1. **State the problem:** A gas initially at pressure $P_1 = 8.20 \times 10^4$ torr is contained in a volume $V_1 = 15.0$ L. It expands into a larger volume $V_2 = 6.00 \times 10^4$ L. We need to find the new pressure $P_2$ after expansion. 2. **Formula used:** For a gas at constant temperature and amount, Boyle's Law applies: $$P_1 V_1 = P_2 V_2$$ This means pressure and volume are inversely proportional. 3. **Rearrange the formula to solve for $P_2$:** $$P_2 = \frac{P_1 V_1}{V_2}$$ 4. **Substitute the known values:** $$P_2 = \frac{8.20 \times 10^4 \times 15.0}{6.00 \times 10^4}$$ 5. **Calculate numerator and denominator:** $$P_2 = \frac{1.23 \times 10^6}{6.00 \times 10^4}$$ 6. **Simplify the fraction:** $$P_2 = \cancel{\frac{1.23 \times 10^6}{6.00 \times 10^4}} = 20.5$$ 7. **Final answer rounded to three significant figures:** $$\boxed{20.5}$$ The new pressure of the gas after expansion is 20.5 torr.