A balloon of helium was put in the freezer at -23^circ C. Its volume at this temperature was 2.5 liters It was removed from the freezer. Eventually, it reached a temperature of 177^circ C What would its volume be at this temperature?(Neglect any force used to stretch the rubber balloon.) square

To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. The formula for Charles's Law is:<br /><br />\[<br />\frac{V_1}{T_1} = \frac{V_2}{T_2}<br />\]<br /><br />where:<br />- \( V_1 \) is the initial volume,<br />- \( T_1 \) is the initial temperature in Kelvin,<br />- \( V_2 \) is the final volume,<br />- \( T_2 \) is the final temperature in Kelvin.<br /><br />First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each Celsius temperature.<br /><br />Initial temperature (\( T_1 \)):<br />\[<br />T_1 = -23^{\circ}C + 273.15 = 250.15 \, K<br />\]<br /><br />Final temperature (\( T_2 \)):<br />\[<br />T_2 = 177^{\circ}C + 273.15 = 450.15 \, K<br />\]<br /><br />Given that the initial volume (\( V_1 \)) is 2.5 liters, we can now solve for the final volume (\( V_2 \)) using Charles's Law:<br /><br />\[<br />\frac{2.5 \, \text{liters}}{250.15 \, K} = \frac{V_2}{450.15 \, K}<br />\]<br /><br />Solving for \( V_2 \):<br /><br />\[<br />V_2 = \frac{2.5 \, \text{liters} \times 450.15 \, K}{250.15 \, K}<br />\]<br /><br />\[<br />V_2 \approx \frac{1125.375}{250.15}<br />\]<br /><br />\[<br />V_2 \approx 4.50 \, \text{liters}<br />\]<br /><br />Therefore, the volume of the balloon at \( 177^{\circ}C \) would be approximately 4.50 liters.