Question

An ideal gas of mass \( M \) in the state \( A \) goes to another state \( B \) via three different processes. If \( Q_1, Q_2 \) and \( Q_3 \) denote the heat absorbed by the gas along the paths 1, 2 and 3 respectively, then

• A. \( Q_1 < Q_2 < Q_3 \)
• B. \( Q_1 < Q_2 = Q_3 \)
• C. \( Q_1 = Q_2 > Q_3 \)
• D. \( Q_1 > Q_2 > Q_3 \)

Hint:

- Same initial & final states $\Rightarrow \Delta U$ same - Compare areas under P–V curve to compare heat

Answer 1

The Correct Option is A

Concept:
First law of thermodynamics: \[ Q = \Delta U + W \]
• \( \Delta U \) depends only on initial and final states → same for all paths
• Heat difference depends on work done \( W \)

Step 1: Compare work done
Work done = area under \( P\text{-}V \) curve From the diagram:
• Path 3 has maximum area → maximum work
• Path 2 has intermediate area
• Path 1 has minimum area \[ W_1 < W_2 < W_3 \]

Step 2: Use first law
Since \( \Delta U \) is same: \[ Q \propto W \] \[ Q_1 < Q_2 < Q_3 \]