The pressure inside the can upon cooling is 0.4 atm.
Explanation:
Given -
Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K
Final Temperature, T2 = 208°C = 208 + 273 K = 481 K
Pressure upon cooling, P2 = ?
Using Gay Lussac's law:
P1/T1 = P2/T2
P2 = P1 X T2 / T1
P2 = 1 atm X 481 / 1181
P2 = 0.4 atm
Therefore, the pressure inside the can upon cooling is 0.4 atm.