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(A) For an isobaric process,the work done $W$ is given by $W = nR\Delta T$ and the heat supplied $Q$ is given by $Q = nC_p\Delta T$.From the graph,at

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$He$

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(A) For an isobaric process,the work done $W$ is given by $W = nR\Delta T$ and the heat supplied $Q$ is given by $Q = nC_p\Delta T$.From the graph,at $Q = 80 \ J$,$W = 32 \ J$.Therefore,the ratio $\frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p} = \frac{32}{80} = \frac{2}{5}$.This implies $5R = 2C_p$.Since $C_p = C_v + R$,we have $5R = 2(C_v + R) = 2C_v + 2R$.$3R = 2C_v \implies C_v = \frac{3}{2}R$.$A$ gas with $C_v = \frac{3}{2}R$ is a monoatomic gas. Among the given options,$He$ (Helium) is a monoatomic gas. Thus,the correct option is $A$.

In the figure shown,the amount of heat supplied to one mole of an ideal gas is plotted on the horizontal axis and the amount of work done by the gas is plotted on the vertical axis. Assuming the process to be isobaric,identify the gas.

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