The volume of a diver's lungs was measured to be 3.6 liters at 1.0 atmosphere of pressure. As the diver descends, assuming that no air is lost, the pressure increases to 3.2 atmospheres. What is the new lung volume? square

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is held constant. 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,<br />- \( V_2 \) is the final volume.<br /><br />Given:<br />- \( P_1 = 1.0 \) atmosphere,<br />- \( V_1 = 3.6 \) liters,<br />- \( P_2 = 3.2 \) atmospheres.<br /><br />We need to find \( V_2 \).<br /><br />Using Boyle's Law:<br /><br />\[ 1.0 \, \text{atm} \times 3.6 \, \text{L} = 3.2 \, \text{atm} \times V_2 \]<br /><br />Solving for \( V_2 \):<br /><br />\[ V_2 = \frac{1.0 \, \text{atm} \times 3.6 \, \text{L}}{3.2 \, \text{atm}} \]<br /><br />\[ V_2 = \frac{3.6}{3.2} \]<br /><br />\[ V_2 = 1.125 \, \text{L} \]<br /><br />Therefore, the new lung volume is \( 1.125 \) liters.