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)$