Assuming we're working with ideal gases We need the total work done by this mol of gas We have two isochoric processes, 1-2 and 3-4 ($V_1=V_2$ and $V_3=V_4$). And two isobaric processes, 2-3 and 4-1 ($P_2=P_3$ and $P_1=P_4), and we know that$T_2=T_4$$W_{12}=0$,$W_{34}=0$(isochoric processes) so, the total work will be$W_T=W_{23} +W_{41}$$W_T=P_2(V_3 -V_2)+P_1(V_1-V_4)$$W_T=P_2V_3 -P_2V_2 +P_1V_1 -P_1V_4$but$P_2V_3=RT_3$,$P_1V_1=RT_1$, and$P_2V_2=P_4V_4=RT_2=RT_4 (because $T_2=T_4$) and from this relations, we get in the work $W_T=RT_3 +RT_1 -2RT_2$ From Gay-Lussac's law $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, and $\frac{P_3}{T_3}=\frac{P_4}{T_4$}$ Multiplying term by term $\frac{P_1P_3}{T_1T_3}=\frac{P_2P_4}{T_2T_4}$ we get $\frac{1}{T_1T_3}=\frac{1}{(T_2)^2}$ so $T_2=(T_1T_3)^\frac{1}{2}$ and susbtituting this into the total work $W_T=RT_3 +RT_1 -2R(T_1T_3)^\frac{1}{2}$ $W_T=R[T_3 + T_1 -2(T_1T_3)^\frac{1}{2}]$ so $W_T=R[(T_3)^\frac{1}{2} -(T_1)^\frac{1}{2}]$
