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+### Statement
+
+$5.6.18.$ [Insert the problem statement]
+
+### Solution
+
+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
+$W_T=R[(T_3)^\frac{1}{2} -(T_1)^\frac{1}{2}]$
+
+#### Answer
+
+[Insert a concise answer or boxed result]