A basketball is contains 1.4 atm in a 20.0°C garage. If the ball is taken outside where the temperature is -5.00°C, what will be the new pressure inside the ball?
Question
A basketball is contains 1.4 atm in a 20.0°C garage. If the ball is taken outside where the temperature is -5.00°C, what will be the new pressure inside the ball?
Solution
To solve this problem, we can use the Gay-Lussac's law which states that the pressure of a gas is directly proportional to its absolute temperature, or P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
Step 1: Convert the temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = °C + 273.15.
T1 = 20.0°C + 273.15 = 293.15 K T2 = -5.00°C + 273.15 = 268.15 K
Step 2: Substitute the known values into the Gay-Lussac's law equation and solve for P2.
P1/T1 = P2/T2 1.4 atm / 293.15 K = P2 / 268.15 K P2 = 1.4 atm * 268.15 K / 293.15 K P2 = 1.28 atm
So, the new pressure inside the ball when it is taken outside where the temperature is -5.00°C will be approximately 1.28 atm.
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