Final answer:
Applying Gay-Lussac's law and the pressure-temperature relationship at constant volume, the final temperature of a gas whose pressure decreased from 1150 torr to 760 torr is calculated to be approximately 49.6 °C. This is option (b).
Step-by-step explanation:
The question pertains to the relationship between gas pressure and temperature at a constant volume, reflecting the principles of the Gay-Lussac's law of gases. To find the final temperature (T2) when the pressure of a gas is decreased, we can use the formula P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, respectively, and P2 is the final pressure. We must first convert the given temperatures to Kelvin by adding 273.15 to the Celsius values.
Original conditions are P1 = 1150 torr and T1 = 75.0 °C + 273.15 = 348.15 K. The final pressure P2 is given as 760 torr. Applying the pressure-temperature relationship yields: 1150 torr / 348.15 K = 760 torr / T2. Solving for T2, we find T2 = (760 torr * 348.15 K) / 1150 torr.
After calculating T2 in Kelvin, we convert it back to Celsius by subtracting 273.15. Through calculation, the final temperature (T2) in Celsius is found to be approximately 49.6 °C.
Therefore, the final temperature of the gas after the pressure decrease is 49.6 °C, corresponding to option (b).