Answer
a. 480 J
b. $\Delta U = 0$
c. 480 J
Work Step by Step
a. Zero work is done during the first stage, because the volume stays constant. Calculate the work in the second step.
$$W=P\Delta V$$
$$=(1.4atm)(\frac{1.01\times10^5Pa}{1atm})(9.3L-5.9L)\frac{1\times10^{-3}m^3}{1L}=480J$$
b. The temperature does not change. $\Delta U = 0$.
c. Calculate the heat flow from $\Delta U=Q-W$.
$$\Delta U =0= Q-W=Q-480J $$
Q=W=480J.
