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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}.

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