3 A sample of neon occupies a volume of 461 mL at STP. What will be the volume of the neon when the pressure is reduced ed to 93.3kPa? Recall standard pressure is 1 atm=760mmHg=101.3kPa. 725 mL 452 mL 550 mL 501 mL Multiple Choice 20 points

To solve this problem, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to the pressure, provided the temperature remains constant. Mathematically, Boyle's Law can be expressed as:<br /><br />\[ P_1V_1 = P_2V_2 \]<br /><br />Where:<br />- \( P_1 \) is the initial pressure<br />- \( V_1 \) is the initial volume<br />- \( P_2 \) is the final pressure<br />- \( V_2 \) is the final volume<br /><br />Given:<br />- Initial volume, \( V_1 = 461 \) mL<br />- Initial pressure, \( P_1 = 101.3 \) kPa<br />- Final pressure, \( P_2 = 93.3 \) kPa<br /><br />We need to find the final volume, \( V_2 \).<br /><br />Using Boyle's Law:<br /><br />\[ 101.3 \, \text{kPa} \times 461 \, \text{mL} = 93.3 \, \text{kPa} \times V_2 \]<br /><br />Solving for \( V_2 \):<br /><br />\[ V_2 = \frac{101.3 \, \text{kPa} \times 461 \, \text{mL}}{93.3 \, \text{kPa}} \]<br /><br />\[ V_2 = \frac{46623.3 \, \text{mL} \cdot \text{kPa}}{93.3 \, \text{kPa}} \]<br /><br />\[ V_2 \approx 501 \, \text{mL} \]<br /><br />Therefore, the volume of the neon when the pressure is reduced to 93.3 kPa will be approximately 501 mL.<br /><br />The correct answer is:<br />\[ \boxed{501 \, \text{mL}} \]