Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab (volume of 180m^3) was 1.1 atm before the storm and 085 atm during the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside? 7,100m^3 230m^3 1,800m^3 190m^3

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, this 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 (1.1 atm),<br />- \( V_1 \) is the initial volume (180 m³),<br />- \( P_2 \) is the final pressure (0.85 atm),<br />- \( V_2 \) is the final volume.<br /><br />We need to find \( V_2 \). Rearranging the equation gives us:<br /><br />\[ V_2 = \frac{P_1 \times V_1}{P_2} \]<br /><br />Substituting the known values:<br /><br />\[ V_2 = \frac{1.1 \, \text{atm} \times 180 \, \text{m}^3}{0.85 \, \text{atm}} \]<br /><br />\[ V_2 = \frac{198 \, \text{m}^3}{0.85} \]<br /><br />\[ V_2 \approx 232.94 \, \text{m}^3 \]<br /><br />The closest answer choice to this calculation is \( 230 \, \text{m}^3 \). Therefore, the laboratory would try to expand to approximately \( 230 \, \text{m}^3 \).