Answer
-78 K
Work Step by Step
The process is adiabatic, so Q=0. There is no heat flow into or out of the gas. Calculate the temperature change from $\Delta U=Q-W$.
$$\Delta U =\frac{3}{2}nR\Delta T = 0-W$$
$$\Delta T=-\frac{2}{3}\frac{W}{nR}$$
$$=-\frac{2(8300J)}{3(8.5mol)(8.314J/(mol\cdot K))}=-78K$$
