JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Heat & Thermodynamics

From the first law, $\Delta Q=\Delta U+W$. For path $B\to C\to D$, both volume and pressure decrease, so temperature decreases ($\Delta U<0$). Also, work done by the gas is negative ($W<0$). Therefore, $\Delta Q<0$, meaning heat flows out (option B). For the complete cycle ABCDA, the area enclosed is positive, indicating net positive work done by the gas (option D). Path $A\to B$ is not isothermal (it's part of a semicircle, not a hyperbola), and work done during $A\to B\to C$ is not zero (the area under the curve is non-zero).