An empty metal can is heated to 908C and sealed. It is then placed in a room to cool to 208C. What is the pressure inside the can upon cooling? Assume that can contains air under ideal conditions.?

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.