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Gay-Lussac’s Law

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Gay-Lussac’s Law The relationship among pressure and temperature, at constant volume, can be mathematically represented by an equation known as Gay-Lussac’ law. P1 = P2 T T2 where: P1 is the initial pressure and P2 is the new pressure. T1 is the initial temperature and T2 is the new temperature. V1 and V2 are the same (constant volume.)

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**Gay-Lussac’s Law 22.4 L At constant volume, when temperature**

is increased, the pressure will increase. At constant volume, when temperature is decreased, the pressure will decrease. 22.4 L

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**Gay-Lussac’s Law P = k T P T (K)**

As the Kelvin temperature of the gas increases, the pressure of the gas increases and vise versa. P = k T P P1 • What temperature is this? P2 • T (K) T2 T1 -273oC

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Gay-Lussac’s Law Ex. (1) If a rigid container of He at STP were cooled to 200. K, then what would be the new pressure in atmospheres? P1V1 P1 = P2V2 P2 T1 T2 (1.00 atm) = (X) (273 K) (200. K) X = 0.733 atm

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Gay-Lussac’s Law Ex. (2) If a rigid container of oxygen gas (O2) at atm and 32.8oC were heated to 50.0oC, what would be new pressure? P1V1 P1 = P2V2 P2 T1 T2 (1.04 atm) = (X) (305.8 K) (323.0 K) X = 1.10 atm

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Gay-Lussac’s Law Ex. (3) What would be the new temperature of a rigid container of a gas at kPa and K if the gas pressure increased to kPa? P1V1 P1 = P2V2 P2 T1 T2 (101.3 kPa) = (115 kPa) (293 K) (X) X = 333 K

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Gay-Lussac’s Law Ex. (4) What would have been the initial temperature of a rigid container of neon (Ne) at 1.00 atm if the final temperature was K and the final pressure was 1.10 atm? P1V1 P1 = P2V2 P2 T1 T2 (1.00 atm) = (1.10 atm) (X) (293 K) X = 266 K

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Column A Column B 1 atm 760 mm Hg kPa 760 mm Hg 760 kPa

Column A Column B 1 atm 760 mm Hg kPa 760 mm Hg 760 kPa

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