Answer
$w_{out}=439$ kJ/kg
Work Step by Step
The exit stagnation temperature $T_{02}$ is determined to be $$
T_{02}=T_{01}\left(\frac{P_{02}}{P_{01}}\right)^{(k-1) / k}=(963.2 \mathrm{~K})\left(\frac{0.1}{0.75}\right)^{(1.33-1) / 1.33}=584.2 \mathrm{~K}
$$ Also, $$
\begin{aligned}
c_p&=k c_v=k\left(c_p-R\right)\\ \longrightarrow \quad c_p & =\frac{k R}{k-1} \\
& =\frac{1.33(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})}{1.33-1} \\
& =1.157 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{aligned}
$$ From the energy balance on the turbine,
$$ -w_{\text {out }}=\left(h_{20}-h_{01}\right)
$$ or, $$
w_{\text {out }}=c_p\left(T_{01}-T_{02}\right)=(1.157 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(963.2-584.2) \mathrm{K}=438.5 \mathrm{~kJ} / \mathrm{kg} \cong 439 \mathrm{~kJ} / \mathrm{k}g
$$
