GAY- LUSSAC'S LAW
|
|
Discovered by Joseph Louis Gay-Lussac in the early 1800's. However, the ChemTeam's knowledge of this is much less sure than concerning Boyle's or Charles' Law.
Gives the relationship between pressure and temperature when volume and amount are held constant. If the temperature of a container is increased, the pressure increases. If the temperature of a container is decreased, the pressure decreases. Why? Suppose the temperature is increased. This means gas molecules will move faster and they will impact the container walls more often. This means the gas pressure inside the container will increase, since the container has rigid walls (volume stays constant). Gay-Lussac's Law is a direct mathematical relationship. This means there are two connected values and when one goes up, the other also increases. The mathematical form of Gay-Lussac's Law is: P ÷ T = k This means that the pressure-temperature fraction will always be the same value if the volume and amount remain constant. Let P1 and T1 be a pressure-temperature pair of data at the start of an experiment. If the temperature is changed to a new value called T2, then the pressure will change to P2. Keep in mind that when volume is not discussed (as in this law), it is constant. That means a container with rigid walls. As with the other laws, the exact value of k is unimportant in our context. It is important to know the PT data pairs obey a constant relationship, but it is not important for us what the exact value of the constant is. Besides which, the value of K would shift based on what pressure units (atm, mmHg, or kPa) you were using. We know this: P1 ÷ T1 = k And we know this: P2 ÷ T2 = k Since k = k, we can conclude that P1 ÷ T1 = P2 ÷ T2. This equation of P1 ÷ T1 = P2 ÷ T2 will be very helpful in solving Gay-Lussac's Law problems. P1 T2 = P2 T1 Notice the similarities to the Charles' Law graphic. This is because both laws are direct relationships. Make sure to convert any Celsius temperature to Kelvin before using it in your calculation. |
Check out this video and answer the problems below1. 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?
2. 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure? 3. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased? 4. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank? 5. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from -100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank? 6. Calculate the final pressure inside a scuba tank after it cools from 1.00 x 103 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm |