To determine whether the aerosol can would explode if heated to 425 °C, we can use the ideal gas law, which relates the pressure, volume, and temperature of a gas. Assuming the volume of the can remains constant, we can use the direct relationship between pressure and temperature as given by the equation
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
where \(P_1\) and \(T_1\) are the initial pressure and temperature, and \(P_2\) and \(T_2\) are the final pressure and temperature, respectively. Temperatures must be in Kelvin.
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Convert the initial temperature from °C to K: \[ T_1 = 25 + 273.15 = 298.15 , \text{K} \]
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Convert the final temperature from °C to K: \[ T_2 = 425 + 273.15 = 698.15 , \text{K} \]
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Set up the knowns:
- \(P_1 = 1.70 , \text{atm}\)
- \(T_1 = 298.15 , \text{K}\)
- \(T_2 = 698.15 , \text{K}\)
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Rearranging the equation to solve for \(P_2\): \[ P_2 = P_1 \times \frac{T_2}{T_1} \]
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Plug in the values: \[ P_2 = 1.70 , \text{atm} \times \frac{698.15 , \text{K}}{298.15 , \text{K}} \]
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Calculate \(P_2\): \[ P_2 \approx 1.70 \times 2.340 \approx 3.98 , \text{atm} \]
Now, compare \(P_2\) with the explosion threshold of 4.00 atm.
Since \(P_2 \approx 3.98 , \text{atm}\), this is below the threshold of 4.00 atm.
Thus, the answer is:
No, because the final pressure is 3.98 atm.