Explanation

(b) Change in internal energy from A to B is
$\mathrm{\Delta}U = \frac{f}{2}nR\mathrm{\Delta}T = \frac{f}{2}(p_{f}V_{f} - p_{i}V_{i})$
$\ \ \ \ \ \ \ \ = \frac{3}{2}\left( 2p_{0} \times 2V_{0} - p_{0} \times V_{0} \right) = \frac{9}{2}p_{0}V_{0}$
Work done in process A to B is equal to the area covered by the graph with volume axis, ie,
$W_{A \rightarrow B} = \frac{1}{2}\left( p_{0} + 2p_{0} \right) \times (2V_{0} - V_{0}) = \frac{3}{2}p_{0}V_{0}\ $
Hence, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{9}{2}p_{0}V_{0} + \frac{3}{2}p_{0}V_{0} = 6p_{0}V_{0}$