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The Pressure Temperature Law

The Pressure Temperature Law. Getting Hot under the Collar since 1802…er, 1702. Such a Caring Scientist. In the same paper from 1802, Joseph Gay-Lussac published another law stating: As the Temperature of a gas Increases, the Pressure Increases proportionally. Joseph Gay-Lussac (1778-1850).

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The Pressure Temperature Law

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  1. The Pressure Temperature Law Getting Hot under the Collar since 1802…er, 1702

  2. Such a Caring Scientist In the same paper from 1802, Joseph Gay-Lussac published another law stating: As the Temperature of a gas Increases, the Pressure Increases proportionally Joseph Gay-Lussac (1778-1850) Pneumatic Chemist # 3…again

  3. Amonton’s / Gay-Lussac’s…NEITHER! • Guillaume Amontons was the first to publish the relationship between pressure and temperature in 1702; however, it languished in relative obscurity for 100 years until Gay-Lussac published it in conjunction with Charles’s Law. • For this reason, it is sometime (erroneously) called Gay-Lussac’s Law. • It is not Amontons’ Law, either.

  4. Can We Agree to Disagree? • Gay-Lussac’s Law: • The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers. • Amontons’ Laws: • The force of friction is directly proportional to the applied load. (Amontons’ 1st Law) • The force of friction is independent of the apparent area of contact. (Amontons’ 2nd Law)

  5. P & T For any given gas, the quotient of its Pressure and absolute Temperature (in Kelvin) is constant If Volume & Mass remain the same P ~ T P / T = C Remember, °C + 273 = K

  6. Hot & Heavy (pressure…) The Pressure/Temp Law: For a fixed amount of an ideal gas kept at a fixed volume, P (pressure) and T (temperature) are directly proportional (when one increases, the other increases, equally). Sorry, no cool gif…

  7. What They Said P ~ T (in Kelvin) P / T = C P2 / T2 = C2 If C = C2 P / T = C = P2 / T2 P1 = P2 T1 T2 300 K (80° F) 1.0 atm 150 K (-189° F) 0.5 atm

  8. Is Gay Smarter than a 10th Grader? A container with a fixed volume is at 20.0°C with a pressure of 63.25 inHg. If the temperature is increased to 110.0°C, what is the new pressure? P1 = 63.25 inHg, T1 = 20°C + 273 = 293 K P2 = ?, T2 = 110°C + 273 = 383 K (63.25 inHg)/(293 K) = P2 / (383 K) P2 = 82.68 inHg

  9. Probably, but He’s Way Dead A container has an internal pressure of 25.6 psi at 15°C. The container will rupture if the internal pressure exceeds 47.0 psi. What is the maximum temperature possible? P1 = 25.6 psi, T1 = 15°C + 273 = 288 K P2 = 47.0 psi, T2 = ? (25.6 psi) / (288 K) = (47.0 psi) / T2 T2 = 529 K = 256°C

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