Final answer:
Using Gay-Lussac's law, the pressure of the gas at 50.0 °C, when heated from 25.0 °C at an initial pressure of 2.80 atm in a sealed and rigid container, is calculated to be 3.03 atm.
Step-by-step explanation:
The question involves a sealed and rigid container of gas that experiences a temperature change and we need to find the new pressure inside the container. We can use Gay-Lussac's law, which states that the pressure of a fixed amount of gas at constant volume is directly proportional to its temperature in Kelvins. Since the volume is constant, and the amount of gas does not change, we can set up the formula P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
First, we convert the temperatures from Celsius to Kelvin: T1 = 25.0 °C + 273.15 = 298.15 K and T2 = 50.0 °C + 273.15 = 323.15 K. Next, we use the formula to calculate P2:
P2 = P1 × (T2 / T1)
P2 = 2.80 atm × (323.15 K / 298.15 K)
P2 = 2.80 atm × 1.08386
P2 = 3.03 atm
Hence, the pressure of the gas at 50.0 °C is 3.03 atm.