To understand what happens to the pressure of the gas inside the aerosol can when it is exposed to a colder temperature, we can refer to Gay-Lussac's Law from the general gas laws. Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
The initial condition is that the can is at 25°C with a pressure of 4.5 atm. When the can is stored outside during a cold night, the temperature will decrease. Given that the volume of the can (350 mL) remains constant and the amount of gas does not change, any decrease in temperature will result in a decrease in pressure, according to Gay-Lussac's Law.
Mathematically, Gay-Lussac's Law is represented as [tex]\( P_1/T_1 = P_2/T_2 \)[/tex], where:
- [tex]\( P_1 \)[/tex] is the initial pressure,
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin,
- [tex]\( P_2 \)[/tex] is the final pressure,
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin.
If we plug the initial conditions into the equation and solve for [tex]\( P_2 \)[/tex], we will see a decrease in temperature results in a decrease in pressure, without needing specific values for the final temperature.
Since we only have to determine whether the pressure will be less, more, or the same—we can conclude without performing any calculations that the pressure will decrease if the temperature decreases, as the question suggests ("...if the can is stored outside in a cold night.").
Therefore, the correct answer is:
A. The pressure of the gas will be less than 4.5 atm.
