If an aerosol can at room temperature (20.0 degree Celsius) and a pressure of 1.0 atm is thrown in a fire. If the pressure increased to 4.10 atm, what is the
temperature of the aerosol can in Kelvin?

-401.06 atm

Explanation:

Applying,

P = (nRT/V-nb)-(an²/V²) Equation 1

Where P = Pressure, R = Universal gas constant, V = molar Volume, T = Temperature in Kelvin, a = gas constant a , b = gas constant b, n = numbers of mole

From the question,

Given: T = 299.8 K, V = 0.8006 L, a = 2.318 L.atm/mol, b = 3.978×10²L/mol

Constant: R = 0.0082 atm.dm³/K.mol

Substitute these values into equation 1

P = [(0.0082×299.8×10.53)/(0.8006-(10.53×397.8)]-[(10.53²×2.318/0.8006²)]

P = (25.89/-4188.0334)-(400.995)

P = -0.0618-400.995

P = -401.06 atm