Question

One mole of an ideal gas is taken from state A to state B by three different processes, (i) ACB (ii) ADB (iii) AEB as shown in the P-V diagram. The heat absorbed by the gas is 

• A. greater in process (ii) than in (i)
• B. the least in process (ii)
• C. the same in (i) and (iii)
• D. less in (iii) than in (ii)

Hint:

For ideal gases: Q = Δ U + W If initial and final states are same, compare heat transfer by comparing the area under the P-V curve.

Answer 1

The Correct Option is D

Step 1: For an ideal gas, change in internal energy depends only on initial and final states: Δ U = constant for all three processes 

Step 2: From first law of thermodynamics: Q = Δ U + W 

Step 3: Hence, heat absorbed depends on the work done W. 

Step 4: Work done is equal to the area under the P-V curve. From the diagram: WADB > WAEB 

Step 5: Therefore, QADB > QAEB ⟹ QAEB < QADB Hence, heat absorbed in process (iii) is less than that in process (ii).