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Gay-Lussacs Law

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The relationship among pressure and temperature, at constant volume, can be mathematically represented by an equation known as Gay- Lussac law. P 1 = P 2 T 1 T 2 where: P 1 is the initial pressure and P 2 is the new pressure. T 1 is the initial temperature and T 2 is the new temperature. V 1 and V 2 are the same (constant volume.) Gay-Lussacs Law

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

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Gay-Lussacs Law As the Kelvin temperature of the gas increases, the pressure of the gas increases and vise versa. P T (K) -273 o C What temperature is this? P = k T P1P1 P2P2 T2T2 T1T1

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Gay-Lussacs 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? P1V1P1V1 P2V2P2V2 T1T1 T2T2 = P1P1 P2P2 (200. K) (X) = (1.00 atm) (273 K) X atm=

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Gay-Lussacs Law Ex. (2) If a rigid container of oxygen gas (O 2 ) at 1.04 atm and 32.8 o C were heated to 50.0 o C, what would be new pressure? P1V1P1V1 P2V2P2V2 T1T1 T2T2 = P1P1 P2P2 (323.0 K) (X) = (1.04 atm) (305.8 K) X 1.10 atm=

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Gay-Lussacs Law Ex. (3) What would be the new temperature of a rigid container of a gas at kPa and 293 K if the gas pressure increased to 115 kPa? P1V1P1V1 P2V2P2V2 T1T1 T2T2 = P1P1 P2P2 (X) (115 kPa) = (101.3 kPa) (293 K) X 333 K=

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Gay-Lussacs 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 293 K and the final pressure was 1.10 atm? P1V1P1V1 P2V2P2V2 T1T1 T2T2 = P1P1 P2P2 (293 K) (1.10 atm) = (1.00 atm) (X) X 266 K=

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