Final answer:
The volume of a weather balloon calibrated at 0.00 °C with a volume of 22.0 L would be approximately 18.6 L at -32.0 °C, assuming pressure is held constant, as per Charles's Law.
Step-by-step explanation:
The question involves calculating the change in volume of a weather balloon when the temperature decreases, with the assumption that the pressure remains constant. This scenario is an application of Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. To find the new volume at -32.0 °C (which is 241 K when converted to Kelvin), we use the formula V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature. Since the initial volume (V1) is 22.0 L and the initial temperature (T1) is 273 K (0 °C in Kelvin), we can solve for the final volume (V2) when the final temperature (T2) is 241 K (-32 °C in Kelvin).
V1/T1 = V2/T2
22.0 L / 273 K = V2 / 241 K
V2 = (22.0 L × 241 K) / 273 K
V2 = 18.6 L (approximately)
The volume of the weather balloon at -32.0 °C would be approximately 18.6 liters, assuming pressure is held constant.