A pressurized can of paint contains a fixed mass of gas at an internal pressure of 1.258 atm at 24.8oC. Suppose the can is able to withstand an internal pressure of 2.410 atm before it explodes. At what temperature, in degrees Celsius, will this can explode? HINT: The can is a rigid container, that means the volume stays constant!
I would use (P1/T1) = (P2/T2), substitute and solve for T2.
P1 = 1.248 atm
T1 = 273 + 24.8 = ?
P2 = 2.410 atm
T2 = ?
We don't need to worry about V because that doesn't change.
To solve this problem, we can use the combined gas law equation, which combines Boyle's law, Charles's law, and Gay-Lussac's law.
The combined gas law equation is given as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume (which stays constant)
T1 = initial temperature
P2 = final pressure (before the can explodes)
T2 = final temperature (we need to solve for this)
Given:
P1 = 1.258 atm
P2 = 2.410 atm
V1 = constant volume
T1 = 24.8 oC (convert to Kelvin: T1 = 24.8 + 273.15 = 297.95 K)
Rearranging the equation, we can solve for T2:
T2 = (P2 * V1 * T1) / (P1 * V2)
Since the volume (V1) is constant, we can set V1 = V2.
Substituting the given values into the equation:
T2 = (2.410 atm * V1 * 297.95 K) / (1.258 atm * V1)
The volume (V1) cancels out, resulting in:
T2 = (2.410 * 297.95) / 1.258
Calculating this equation, we find that:
T2 ≈ 576.4 K
Converting this temperature back to degrees Celsius:
T2 ≈ 576.4 - 273.15
Therefore, the can will explode at a temperature of approximately 303.25 oC.
To determine the temperature at which the can will explode, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = number of moles of gas
R = gas constant
T = temperature
Since the volume of the can is constant, we can rewrite the equation as:
P1/T1 = P2/T2
Where:
P1 = initial pressure
T1 = initial temperature
P2 = final pressure (pressure at which the can will explode)
T2 = final temperature (temperature at which the can will explode)
Plugging in the given values:
P1 = 1.258 atm
T1 = 24.8°C + 273.15 (convert Celsius to Kelvin)
P2 = 2.410 atm
We rearrange the equation to solve for T2:
T2 = (P2 * T1) / P1
Substituting the values:
T2 = (2.410 atm * (24.8°C + 273.15 K)) / 1.258 atm
Calculating this, we get:
T2 ≈ 292.55 K
Converting back to degrees Celsius:
T2 ≈ 292.55 K - 273.15 ≈ 19.40°C
Therefore, the can will explode at approximately 19.40°C.