Water vapor inside an aluminum can has a pressure of 1.00 atm at a temperature of 120.°С. What is the pressure of the gas when the
temperature is dropped to 100.°C?*

P2 = 0.95atm

Explanation:

Using Gay Lussac's equation as follows

P1/T1 = P2/T2

Where;

P1 = initial pressure

T1 = initial temperature

P2 = final pressure

T2 = final temperature

According to the information provided in this question; P1 = 1.00atm, T1 = 120°C, P2 = ?, T2 = 100°C

We need to convert temperature in °C to K

T1 (K) = 120°C + 273K = 393K

T2 (K) = 100°C + 273K = 373K

Hence;

P1/T1 = P2/T2

1.00/393 = P2/373

373 × 1 = 393 × P2

373 = 393P2

P2 = 373/393

P2 = 0.949

P2 = 0.95atm

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