Final answer:
The final pressure of a gas initially at 80 kPa and 100 K, when heated to 400 K at constant volume, can be determined using Gay-Lussac's Law to be 320 kPa.
Step-by-step explanation:
The student asked about the final pressure of a gas when its temperature is raised from 100 K to 400 K while being kept at constant volume. Since no volume change is mentioned, we can use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature when volume is held constant. Thus, the final pressure can be found using the equation P1/T1 = P2/T2, where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature.
Substituting into the formula, we get P2 = (P1 × T2) / T1. Given that P1 = 80 kPa and T1 = 100 K, while T2 = 400 K, we can calculate the final pressure as follows: P2 = (80 kPa × 400 K) / 100 K = 320 kPa. Therefore, the final pressure of the gas when heated to 400 K is 320 kPa.