An empty aerosol-spray can at room temperature (20 ◦C) is thrown into an incinerator where the temperature reaches 458◦C. If the gas inside the empty container was initially at a pressure of 1 atm, what pressure did it reach inside the incinerator? Assume the gas was at constant volume and the can did not explode.
Answer in units of atm.
To solve this problem, we can make use of the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
In this case, the volume of the empty aerosol-spray can is kept constant, so we can rearrange the equation as:
P1/T1 = P2/T2
P1 = initial pressure of the gas (1 atm)
T1 = initial temperature of the gas (20 ◦C = 20 + 273 = 293 K)
P2 = pressure of the gas inside the incinerator (what we need to find)
T2 = temperature inside the incinerator (458 ◦C = 458 + 273 = 731 K)
Plugging in the values, we get:
1 atm / 293 K = P2 / 731 K
To find P2, we can cross-multiply and solve for it:
P2 = (1 atm * 731 K) / 293 K
P2 ≈ 2.49 atm
Therefore, the pressure reached inside the incinerator is approximately 2.49 atm.
(P1/T1) = (P2/T2)
Substitute and solve for P2. Remember T MUST be in kelvin.