Answer
$\eta=89.5\%$
Work Step by Step
For isentropic processes on ideal gases:
$\dfrac{T_{2,s}}{T_1}=\left(\dfrac{P_{2,s}}{P_1}\right)^\dfrac{k-1}{k}$
Given $T_1=300\ K,\ P_1=100\ kPa,\ k=1.260,\ P_{2,s}=600\ kPa$:
$T_{2,s}=434.2\ K$
Efficiency:
$\eta=\dfrac{w_s}{w_a}=\dfrac{h_{2,s}-h_1}{h_{2,a}-h_1}=\dfrac{c_p(T_{2,s}-T_1)}{c_p(T_{2,a}-T_1)}$
$\eta=\dfrac{T_{2,s}-T_1}{T_{2,a}-T_1}$
Given $T_{2,a}=450\ K$
$\eta=89.5\%$
