha T) \text { for } A B\end{array}$
For $B-C \because P=$ constant $(v \propto T)$ for $B C$
$(V \propto T)$ (constant line parallel to $T($ temperature)) axis.
For $C D$ it will be similar to $A B$
and $DA$ It will be simicar to $BC$
$\therefore$ considering all above constraints option $(A)$ is the correct graph
For $B-C \because P=$ constant $(v \propto T)$ for $B C$
$(V \propto T)$ (constant line parallel to $T($ temperature)) axis.
For $C D$ it will be similar to $A B$
and $DA$ It will be simicar to $BC$
$\therefore$ considering all above constraints option $(A)$ is the correct graph
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