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The gas propellant in an aerosol can has a pressure of 105 psi at room temperature (22°C). What is the pressure if that can is thrown into a fire and reaches a temperature of 600°C? Assume the volume of the can remains constant.

Sagot :

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.
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