Final answer:
Using the combined gas law and converting the increased pressure from 800 mmHg to atm, the temperature at which nitrogen gas would have a volume of 6 mL is calculated to be approximately 436.05 K, which is not among the provided options. However, the closest match from the options would be 400 K.
Step-by-step explanation:
To calculate the temperature at which nitrogen gas will have a volume of 6 mL when the pressure is increased to 800 mmHg, we must use the combined gas law, which is derived from Boyle's Law, Charles's Law, and Gay-Lussac's Law. The combined gas law is expressed as P1V1/T1 = P2V2/T2, where P stands for pressure, V for volume, and T for temperature.
First, standard temperature and pressure (STP) are defined as 273 K and 1 atm (760 mmHg) respectively. To solve this problem, we need to make sure all units are consistent. Since the initial pressure is at STP (1 atm), we can convert 800 mmHg to atmospheres:
(800 mmHg / 760 mmHg per atm) = 1.05 atm (approximately)
Next, rearranging the combined gas law formula to solve for T2 gives us:
(P1V1T2 = P2V2T1)
Then plugging in the relevant values:
(1 atm * 4 mL * T2) / (1.05 atm * 6 mL) = 273 K
T2 = (1.05 atm * 6 mL * 273 K) / (1 atm * 4 mL)
T2 = 436.05 K
However, this result is not one of the options given. We would need more context to provide a possible answer from the provided options. But based on the principle of the combined gas law, the closest temperature option to our calculation would be D) 400 K, acknowledging the simplified assumptions made.