Answer
$P_{1}=30.0\text{ MPa}$
$Q_{in}=0.706\text{ kJ}$
$m_{}=0.00296\text{ kg}$
Work Step by Step
(a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process, which is state 1. (Table A-17) $$
\begin{aligned}
& T_1=1200 \mathrm{~K} \longrightarrow P_{r_1}=238 \\
& T_4=350 \mathrm{~K} \longrightarrow P_{r_4}=2.379 \quad(\text { Table A-17) } \\
& P_1=\frac{P_{r_1}}{P_{r_4}} P_4=\frac{238}{2.379}(300\ \mathrm{kPa})=30,013\ \mathrm{kPa} \cong \mathbf{3 0 . 0}\ \mathbf{M P a}=P_{\max }
\end{aligned}
$$ (b) The heat input is determined from $$
\begin{aligned}
& \eta_{\text {th }}=1-\frac{T_L}{T_H}=1-\frac{350 \mathrm{~K}}{1200 \mathrm{~K}}=70.83 \% \\
& Q_{\text {in }}=W_{\text {net,out }} / \eta_{\text {th }}=(0.5 \mathrm{~kJ}) /(0.7083)=0.706\ \mathbf{k J}
\end{aligned}
$$ (c) The mass of air is $$
\begin{aligned}
s_4-s_3 & =\left(s_4^{\circ}-s_3^{\circ}\right)^{\oplus 0}-R \ln \frac{P_4}{P_3}=-(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{300 \mathrm{kPa}}{150 \mathrm{kPa}} \\
& =-0.199 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}=s_1-s_2 \\
w_{\text {net,out }} & =\left(s_2-s_1\right)\left(T_H-T_L\right)=(0.199 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1200-350) \mathrm{K}=169.15 \mathrm{~kJ} / \mathrm{kg} \\
m & =\frac{W_{\text {net,out }}}{w_{\text {net,out }}}=\frac{0.5 \mathrm{~kJ}}{169.15 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{0 . 0 0 2 9 6}\ \mathbf{k g}
\end{aligned}
$$
