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Gas Laws

Gas Laws. States of matter unit note #2. Physical Characteristics of Gases. Gases have M ass L ow density Compressibility: the most compressible state of matter Diffusion & effusion Diffusion: different gases mix evenly and completely due to their random motion

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Gas Laws

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  1. Gas Laws States of matter unit note #2

  2. Physical Characteristics of Gases Gases have • Mass • Low density • Compressibility: the most compressible state of matter • Diffusion & effusion • Diffusion: different gases mix evenly and completely due to their random motion • Effusion: gas particle pass through a tiny opening under pressure • Pressure

  3. Pressure • Pressure (P) is defined as the force per unit area on a surface • The pressure exerted by a gas depends on • Volume • Temperature • The number of molecules present • A barometer is a device used to measure atmospheric pressure • Units: mmHg (760 at sea level), atm (1 at sea level), kPa (101.325 at sea level), Torr (760 at sea level) • STP: Standard Temp and Pressure (1 atm and 0 oC)

  4. The Gas Laws • Boyle’s Law: Pressure-Volume Relationship • The volume (V) of a fixed mass of gas varies inversely with the pressure (P) at constant temperature (T) • P1V1 = P2V2 = constant

  5. Boyle’s Law: P1V1 = P2V2 • If P1 = 0.947 atm, V1 = 150 mL, P2 = 0.987 atm, then V2 = ? V2 = P1V1 / P2 = 0.947 atm X 150 mL / 0.987 atm = 144 mL

  6. Charles’s Law: Volume-Temperature Relationship • The volume (V) of a fixed mass of gas at constant pressure (P) varies directly with the Kelvin temperature (T in K) • Kelvin temperature: T(K) = 273 + t(oC) • V1/T1 = V2/T2 = constant

  7. Charles’s Law: V1/T1 = V2/T2 • If V1 = 752 mL, T1 = 25oC, T2 = 50oC, then V2 = ? T1 = 273 + 25 = 298 K, T2= 273 + 50 = 323 K V2 = T2 x V1/T1 = 323 K x 752 mL / 298 K = 815 mL

  8. Gay-Lussac’s Law: Pressure-Temperature Relationship • The pressure (P) of a fixed mass of gas at constant volume (V) varies directly with the Kelvin temperature (T in K) • P1/T1 = P2/T2 = constant

  9. Gay-Lussac’s Law: P1/T1 = P2/T2 • P1 = 3.00 atm, t1 = 25 oC, t2 = 52 oC, then P2 = ? T1 = 273 + 25 = 298 K, T2 = 273 + 52 = 325 K P2 = T2 x P1/T1 = 325 K x 3.00 atm / 298 K = 3.27 atm

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