Answer
$η_{II}=67.6\%$
Work Step by Step
The reversible work input to the compressor is determined from $$
w_{n e v}=h_2-h_1-T_0\left(s_2-s_1\right)=767.4\ \mathrm{Btu} / \mathrm{lbm}-(537\ \mathrm{R})(0.4628\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})=518.8\ \mathrm{Btu} / \mathrm{lbm}
$$ The second-law efficiency of the compressor is $$
\eta_{\mathrm{II}}=\frac{w_{\text {rev }}}{w_{\text {natual }}}=\frac{518.8}{767.4}=0.676=\mathbf{6 7 . 6 \%}
$$
