One mole of a gas at 300 K has some translational kinetic energy in the random motion of its…

Step 1. The gas. $E = \tfrac32 \times 1\ \text{mol} \times 8.314\ \text{J/(mol K)} \times 300\ \text{K} = 3741$ J.

Step 2. The ball. $\tfrac12 mv^2 = E$. We want $v$, so $v = \sqrt{\dfrac{2 \times 3741\ \text{J}}{1.0\ \text{kg}}} = 86.5$ m/s.