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.
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.
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.
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.