A balloon was partially filled with helium gas at room temperature. It occupied 4.0 liters of volume at 700.0 mm Hg atmospheric pressure . When the balloon was released, it traveled upward until it burst at 99 mm Hg atmospheric pressure . (Neglect any force exerted to stretch the rubber balloon.) What was the volume of the balloon when it burst? square

To solve this problem, we can use Boyle's Law, which states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional. Mathematically, Boyle's Law is expressed as:<br /><br />\[ P_1 \times V_1 = P_2 \times V_2 \]<br /><br />where:<br />- \( P_1 \) is the initial pressure,<br />- \( V_1 \) is the initial volume,<br />- \( P_2 \) is the final pressure, and<br />- \( V_2 \) is the final volume.<br /><br />Given:<br />- \( P_1 = 700.0 \) mm Hg,<br />- \( V_1 = 4.0 \) liters,<br />- \( P_2 = 99 \) mm Hg.<br /><br />We need to find \( V_2 \).<br /><br />Using Boyle's Law:<br /><br />\[ 700.0 \, \text{mm Hg} \times 4.0 \, \text{liters} = 99 \, \text{mm Hg} \times V_2 \]<br /><br />Now, solve for \( V_2 \):<br /><br />\[ V_2 = \frac{700.0 \, \text{mm Hg} \times 4.0 \, \text{liters}}{99 \, \text{mm Hg}} \]<br /><br />\[ V_2 = \frac{2800.0}{99} \]<br /><br />\[ V_2 \approx 28.28 \, \text{liters} \]<br /><br />Therefore, the volume of the balloon when it burst was approximately 28.28 liters.