Answer
(a)$W = 0 $
(b) $T_b = 2T_a$
(c) $ U_b = 700 J + U_a$
Work Step by Step
(a) The work done is calculated by
$W = p(\Delta V)$ and in the figure, there is no change in volume so $\Delta V = 0$
$W = p(0)$
$W = 0 $
(b) The relationship between P and T is $pV = nRT$. During this process, heat is added and the pressure doubles in constant volume.
$\frac{P}{T} = \frac{nR}{V}$ are constant so when P doubles, T also doubles $T_b = 2T_a$
(c) The change in internal energy is
$\Delta U = \Delta Q - W$
$\Delta U = 700J - 0 $
$\Delta U = 700 J$
The internal energy at a and b is added together to result in $\Delta u$
$\Delta U = U_b - U_a$
$700 J = U_b - U_a$
$ U_b = 700 J + U_a$
