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### Statement
−$5.6.14.$ [Insert the problem statement]
+$5.6.14.One mole of gas participating in the process, the graph of which is shown in
+the figure, passes through states 1, 2, and 3 sequentially. The internal energy
+of the gas is proportional to the temperature (U = cT). Find the amount of
+heat absorbed by the gas in this process.
+![Снимок экрана 2025-06-18 160955.png|459x311, 50%](../../img/5.6.14/Снимок экрана 2025-06-18 160955.png)
### Solution
Assuming we're working with ideal gases\
−We have process 1-2 (isochoric), $W_1-2=0$, and for first law of thermodynamics\
−$Q_1-2=Delta U_1-2$\
−$Q_1-2=c(T_2 - T_1)$\
+We have process 1-2 (isochoric), $W_{12}=0$, and for first law of thermodynamics\
+$Q_{12}=\Delta U_{12}$\
+$Q_{12}=c(T_2 - T_1)$\
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 $Q_a=Q_1-2 +Q_2-3$\
−$Q_a=c(T_2 -T1) + c(T_3 -T_2)+P_2(V_2 -V_1)$\
+$Q_{23}=\Delta U_{23} + W_{23}$ , where $W_{23}=P_2(V_2 -V_1)$, $P_2=P_3$\
+$Q_{23}=c(T_3 - T_2) + P_2(V_2 -V_1)$\
+The amount of heat absorbed by the gas in this whole process is\
+$Q_a=Q_{12 }+Q_{23}$\
+$Q_a=c(T_2 -T_1) + c(T_3 -T_2)+P_2(V_2 -V_1)$\
$Q_a=c(T_3 -T_1)+P_2(V_2 -V_1)$\
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\
−$Q_a=\frac{c(P_2V_2 -P_1V_1}{R} + P_2(V_2 -V_1)$
+$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} + P_2(V_2 -V_1)$
#### Answer
−
−[Insert a concise answer or boxed result]
+$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} +P_2(V_2 -V_1)$