Answer
$w_{out}=−140\text{ kJ/kg}$
Work Step by Step
For an isentropic process of an ideal gas with constant specific heats, the work is expressed as $$
\begin{aligned}
w_{\text {out }} & =\int_1^2 P d v=P_1 v_1^k \int_1^2 v^{-k} d v \\
& =\frac{P_1 v_1^k}{1-k}\left(v_2^{1-k}-v_1^{1-k}\right)=\frac{P_1 v_1^k}{1-k}\left[\left(\frac{v_2}{v_1}\right)^{1-k}-1\right]
\end{aligned}
$$ since $P_1 v_1^k=P v^k$ for an isentropic process. Also, $$
\begin{aligned}
P_1 v_1 & =R T_1 \\
\left(v_2 / v_1\right)^k & =P_1 / P_2
\end{aligned}
$$ Substituting, we obtain $$
\begin{aligned}
w_{\text {out }} & =\frac{R_u T_1}{M(1-k)}\left[\left(\frac{P_2}{P_1}\right)^{(k-1) / k}-1\right] \\
& =\frac{(8.314 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K})(288 \mathrm{~K})}{(32 \mathrm{~kg} / \mathrm{kmol})(1-1.35)}\left[\left(\frac{700 \mathrm{kPa}}{100 \mathrm{kPa}}\right)^{(1.351) / 1.35}-1\right] \\
& =-140 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The negative sign shows that the work is done on the system.
