Q 11-11-026NEETJEE MainMedium
An ideal gas is taken round the cycle shown by a rectangle on the P-V diagram with sides $\Delta P = 2 \times 10^5$ Pa and $\Delta V = 3 \times 10^{-3}\ \text{m}^3$, traversed clockwise. The net heat absorbed by the gas in one cycle is
Answer: (A) $600$ J
Concept. Over a cycle $\Delta U = 0$, so $Q = W =$ enclosed area (positive when clockwise).
Step 1. Area. $2 \times 10^5 \times 3 \times 10^{-3} = 600$ J.
Answer. $600$ J (Option A).
Solution by Sreeraj P, M.Sc Physics