Answer
$T_2 = 284.8 K = 11.65^oC$
Work Step by Step
Temperature of the final air mass is
$T_2 = T_1 (\frac{p_2}{p_1})^{\gamma -1/\gamma} $
Assume that air is an ideal gas, with $\gamma = 1.40$.
$T_2 = (299.15 K) (\frac{0.850\times 10^5 Pa}{1.01 \times 10^5 Pa})^{0.40/1.40} $
$T_2 = 284.8 K = 11.65^oC$
