- The problem uses Boyle's Law: \(P_1 V_1 = P_2 V_2\).
- Substitute the given values: \(P_1 = 3.2 \text{ atm}\), \(V_1 = 2.0 \text{ L}\), and \(P_2 = 1.0 \text{ atm}\).
- Solve for \(V_2\): \(V_2 = \frac{3.2 \times 2.0}{1.0} = 6.4 \text{ L}\).
- The final answer is \(\boxed{6.4 \text{ L}}\).
Explanation
- Problem Analysis
The problem asks for the final volume \(V_2\) of an oxygen container, given that the initial volume \(V_1\) is \(2.0 \text{ L}\), the initial pressure \(P_1\) is \(3.2 \text{ atm}\), and the desired final pressure \(P_2\) is \(1.0 \text{ atm}\). We are instructed to use Boyle's Law, represented by the equation \(P_1 V_1 = P_2 V_2\), which describes the relationship between pressure and volume for a gas at a constant temperature.
- Calculation of Final Volume
To find the final volume \(V_2\), we rearrange the given formula \(P_1 V_1 = P_2 V_2\) to solve for \(V_2\):
\[V_2 = \frac{P_1 \times V_1}{P_2}\]
Substituting the known values into this equation, we get:
\[V_2 = \frac{3.2 \text{ atm} \times 2.0 \text{ L}}{1.0 \text{ atm}}\]
Performing the calculation:
\[V_2 = \frac{6.4 \text{ atm} \cdot \text{L}}{1.0 \text{ atm}} = 6.4 \text{ L}\]
Therefore, the necessary volume is \(6.4 \text{ L}\).
- Conclusion and Result
Comparing our result \(V_2 = 6.4 \text{ L}\) with the given options (0.625 L, 0.64 L, 6.4 L, 64 L), we find that the correct volume is \(6.4 \text{ L}\). Thus, the final answer is \(\boxed{6.4 \text{ L}}\).
Examples
Boyle's Law is commonly used in scuba diving to understand how air volume in a diver's equipment changes as the pressure increases with depth. For example, if a gas bubble at a certain volume is brought to the surface, the decrease in pressure causes the volume to expand significantly. Understanding these relationships ensures safety and proper management of breathing gas during underwater activities.