1. We are given that the initial pressure $p_1 = 500000$ Pa and initial volume $V_1 = 10$ cm$^3$. The atmospheric pressure is $p_2 = 100000$ Pa. 2. Using Boyle's Law, which states $p_1 V_1 = p_2 V_2$, solve for $V_2$ when pressure changes to atmospheric pressure: $$V_2 = \frac{p_1 V_1}{p_2} = \frac{500000 \times 10}{100000} = 50\ \text{cm}^3$$ 3. Next, calculate the volume when pressure reduces to $12500$ Pa using the same formula: $$V_2 = \frac{p_1 V_1}{p_2} = \frac{500000 \times 10}{12500} = 400\ \text{cm}^3$$ 4. Finally, we calculate the pressure $p_2$ if the volume decreases to $2.0$ cm$^3$: $$p_2 = \frac{p_1 V_1}{V_2} = \frac{500000 \times 10}{2.0} = 2500000\ \text{Pa}$$ **Final answers:** - Volume at atmospheric pressure: $50$ cm$^3$ - Volume at $12500$ Pa: $400$ cm$^3$ - Pressure if volume decreases to $2.0$ cm$^3$: $2500000$ Pa