Final answer:
The pressure of the basketball when taken from a 24.0°C garage to the outside temperature of -1.00°C will decrease to approximately 1.74 atm, calculated using the Ideal Gas Law, specifically Charles's Law, and assuming constant volume.
Step-by-step explanation:
The question pertains to the behavior of gas pressure in a basketball as it undergoes a temperature change. We can use the Ideal Gas Law and in particular, Charles's Law, which relates volume and temperature, to calculate the new pressure. The original pressure inside the basketball is 1.90 atm at 24.0°C (297 K) in the garage. We need to find the new pressure outside where the temperature is -1.00°C (272 K). Since the volume of the basketball does not change significantly, we can assume it to be constant.
Charles's Law states that for a given amount of gas at constant pressure, the volume is directly proportional to its temperature in kelvins (V/T = constant). In this case, we keep the volume constant, not the pressure, so the law slightly modifies to P1/T1 = P2/T2, where P is pressure and T is temperature. Plugging in the values and solving for P2 gives us:
P2 = P1 * (T2 / T1)
P2 = 1.90 atm * (272 K / 297 K)
P2 ≈ 1.74 atm
Therefore, the pressure of the basketball outside at -1.00°C is approximately 1.74 atm.