The figure shows the p-V plot of an ideal gas taken through a cycle...

(A) $p-V$ graph is not rectangular hyperbola. Therefore, process $A-B$ is not isothermal. (B) In process $B C D$, product of $p V$ (therefore temperature and intemal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, $Q$ will be negative or heat will flow out of the gas. (C) $W_{A B C}=$ positive (D) For clockwise cycle on $p-V$ diagram with $P$ on $y$-axis, net work done is positive.