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en/5.6.14.md +25 −0
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+### Statement
+
+$5.6.14.$ [Insert the problem statement]
+
+### Solution
+
+Assuming we're working with ideal gases\
+We have process 1-2 (isochoric), $W_1-2=0$, and for first law of thermodynamics\
+$Q_1-2=Delta U_1-2$\
+$Q_1-2=c(T_2 - T_1)$\
+and for process 2-3 (isobaric), we'll have\
+$Q_2-3=Delta U_2-3 + W_2-3$ , where $W_2-3=P_2(V_2 -V_1)$\
+$Q_2-3=c(T_3 - T_2) + P_2(V_2 -V_1)\
+The amount of heat absorbed by the gas in this whole process is $Q_a=Q_1-2 +Q_2-3$\
+$Q_a=c(T_2 -T1) + c(T_3 -T_2)+P_2(V_2 -V_1)$\
+$Q_a=c(T_3 -T_1)+P_2(V_2 -V_1)$\
+by the ideal gas law, we have\
+$P_1V_1=RT_1$, and $P_2V_2=RT_3$, from this, we get\
+$T_3 -T_1=\frac{P_2V_2 -P_1V_1}{R}$\
+and substituting this into Q_a, we get\
+$Q_a=\frac{c(P_2V_2 -P_1V_1}{R} + P_2(V_2 -V_1)$
+
+#### Answer
+
+[Insert a concise answer or boxed result]