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NEET PHYSICSThermodynamicsEasy

Question

A thermodynamic system is taken through the cycle ABCD as shown in the figure. Heat rejected by the gas during the cycle is:

A

2PV

B

4PV

C

1/2 PV

D

PV

Step-by-Step Solution

For a cyclic process, the system returns to its initial state, so the change in internal energy (ΔU\Delta U) is zero . According to the First Law of Thermodynamics, ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W, which implies ΔQ=ΔW\Delta Q = \Delta W . The net work done (ΔW\Delta W) in a cyclic process is equal to the area enclosed by the cycle on the P-V diagram .

Assuming the standard figure for this AIPMT 2012 question (a rectangle with limits typically from PP to 2P2P and VV to 2V2V, or similar dimensions yielding an area of PVPV): Area = ΔP×ΔV=(2PP)×(2VV)=P×V=PV\Delta P \times \Delta V = (2P - P) \times (2V - V) = P \times V = PV.

The question asks for heat rejected. In a cyclic process, if the cycle is traced counter-clockwise, work is done on the system (negative work), and heat is rejected by the system (negative heat). The magnitude of this heat is equal to the area of the loop, which is PVPV.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermodynamicsthermodynamicsystemthroughfigurerejected

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