If a balloon containing 1,200 L of gas at 25^circ C and 760 mm Hg pressure ascends to an altitude where the pressure is 380 mm Hg and the temperature is 54^circ C the volume will be (Be sure to use the correct number of significant figures.) 2,200 L 3.8times 10^-4L 2,600 L 660 L

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law is given by:<br /><br />\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]<br /><br />where:<br />- \( P_1 \) and \( P_2 \) are the initial and final pressures,<br />- \( V_1 \) and \( V_2 \) are the initial and final volumes,<br />- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.<br /><br />First, we need to convert the temperatures from Celsius to Kelvin:<br />\[ T_1 = 25^{\circ}C + 273.15 = 298.15 \, K \]<br />\[ T_2 = 54^{\circ}C + 273.15 = 327.15 \, K \]<br /><br />Next, we plug in the known values into the combined gas law equation:<br />\[ P_1 = 760 \, \text{mm Hg} \]<br />\[ V_1 = 1200 \, L \]<br />\[ T_1 = 298.15 \, K \]<br />\[ P_2 = 380 \, \text{mm Hg} \]<br />\[ T_2 = 327.15 \, K \]<br /><br />We need to find \( V_2 \). Rearrange the combined gas law to solve for \( V_2 \):<br />\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]<br /><br />Substitute the values into the equation:<br />\[ V_2 = \frac{(760 \, \text{mm Hg})(1200 \, L)(327.15 \, K)}{(380 \, \text{mm Hg})(298.15 \, K)} \]<br /><br />Calculate the result:<br />\[ V_2 = \frac{760 \times 1200 \times 327.15}{380 \times 298.15} \]<br />\[ V_2 = \frac{298,776,000}{113,297} \]<br />\[ V_2 \approx 2637.5 \, L \]<br /><br />Considering significant figures, the initial volume (1200 L) has 2 significant figures, so the final answer should also be reported with 2 significant figures:<br />\[ V_2 \approx 2600 \, L \]<br /><br />Therefore, the correct answer is:<br />\[ \boxed{2600 \, L} \]