Answer
See explanation
Work Step by Step
(a) Work done by the gas, During the initial expansion, state $1 \rightarrow 2$
$W = p ΔV = nR ΔT $
$W =(0.150 mol)(8.3145 J/mol ⋅K)(−150 K)$
$W = -187 J$
$Q = nCpΔT$
$Q = (0.150 mol)(29.07 J/mol ⋅K)(−150 K)$
$Q = -654 J$
$\Delta U = Q - W$
$\Delta U = -654 J - (-187 J)$
$\Delta U = -467 J $
(b) $W = - \frac{nR \Delta T (\frac{V_1}{V_2})^{(𝛾 −1)}}{(𝛾 −1)}$
$W = - \frac{(0.150 mol) (8.3145 J/mol.K) (150K) (1 - \frac{V_1}{2V_1})^{(1.40 −1)}}{(1.40 −1)}$
$W = - \frac{(0.150 mol) (8.3145 J/mol.K) (150K) (1 - \frac{1}{2}^{0.40})}{(0.40)}$
$W = -113 J$
$ Q = 0 $ in adiabatic process.
$\Delta U = Q - W$
$\Delta U = 0 - (-113 J) $
$\Delta U = 113 J$
(c) During the final heating, the volume is kept constant. Hence
$W = o$
$Q = nCvΔT$
$Q = (0.150 mol)(20.76 J/mol ⋅K)(300 K −113.7 K)$
$Q = 580 J $
$\Delta U = Q −W $
$\Delta U = Q −0 $
$\Delta U = 580 J$
