Sure! To solve this problem, we'll use the Ideal Gas Law, which states that for a given amount of gas at constant volume, the pressure of the gas is directly proportional to its temperature.
The relationship can be expressed as:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the initial pressure (105 psi)
- [tex]\( T_1 \)[/tex] is the initial temperature
- [tex]\( P_2 \)[/tex] is the final pressure
- [tex]\( T_2 \)[/tex] is the final temperature
First, let's convert the temperatures from Celsius to Kelvin because the Ideal Gas Law requires temperatures in Kelvin.
1. Initial temperature, [tex]\( T_1 \)[/tex]:
- Room temperature is 22°C.
- To convert to Kelvin: [tex]\( T_1 = 22 + 273.15 = 295.15 \)[/tex] K
2. Final temperature, [tex]\( T_2 \)[/tex]:
- Fire temperature is 600°C.
- To convert to Kelvin: [tex]\( T_2 = 600 + 273.15 = 873.15 \)[/tex] K
Now, we can use the proportion to find the final pressure [tex]\( P_2 \)[/tex]:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
To isolate [tex]\( P_2 \)[/tex], we rearrange this equation:
[tex]\[ P_2 = P_1 \times \frac{T_2}{T_1} \][/tex]
Substituting the known values:
[tex]\[ P_2 = 105 \, \text{psi} \times \frac{873.15 \, \text{K}}{295.15 \, \text{K}} \][/tex]
When calculated, this yields:
[tex]\[ P_2 = 310.62 \, \text{psi} \][/tex]
Thus, if the aerosol can is thrown into a fire and reaches a temperature of 600°C, the pressure inside the can will increase to approximately 310.62 psi.
