Final answer:
Using Gay-Lussac's Law, the final pressure of a gas sample with a volume of 5.3L when the temperature rises from 27°C to 76°C while keeping the volume constant is calculated to be approximately 841.29 mmHg.
Step-by-step explanation:
The question relates to the behavior of gases under different conditions of temperature and pressure, which can be addressed using Gay-Lussac's Law. This law states that for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature. So, to find the final pressure of the gas when the temperature rises, we can use the formula derived from Gay-Lussac's law:
P1/T1 = P2/T2
Where:
P1 is the initial pressure of the gas (725 mmHg)
T1 is the initial absolute temperature in Kelvin (27°C + 273.15 = 300.15 K)
P2 is the final pressure of the gas
T2 is the final absolute temperature in Kelvin (76°C + 273.15 = 349.15 K)
By substituting the known values into the equation and solving for P2, we get:
725 mmHg / 300.15 K = P2 / 349.15 K
P2 = (725 mmHg * 349.15 K) / 300.15 K
P2 ≈ 841.29 mmHg
So, the final pressure of the gas sample when the temperature rises to 76°C is approximately 841.29 mmHg.