Answer
$\operatorname{COP}_{\mathrm{R}} =0.503$
Work Step by Step
From the isentropic relations, $$
\begin{aligned}
& T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(243 \mathrm{~K})(7)^{0.4 / 1.4}=423.7 \mathrm{~K} \\
& T_4=T_6=T_3\left(\frac{P_4}{P_3}\right)^{(k-1) / k}=(288 \mathrm{~K})(7)^{0.4 / 1.4}=502.2 \mathrm{~K} \\
& T_8=T_7\left(\frac{P_8}{P_7}\right)^{(k-1) / k}=(288 \mathrm{~K})\left(\frac{1}{7 \times 7 \times 7}\right)^{0.4 / 1.4}=54.3 \mathrm{~K}
\end{aligned}
$$ The COP of this ideal gas refrigeration cycle is determined from $$
\begin{aligned}
\operatorname{COP}_{\mathrm{R}} & =\frac{q_L}{w_{\text {net,in }}}=\frac{q_L}{w_{\text {comp,in }}-w_{\text {turb,out }}} \\
& =\frac{h_1-h_8}{\left(h_2-h_1\right)+\left(h_4-h_3\right)+\left(h_6-h_5\right)-\left(h_7-h_8\right)} \\
& =\frac{T_1-T_8}{\left(T_2-T_1\right)+2\left(T_4-T_3\right)-\left(T_7-T_8\right)} \\
& =\frac{243-54.3}{(423.7-243)+2(502.2-288)-(288-54.3)} \\
& =\mathbf{0 . 5 0 3}
\end{aligned}
$$
