# Gay-Lussacs Law The Third Gas Law. Objectives When you complete this presentation, you will be able to state Gay-Lussacs Law in terms of pressure and.

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Gay-Lussacs Law The Third Gas Law

Objectives When you complete this presentation, you will be able to state Gay-Lussacs Law in terms of pressure and temperature use Gay-Lussacs Law to calculate pressure and temperature values when a gas is kept at a constant volume.

Introduction This law was not discovered by Joseph Louis Gay-Lussac. He was actually working on measurements related to Charless 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-Lussacs Law.

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.

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.

Application We can write Gay-Lussacs 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.

Application When we are trying to solve a Gay-Lussacs 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 )

Example 1 – Finding P 2 A 3.00 L flask of oxygen gas has a pressure of 1.15 atm at a temperature of 308 K. What is the pressure when the temperature is raised to 373 K? P 1 = 1.15 atm T 1 = 308 K P 2 = ? atm T 2 = 373 K P 2 = P 1 (T 2 /T 1 ) = (1.15 atm)[(373 K)/(308 K)]= 1.39 atm

Practice Problems – Finding P 2 1.A sample of nitrogen gas has a pressure of 1.00 atm at a temperature of 298 K. What is the pressure at 350 K? 2.A sample of hydrogen gas has a pressure of 0.450 atm at 350 K. What is the pressure at 600 K? 3.A flask of oxygen gas has a pressure of 25.0 atm at 650 K. What is the pressure at 250 K? 4.A tank of neon gas has a pressure of 0.125 atm at 250 K. What is the pressure at 750 K? 1.17 atm 0.771 atm 10.4 atm 0.375 atm

Example 2 – Finding P 1 A tank was pumped full of air at a temperature of 313 K. What was the original pressure if the pressure in the tank is 260 kPa when the temperature is lowered to 263 K? P 1 = ? kPa T 1 = 313K P 2 = 260 kPa T 2 = 263 K P 1 = P 2 (T 1 /T 2 ) = (260 kPA)[(313 K)/(263 K)]= 309 kPa

Practice Problems – Finding P 1 1.A flask of nitrogen gas has a pressure of 2.50 atm at 350 K. What was the pressure at 250 K? 2.A sample of hydrogen gas has a pressure of 32.4 kPa at 400 K. What was the pressure at 600 K? 3.A flask of neon gas has a pressure of 143 mm Hg at 125 K. What was the pressure at 298 K? 4.A tank of argon gas has a pressure of 245 atm at 300 K. What was the pressure at 350 K? 1.79 atm 48.6 kPa 341 mm Hg 286 atm

Example 3 – Finding T 2 A gas collecting tube held hydrogen gas at 0.995 atm and 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 P 2 = 0.845 atm V 2 = ? K T 2 = T 1 (P 2 /P 1 ) = (298 K)[(0.845 atm)/(0.995 atm)]= 253 K

Practice Problems – Finding T 2 1.A xenon gas sample had a pressure of 0.972 atm at 298 K. What is the temperature at 4.50 atm? 2.A sample of helium gas had a pressure of 322 kPa at 310 K. What is the temperature at 154 kPa? 3.A flask of chlorine gas had a pressure of 847 mm Hg at 432 K. What is the temperature at 760 mm Hg? 4.A tank of radon gas had a pressure of 4.25 atm at 301 K. What is the temperature at 4.75 atm? 1,380 K 148 K 387 K 334 K

Example 4 – Finding T 1 A tank held nitrogen gas at 784 mm Hg. When the temperature of the flask is set at 330 K, the pressure is 642 mm Hg. What was the initial 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 ) = (330 K)[(784 mm Hg)/(642 mmHg)]= 1.39 atm

Practice Problems – Finding T 1 1.A carbon dioxide gas sample has a pressure of 1.25 atm at 298 K. What was the temperature at 1.00 atm? 2.A sample of fluorine gas has a pressure of 98.3 kPa at 310 K. What was the temperature at 85.3 kPa? 3.A flask of chlorine gas has a pressure of 675 mm Hg at 425 K. What was the temperature at 473 mm Hg? 4.A tank of air has a pressure of 240 atm at 301 K. What was the temperature at 250 atm? 238 K 269 K 298 K 314 K

Summary Gay-Lussacs 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

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