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(A) From the graph,$V \propto T$,which implies the process is isobaric (constant pressure).For an isobaric process,the heat absorbed is given by $dQ =

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$\frac{2}{5}$

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(A) From the graph,$V \propto T$,which implies the process is isobaric (constant pressure).For an isobaric process,the heat absorbed is given by $dQ = nC_p dT$,where $C_p$ is the molar heat capacity at constant pressure.For a monatomic gas,the degrees of freedom $f = 3$. Thus,$C_p = \frac{f+2}{2}R = \frac{3+2}{2}R = \frac{5}{2}R$.Therefore,$dQ = n \left( \frac{5}{2}R \right) dT$.The work done by the gas is $dW = PdV$. Since $PV = nRT$,for a constant pressure process,$PdV = nRdT$.The required ratio is $\frac{dW}{dQ} = \frac{nRdT}{n(\frac{5}{2}R)dT} = \frac{1}{5/2} = \frac{2}{5}$.

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

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