Final answer:
The final temperature of the air in a 3.00 L sealed tank that cools from 3.00 atm and 20.0°C to 1.00 atm is 97.72 K, calculated using the combined gas law with pressure and temperature variables.
Step-by-step explanation:
The student asked about the final temperature of the air in a sealed tank that is cooled until the pressure changes from 3.00 atm to 1.00 atm. To solve this, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas: P1×V1/T1 = P2×V2/T2. Since the volume of the tank is constant and there's no mention of a change in the amount of air in the tank, we assume the quantity of gas remains constant. Therefore, the equation simplifies to P1/T1 = P2/T2.
First, we need to convert temperatures to Kelvin: T1 = 20.0 + 273.15 = 293.15 K.
Now, we just need to solve for the final temperature T2:
T2 = (P2×T1)/P1 = (1.00 atm × 293.15 K)/3.00 atm = 97.72 K.
Therefore, the air in the tank cools down to a final temperature of 97.72 K when the pressure decreases to 1.00 atm.