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Gay-Lussac’s Law

Gay-Lussac’s Law. 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 . P 1 = P 2 T 1 T 2 where:

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Gay-Lussac’s Law

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  1. Gay-Lussac’s Law

  2. 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 T1 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.)

  3. Gay-Lussac’s Law 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

  4. Gay-Lussac’s Law • 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

  5. 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) = 0.733 atm X

  6. Gay-Lussac’s Law Ex. (2) If a rigid container of oxygen gas (O2) at 1.04 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) = 1.10 atm X

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

  8. 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 293 K and the final pressure was 1.10 atm? = P1V1 P1 P2V2 P2 T1 T2 = (1.00 atm) (1.10 atm) (X) (293 K) = 266 K X

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