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, will the pressure increase or decrease?

To answer this question, we need to use the Gay-Lussac's law which states that the pressure of a gas is directly proportional to its absolute temperature, when the volume is held constant.

Step 1: Convert the temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = °C + 273.15.

For the garage temperature: K = 20.0°C + 273.15 = 293.15 K For the outside temperature: K = -5.00°C + 273.15 = 268.15 K

Step 2: Apply Gay-Lussac's law. The formula is P1/T1 = P2/T2, where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature.

We know P1 (1.4 atm), T1 (293.15 K), and T2 (268.15 K), and we need to find P2.

Rearranging the formula to find P2 gives us P2 = P1 * (T2/T1).

Step 3: Substitute the known values into the equation.

P2 = 1.4 atm * (268.15 K / 293.15 K) = 1.28 atm

So, the pressure of the basketball will decrease when it is taken from the garage to the outside.