Answer
a. 0 J
b. -465 kJ
Work Step by Step
a. Calculate the work done by a gas at constant pressure from equation 15–3.
$$W=P\Delta V=0$$
The container has rigid walls and the volume did not change.
b. Calculate the change in internal energy from $\Delta U=Q-W$
$$\Delta U = (-465kJ)-0=-465kJ $$
