Answer
$\eta_{\text {th }}=1070\text{ R}$
$m_{}=0.03785\text{ lbm}$
$P_{1}=250\text{ psia}$
Work Step by Step
From the thermal efficiency relation,
$$
\eta_{\text {th }}=\frac{W_{\text {net }}}{Q_{\text {in }}}=1-\frac{T_L}{T_H} \longrightarrow \frac{2.5 \mathrm{Btu}}{5 \mathrm{Btu}}=1-\frac{535 \mathrm{R}}{T_H} \longrightarrow T_H=1070\ \mathrm{R}
$$ State 3 may be used to determine the mass of air in the system, $$
m=\frac{P_3 V_3}{R T_3}=\frac{(15 \mathrm{psia})\left(0.5 \mathrm{ft}^3\right)}{\left(0.3704 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{lbm} \cdot \mathrm{R}\right)(535 \mathrm{R})}=\mathbf{0 . 0 3 7 8 5\ \mathrm { lbm }}
$$ The maximum pressure occurs at state 1, $$
P_1=\frac{m R T_1}{V_1}=\frac{(0.03785 \mathrm{lbm})\left(0.3704 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{lbm} \cdot \mathrm{R}\right)(1070 \mathrm{R})}{0.06 \mathrm{ft}^3}=\mathbf{2 5 0}\ \mathbf{p s i a}
$$
