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- Chemistry - Gas Laws (notes)

- Chemistry - Gas Laws (notes). Boyle’s Law. When V. then P. -> P x V is a constant. “P and V hold an inverse relationship”. P x V = k or P 1 V 1 = P 2 V 2. Mnemo: “ ”. Charles’ Law. “T and V hold a direct relationship”. When T.

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- Chemistry - Gas Laws (notes)

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  1. - Chemistry - Gas Laws (notes)

  2. Boyle’s Law When V then P -> P x V is a constant “P and V hold an inverse relationship” P x V = k or P1V1 = P2V2 Mnemo: “ ”

  3. Charles’ Law “T and V hold a direct relationship” When T then V -> V / T is a constant V / T = k or V1/T1 = V2/T2 Mnemo: “Charles has Direct TV”

  4. Gay-Lussac’s Law “T and P hold a direct relationship” When T then P -> P / T is a constant P / T = k or P1/T1 = P2/T2 Mnemo: “ ”

  5. Keep constant: Play with: Law: P, V Boyle: T PV = k T, V Charles: T / V = k P V T, P Gay-Lussac: T / P = k

  6. Example: A gas @ 1 atm pressure occupies a volume of 4L. What will be the pressure if you reduce the volume to 2L? P1= 1 atm V1= 4L P2= ? V2= 2L Formula: Law: P1 x V1 = P2 x V2 P2 = P1 x V1 / V2 Boyle’s Isolate P2: = 1 atm x 4L / 2L = 2 atm Add to your Tabs foldable.

  7. Example 2: A gas @ 303K temperature occupies a volume of 3L. What will be the volume if you change the temperature to 400K? T1= 303 K V1= 3L T2= 400K V2= ? Formula: Law: V1 / T1 = V2 / T2 V2 = V1 x T2 / T1 Charles’ Isolate V2: = 3L x 400K / 303K = 3.96 L Add to your Tabs foldable.

  8. Example 3: A gas @ 273K temperature has a pressure of 590 mmHg. What will be the pressure if you change the temperature to 373K? P2= ? T1= 273 K P1= 590mmHg T2= 373K Formula: Law: P1 / T1 = P2 / T2 P2 = P1 x T2 / T1 Gay-Lussac’s Isolate P2: = 590mmHg x 373K / 273K = 806.1 mmHg Add to your Tabs foldable.

  9. Front Tab: Your name, period, date “Unit 5: Gas Laws” Then: Units of P, V, and T BOYLE’S Law CHARLES’ Law GAY-LUSSAC’S Law Include, for each gas law: Description Formulas Mnemo device Example problem

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