Answer
(a) $W=3P_iV_i$
(b) $\Delta U=\frac{15}{2}P_iV_i$
(c) $Q= \frac{21}{2}P_iV_i$
Work Step by Step
(a) We know that
$W=\frac{1}{2}(3V_i-V_i)(P_i+2P_i)$
$W=\frac{1}{2}\times 6P_iV_i$
$W=3P_iV_i$
(b) We know that
$\Delta U=\frac{3}{2}nRT_f-\frac{3}{2}nRT_i$
$\implies \Delta U=\frac{3}{2}P_fV_f-\frac{3}{2}P_iV_i$
$\Delta U=\frac{3}{2}(P_fV_f-P_iV_i)$
$\Delta U=\frac{3}{2}[(2P_i)(3V_i)-P_iV_i]$
$\Delta U=\frac{15}{2}P_iV_i$
(c) $Q=\Delta U + W= \frac{21}{2}P_iV_i$
