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NEET generalGeneralMedium

Question

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

A

1/3

B

2/3

C

2/5

D

2/7

Step-by-Step Solution

Given process is isobaric. dQ=nCpdTdQ = n C_p dT. For a monatomic gas, Cp=52RC_p = \frac{5}{2}R. So, dQ=n(52R)dTdQ = n(\frac{5}{2}R)dT. Work done dW=PdV=nRdTdW = P dV = nR dT. The required ratio is dWdQ=nRdTn(52R)dT=25\frac{dW}{dQ} = \frac{nRdT}{n(\frac{5}{2}R)dT} = \frac{2}{5}.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. 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.

generalvolumemonatomicvariestemperatureabsorbed

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