Final answer:
Using Gay-Lussac's Law, the new pressure of nitrogen gas when cooled from 539K to 211K (converted from -62°C) at constant volume is found to be approximately 2.57 kPa.
Step-by-step explanation:
The question involves the application of Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature, provided that the volume and the amount of gas remain constant. In this scenario, we want to find the new pressure of nitrogen gas when the temperature changes from 539K to -62°C without any change in volume.
First, convert -62°C to Kelvin, which is 211K. Then apply the formula derived from Gay-Lussac's Law, P1/T1 = P2/T2, where P1 is the initial pressure (6.58 kPa), T1 is the initial temperature (539K), P2 is the final pressure, and T2 is the final temperature (211K). Solving for P2, we get:
P2 = (P1 x T2) / T1 = (6.58 kPa x 211K) / 539K
After calculation, P2 is found to be approximately 2.57 kPa.