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The figure shows the $P-V$ plot of an ideal gas taken through a cycle $ABCDA$. The part $ABC$ is a semi-circle and $CDA$ is half of an ellipse. Then,
$(A)$ the process during the path $\mathrm{A} \rightarrow \mathrm{B}$ is isothermal
$(B)$ heat flows out of the gas during the path $\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}$
$(C)$ work done during the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$ is zero
$(D)$ positive work is done by the gas in the cycle $ABCDA$
$(B,D)$
$(C,D)$
$(A,B)$
$(A,C)$
Solution
The correct options are
$B$ heat flows out of the gas during the path $B \rightarrow C \rightarrow D$.
D positive work is done by the gas in the cycle ABCDA.
Isolthermal process is represented by straight line on $PV$-diagram.
If $W$ is work done by the gas, $\Delta Q=\Delta U+W$
For process $B \rightarrow C \rightarrow D$
$\Delta U$ is negative and $W$ by gas is also negative, so $\Delta Q$ is also negative, hence heat flows out of gas during this process.
A to $B$, work done by gas is +ve. $B$ to $C$ work done by gas is -ve. But +ve work is more, so net work is not zero in process $A \rightarrow B \rightarrow C$
Area enclosed by $P-V$ graph is equal to positive work done by gas in cyclic process.
