Answer
(a) $W = 738 J$
$Q = 2589 J$
$\Delta U = 1851 J $
(b) $W = 0 $
$Q = -1850 J$
$\Delta U = -1850 J$
(c) $\Delta U = 0 $
Work Step by Step
(a) During the initial expansion, state $1 \rightarrow 2$
$W = p ΔV = nR ΔT $
$W = (0.250 mol)(8.3145 J/mol.K)(710 K − 355 K)$
$W = 738 J$
$Q = nCpΔT$
$Q = (0.250 mol)(29.17 J/mol ⋅K)(710 K − 355 K)$
$Q = 2589 J$
$\Delta U = Q - W$
$\Delta U = 2589 J - 738 J$
$\Delta U = 1851 J $
(b) During the final cooling, the process is isochoric which means the volume remains the same.
$W = p\Delta V$
$W = p (0)$
$W = 0 $
$Q = nCvΔT$
$Q = (0.250 mol)(20.85 J/mol.K)(355 K − 710 K) $
$Q = -1850 J$
$\Delta U = Q - W$
$\Delta U = -1850 J - 0$
$\Delta U = -1850 J$
(c) The internal energy change during the isothermal compression. For this process, there is no change of temperature.
$\Delta U = Q = nCvΔT$
$\Delta U = nCv (0)$
$\Delta U = 0 $