1. Gay-Lussac’s Law The Third Gas Law

2. Introduction This law was not discovered by Joseph Louis Gay- Lussac. He was actually working on measurements related to Charles’s Law. This relationship between pressure and temperature was actually discovered by another French chemist, Guillaume Amontons in about 1702. However, it is still called Gay-Lussac’s Law.

3. Introduction Amontons noticed that there was a relationship between the pressure of a gas and the temperature of that gas when volume was held constant. He noticed that pressure and temperature were directly related. As the temperature increased, the pressure increased. As the temperature decreased, the pressure decreased.

4. Introduction This behavior would be expected from the assumptions of the kinetic theory. As the temperature increases, the average speed of the gas particles also increases. This causes the collisions with the walls of the container to be more forceful. More force over the same area (remember, volume is constant) gives more pressure.

5. Application We can write Gay-Lussac’s law two different ways: P/T = k or P = kT, where "k" is a constant. P 1 /T 1 = P 2 /T 2 We most often use the second notation to solve problems.

6. Application When we are trying to solve a Gay-Lussac’s law problem, we will need to know three of the four variables. For P 1 /T 1 = P 2 /T 2 we can solve for: P 1 = P 2 (T 1 /T 2 ) T 1 = T 2 (P 1 /P 2 ) P 2 = P 1 (T 2 /T 1 ) T 2 = T 1 (P 2 /P 1 )

7. Example 1 A 3.00 L flask of oxygen gas has a pressure of 115 kPa at a temperature of 35.0°C (308 K). What is the pressure when the temperature is raised to 100°C (373 K)? P 1 = 115 kPa T 1 = 308 K = 35°C = 95°F P 2 = ? kPa T 2 = 373 K = 100°C = 212°F P 2 = P 1 (T 2 /T 1 ) P 2 = (115 kPa)(373 K/308 K) P 2 = (115 kPa)(1.21) = 139 kPa

8. Example 2 A tank was pumped full of air at a temperature of 40.0°C (313 K). What was the original pressure if the pressure in the tank is 260 kPa when the temperature is lowered to - 10.0°C (263 K)? P 1 = ? kPa T 1 = 313 K = 40°C = 104°F P 2 = 260 kPa T 2 = 263 K = -10°C = 14°F P 1 = P 2 (T 1 /T 2 ) P 1 = (260 kPa)(313 K/263 K) P 1 = (260 kPa)(1.19) = 309 kPa

9. Example 3 A gas collecting tube held hydrogen gas at 0.995 atm and 25.0°C (298 K). What is the temperature of the gas if the pressure in the tube is 0.845 atm? P 1 = 0.995 atm T 1 = 298 K = 25°C = 77°F P 2 = 0.845 atm T 2 = ? K T 2 = T 1 (P 2 /P 1 ) T 2 = (298 K)(0.845 atm/0.995 atm) T 2 = (298 K)(0.849) = 253 K = -20°C = 4°F

10. Example 4 A tank held nitrogen gas at 784 mm Hg. When the temperature of the flask is set at 57.0°C (330 K), the pressure is 642 mm Hg. What was the original temperature of the tank? P 1 = 784 mm Hg T 1 = ? K P 2 = 642 mm Hg T 2 = 330 K T 1 = T 2 (P 1 /P 2 ) T 1 = (330 K)(784 mm Hg/642 mm Hg) T 1 = (330 K)(1.22) = 403 K = 130°C = 266°F

11. Summary Gay-Lussac’s Law: At a constant volume, the pressure of a gas is directly proportional to its temperature. Equations: P/T = k or P = kT, where k is a constant P 1 /T 1 = P 2 /T 2