Solution

Assuming we're working with ideal gases
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
$Q_{23}=\Delta U_{23} + W_{23}$ , 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
$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} + P_2(V_2 -V_1)$