A pump contains 0.5 L of air at 203 kPa You draw back on the piston of the pump, expanding the volume until the pressure reads 508 kPa. What is the new volume of air in the pump? P_(1)V_(1)=P_(2)V_(2) P_(1)=203kPa V_(1)=0.5L P_(2)=50.8kPa V_(2)=? Hint: there is an example problem in your student notes. (Boyle's Law) 1.0 L 0.2 L 2.0 L 4.0 L

To solve this problem, we can use Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure and volume are inversely proportional. Mathematically, it is expressed as:<br /><br />\[ P_1 V_1 = P_2 V_2 \]<br /><br />Given:<br />- \( P_1 = 203 \text{ kPa} \)<br />- \( V_1 = 0.5 \text{ L} \)<br />- \( P_2 = 50.8 \text{ kPa} \)<br />- \( V_2 = ? \)<br /><br />We need to find the new volume \( V_2 \). Rearrange the equation to solve for \( V_2 \):<br /><br />\[ V_2 = \frac{P_1 V_1}{P_2} \]<br /><br />Substitute the given values into the equation:<br /><br />\[ V_2 = \frac{203 \text{ kPa} \times 0.5 \text{ L}}{50.8 \text{ kPa}} \]<br /><br />Calculate the value:<br /><br />\[ V_2 = \frac{101.5 \text{ kPa} \cdot \text{L}}{50.8 \text{ kPa}} \]<br /><br />\[ V_2 \approx 2.0 \text{ L} \]<br /><br />So, the new volume of air in the pump is approximately 2.0 liters.<br /><br />Therefore, the correct answer is:<br /><br />\[ \boxed{2.0 \text{ L}} \]