Answer
$W_{out}=781.3\text{ kJ/kg}$
Work Step by Step
The $C_p$ and $\mathrm{k}$ values of this equimolar mixture are determined from $$
\begin{aligned}
M_m & =\sum y_i M_i=y_{\mathrm{He}} M_{\mathrm{He}}+y_{\mathrm{Ar}} M_{\mathrm{Ar}}=0.5 \times 4+0.5 \times 40=22 \mathrm{~kg} / \mathrm{kmol} \\
\mathrm{mf}_i & =\frac{m_i}{m_m}=\frac{N_i M_i}{N_m M_m}=\frac{y_i M_i}{M_m} \\
c_{p, m} & =\sum \mathrm{mf}_i c_{p, i}=\frac{y_{\mathrm{He}} M_{\mathrm{He}}}{M_m} c_{p, \mathrm{He}}+\frac{y_{\mathrm{Ar}} M_{\mathrm{Ar}}}{M_m} c_{p, \mathrm{Ar}} \\
& =\frac{0.5 \times 4 \mathrm{~kg} / \mathrm{kmol}}{22 \mathrm{~kg} / \mathrm{kmol}}(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})+\frac{0.5 \times 40 \mathrm{~kg} / \mathrm{kmol}}{22 \mathrm{~kg} / \mathrm{kmol}}(0.5203 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \\
& =0.945 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{aligned}
$$ and
$k_{\mathrm{m}}=1.667$ since $k=1.667$ for both gases.
Therefore, the He-Ar mixture can be treated as a single ideal gas with the properties above. For isentropic processes, $$
T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(1300 \mathrm{~K})\left(\frac{200 \mathrm{kPa}}{2500 \mathrm{kPa}}\right)^{0.667 / 1.6 \overline{6}}=473.2 \mathrm{~K}
$$ From an energy balance on the turbine, $$
\begin{aligned}
\dot{E}_{\text {in }}-\dot{E}_{\text {out }} & =\Delta \dot{E}_{\text {system }} 0(\text { steady) }=0 \\
\dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\
h_1 & =h_2+w_{\text {out }} \\
w_{\text {out }} & =h_1-h_2 \\
w_{\text {out }} & =c_p\left(T_1-T_2\right)=(0.945 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1300-473.2) \mathrm{K}\\&=781.3 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
