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Gay-Lussac’s Law The Third Gas Law Objectives • When you complete this presentation, you will be able to • state Gay-Lussac’s Law in terms of pressure and temperature • use Gay-Lussac’s 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 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. 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-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. 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) Example 1 – Finding P2 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? P1 = 1.15 atm T1 = 308 K P2 = ? atm T2 = 373 K P2 = P1(T2/T1) = (1.15 atm)[(373 K)/(308 K)] = 1.39 atm Practice Problems – Finding P2 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? 1.17 atm 2. A sample of hydrogen gas has a pressure of 0.450 atm at 350 K. What is the pressure at 600 K? 0.771 atm 3. A flask of oxygen gas has a pressure of 25.0 atm at 650 K. What is the pressure at 250 K? 10.4 atm 4. A tank of neon gas has a pressure of 0.125 atm at 250 K. What is the pressure at 750 K? 0.375 atm Example 2 – Finding P1 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? P1 = ? kPa T1 = 313K P2 = 260 kPa T2 = 263 K P1 = P2(T1/T2) = (260 kPA)[(313 K)/(263 K)] = 309 kPa Practice Problems – Finding P1 1. A flask of nitrogen gas has a pressure of 2.50 atm at 350 K. What was the pressure at 250 K? 1.79 atm 2. A sample of hydrogen gas has a pressure of 32.4 kPa at 400 K. What was the pressure at 600 K? 48.6 kPa 3. A flask of neon gas has a pressure of 143 mm Hg at 125 K. What was the pressure at 298 K? 341 mm Hg 4. A tank of argon gas has a pressure of 245 atm at 300 K. What was the pressure at 350 K? 286 atm Example 3 – Finding T2 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? P1 = 0.995 atm T1 = 298 K P2 = 0.845 atm V2 = ? K T2 = T1(P2/P1) = (298 K)[(0.845 atm)/(0.995 atm)] = 253 K Practice Problems – Finding T2 1. A xenon gas sample had a pressure of 0.972 atm at 298 K. What is the temperature at 4.50 atm? 1,380 K 2. A sample of helium gas had a pressure of 322 kPa at 310 K. What is the temperature at 154 kPa? 148 K 3. A flask of chlorine gas had a pressure of 847 mm Hg at 432 K. What is the temperature at 760 mm Hg? 387 K 4. A tank of radon gas had a pressure of 4.25 atm at 301 K. What is the temperature at 4.75 atm? 334 K Example 4 – Finding T1 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? P1 = 784 mm Hg T1 = ? K P2 = 642 mm Hg T2 = 330 K T1 = T2(P1/P2) = (330 K)[(784 mm Hg)/(642 mmHg)]= 1.39 atm Practice Problems – Finding T1 1. A carbon dioxide gas sample has a pressure of 1.25 atm at 298 K. What was the temperature at 1.00 atm? 238 K 2. A sample of fluorine gas has a pressure of 98.3 kPa at 310 K. What was the temperature at 85.3 kPa? 269 K 3. A flask of chlorine gas has a pressure of 675 mm Hg at 425 K. What was the temperature at 473 mm Hg? 298 K 4. A tank of air has a pressure of 240 atm at 301 K. What was the temperature at 250 atm? 314 K 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