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Gay-Lussac’s Law

Gay-Lussac’s Law. The Third Gas Law. Introduction. This law was not discovered by Joseph Louis Gay-Lussac. He was actually working on measurements related to Charles’s Law.

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Gay-Lussac’s Law

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  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. • P1/T1 = P2/T2 • 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 P1/T1 = P2/T2 we can solve for: • P1 = P2(T1/T2) • T1 = T2(P1/P2) • P2 = P1(T2/T1) • T2 = T1(P2/P1)

  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)? P1 = 115 kPa T1 = 308 K = 35°C = 95°F P2 = ? kPa T2 = 373 K = 100°C = 212°F P2 = P1(T2/T1) P2 = (115 kPa)(373 K/308 K) P2 = (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)? P1 = ? kPa T1 = 313 K = 40°C = 104°F P2 = 260 kPa T2 = 263 K = -10°C = 14°F P1 = P2(T1/T2) P1 = (260 kPa)(313 K/263 K) P1 = (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? P1 = 0.995 atm T1 = 298 K = 25°C = 77°F P2 = 0.845 atm T2 = ? K T2 = T1(P2/P1) T2 = (298 K)(0.845 atm/0.995 atm) T2 = (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? P1 = 784 mm Hg T1 = ? K P2 = 642 mm Hg T2 = 330 K T1 = T2(P1/P2) T1 = (330 K)(784 mm Hg/642 mm Hg) T1 = (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 • P1/T1 = P2/T2

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