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NEET PHYSICSMedium

An ideal gas goes from state AA to state BB via three different processes, as indicated in the PP-VV diagram. If Q1,Q2,Q3Q_1, Q_2, Q_3 indicates the heat absorbed by the gas along the three processes and ΔU1,ΔU2,ΔU3\Delta U_1, \Delta U_2, \Delta U_3 indicates the change in internal energy along the three processes respectively, then:

A

Q1>Q2>Q3Q_1>Q_2>Q_3 and ΔU1=ΔU2=ΔU3\Delta U_1=\Delta U_2=\Delta U_3

B

Q3>Q2>Q1Q_3>Q_2>Q_1 and ΔU1=ΔU2=ΔU3\Delta U_1=\Delta U_2=\Delta U_3

C

Q1=Q2=Q3Q_1=Q_2=Q_3 and ΔU1>ΔU2>ΔU3\Delta U_1>\Delta U_2>\Delta U_3

D

Q3>Q2>Q1Q_3>Q_2>Q_1 and ΔU1>ΔU2>ΔU3\Delta U_1>\Delta U_2>\Delta U_3

Step-by-Step Solution

Internal energy (UU) is a state function, meaning its change depends only on the initial and final states, and is independent of the path taken. Since all three processes go from the same initial state AA to the same final state BB, the change in internal energy is the same for all three processes. Therefore, ΔU1=ΔU2=ΔU3\Delta U_1 = \Delta U_2 = \Delta U_3.

According to the first law of thermodynamics, Q=ΔU+WQ = \Delta U + W. The work done (WW) is equal to the area under the PP-VV curve. From the standard diagram for this problem, process 1 has the maximum area under the curve, followed by process 2, and then process 3. Thus, W1>W2>W3W_1 > W_2 > W_3. Since ΔU\Delta U is the same for all paths, the heat absorbed (QQ) will be directly proportional to the work done. Therefore, Q1>Q2>Q3Q_1 > Q_2 > Q_3.

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