Final answer:
Using Gay-Lussac's law, the likely pressure of the gas when the temperature is increased from 300 K to 600 K is 2 atm, since pressure is directly proportional to temperature at constant volume.
Step-by-step explanation:
The question asks about the behavior of a gas when its temperature is increased in a closed system, which involves the principles of gas laws in physics.
Given that the initial pressure (P1) of the gas is 1 atm at a temperature (T1) of 300 K in a 4 L container, we are asked to find the new pressure (P2) when the temperature is increased to 600 K, assuming the volume remains constant. To solve this, we use Gay-Lussac's law, which states that the pressure of a gas at constant volume is directly proportional to its absolute temperature (P/T = constant).
By setting up the equation P1/T1 = P2/T2, we can calculate the new pressure P2 as follows:
- P1 = 1 atm
- T1 = 300 K
- P2 = ?
- T2 = 600 K
Substituting the known values into the equation:
(1 atm / 300 K) = (P2 / 600 K)
Therefore, P2 = (1 atm × 600 K) / 300 K = 2 atm.
The likely pressure of the gas at 600 K, when the temperature is doubled from 300 K, would be 2 atm, making the correct answer B) 2 atm.