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(B) Let the initial state be $1$ with temperature $T_1$. Points $2$ and $3$ lie on the same isothermal curve,so $T_2 = T_3 = T_f$.According to the fir

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$Q_1 < Q_2$

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(B) Let the initial state be $1$ with temperature $T_1$. Points $2$ and $3$ lie on the same isothermal curve,so $T_2 = T_3 = T_f$.According to the first law of thermodynamics,$Q = \Delta U + W$.Since the change in internal energy $\Delta U$ depends only on the change in temperature,$\Delta U_1 = nC_v(T_f - T_1)$ and $\Delta U_2 = nC_v(T_f - T_1)$. Thus,$\Delta U_1 = \Delta U_2$.Work done $W$ is equal to the area under the $P-V$ curve. From the diagram,the area under the path $1-3$ is greater than the area under the path $1-2$,so $W_2 > W_1$.Since $Q = \Delta U + W$ and $\Delta U_1 = \Delta U_2$,the process with greater work done will have greater heat transfer.Therefore,$Q_2 > Q_1$ or $Q_1 < Q_2$.

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