Answer
$P_{1}=6524\text{ kPa}$
$Q_{in}=0.706\text{ kJ}$
$m=0.000409\text{ kg}$
Work Step by Step
(a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process, which is state 1 $$
\frac{T_1}{T_4}=\left(\frac{P_1}{P_4}\right)^{(k-1) / k}
$$ or, $$
P_1=P_4\left(\frac{T_1}{T_4}\right)^{k /(k-1)}=(300 \mathrm{kPa})\left(\frac{1200 \mathrm{~K}}{350 \mathrm{~K}}\right)^{1.667 / 0.66}=\mathbf{6 5 2 4} \mathbf{~ k P a}
$$ (b) The heat input is determined from $$
\begin{aligned}
& \eta_{\text {th }}=1-\frac{T_L}{T_H}=1-\frac{350 \mathrm{~K}}{1200 \mathrm{~K}}=70.83 \% \\
& Q_{\text {in }}=W_{\text {net,out }} / \eta_{\text {th }}=(0.5 \mathrm{~kJ}) /(0.7083)=\mathbf{0 . 7 0 6}\ \mathbf{k J}
\end{aligned}
$$ (c) The mass of helium is determined from $$
\begin{aligned}
s_4-s_3 & =c_p \ln {\frac{T_4}{T_3}}^{\circ 0}-R \ln \frac{P_4}{P_3}=-(2.0769 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{300 \mathrm{kPa}}{150 \mathrm{kPa}} \\
& =-1.4396 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}=s_1-s_2 \\
w_{\text {net,out }} & =\left(s_2-s_1\right)\left(T_H-T_L\right)=(1.4396 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1200-350) \mathrm{K}=1223.7 \mathrm{~kJ} /kg \\
m & =\frac{W_{\text {net,out }}}{w_{\text {net,out }}}=\frac{0.5 \mathrm{~kJ}}{1223.7 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{0 . 0 0 0 4 0 9}\ \mathbf{k g}
\end{aligned}
$$
