gaslawss
DownloadTélécharger
Actions
Vote :
ScreenshotAperçu
Informations
Catégorie :Category: nCreator TI-Nspire
Auteur Author: rylanf
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.81 Ko KB
Mis en ligne Uploaded: 26/03/2025 - 10:55:27
Uploadeur Uploader: rylanf (Profil)
Téléchargements Downloads: 3
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a4547646
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.81 Ko KB
Mis en ligne Uploaded: 26/03/2025 - 10:55:27
Uploadeur Uploader: rylanf (Profil)
Téléchargements Downloads: 3
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a4547646
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
<<
U10 Gases.pptx Units of Note for Gas Laws Pressure " atmosphere (atm), Torricelli (torr), or millimeters of Hg (mmHg) " 1atm = 760torr = 760mmHg Volume " liter (L) or milliliter (mL) " 1L = 1000mL Temperature " Kelvin (K) or Celsius (C) " K = C + 273 *STP = standard temperature and pressure = 273K and 1.00atm Daltons law of partial pressures Total pressure exerted by all gases in a mixture is the sum of all partial pressures of individual gases in the mixture p= pressure exerted by a given gas n= dedsignation for a specific gas p1= pressure of first gas Partial Pressure Example A sample of gas X evaporates over water in a closed system. What is the pressure of gas X if the total pressure is 780.0 torr and water vapor pressure is 1.00 atm? 780.05a55_5_ 15N5a5Z/ 7605a55_5_ = 1.035N5a5Z 5a55a5N5Y 1.035N5a5Z 5a55a5N5Y 1.005N5a5Z 55S 5d5N5a5R5_ = 0.03005N5a5Z 55S 5T5N5` Mole Ratio of Gases The mole ratio of a specific gas is described by& 5K5V = 5[5V/ 5[5a55a5N5Y = 5C5V /5C5a55a5N5Y = 5I5V /5I5a55a5N5Y 5K5V = mole ratio of a given gas 5[5V = moles of a given gas 5C5V = pressure of a given gas 5I5V = volume of a given gas *Mole ratios are used as conversion factors for mixtures of gases Boyles Law Robert Boyle (1627-1691) Volume of a gas inversely proportional to the Pressure exerted by the gas. 5C15I1 = 5C25I2 As volume and pressure are inversely proportional, as volume increases, pressure must decrease by the same factor. Therefore, since volume doubled, pressure must be reduced by half 5C2 = 5C15I1/ 5I2 = (0.5005N5a5Z)(1.005?)/ (5.005?) = 0.1005N5a Charles Law Jaques Charles (1746-1823) Volume of a gas is directly proportional to the Temperature of a gas 5I1/ 5G1 = 5I2/ 5G2 Since volume and temperature are directly proportional, if temperature increases the volume of the gas must also increase by the same factor. In this case, the temperature was doubled, therefore the volume must double as well 5I2 = 5I15G2/ 5G1 = (2.325?)(3485>)/ (3135>) = 2.58 Guy-Lussacs Law Guy-Lussacs Law Joseph Guy-Lussac (1778-1850) Pressure of a gas is directly proportional to the Temperature of a gas 5C1/ 5G1 = 5C2/ 5G2 Since pressure and temperature are directly proportional, any increase in temperature would cause a corresponding increase in pressure. As the temperature doubles, the pressure must also double 5C2 = 5C15G2/ 5G1 = (5.005N5a5Z)(2635>)/ (2985>) = 4.415N5a Avogadros Law Amedeo Avogadro (1776-1856) Volume of gas is directly proportional to the number of Moles of a gas 5I1/ 5[1 = 5I2/ 5[2 As volume and the number of moles of a gas are directly proportional, reducing the number of moles by half would also reduce the volume by half. The vessel would contract to half its current volume, or 2.5L A balloon is filled to a volume of 1.90L with 0.0920mol He gas. If 0.0210mol of additional He is added to the balloon, while temperature and pressure and held constant, what is the new volume of the balloon? 5I2 = 5I15[2/ 5[1 = (1.905?)(0.09205Z55Y 5;5R + 0.02105Z55Y 5;5R)/ 0.09205Z55Y 5;5R = 2.33 Combined Gas Law Combine the Laws of Boyle, Charles, and Guy-Lussac P is inversely proportional to V. P and V are directly proportional to T. 5C15I1/ 5G1 = 5C25I2/ 5G2 A gas at 1.10atm and 303K fills a flexible container with an internal volume of 2.00L. If the temperature is raised to 353K and the pressure increases to 4.40atm, what is the new volume? 5I2 = 5C15I15G2 5G15C2 = (1.105N5a5Z)(2.005?)(3535>)/ (3035>)(4.405N5a5Z) = 0.583 Ideal Gas Law Gas that behaves perfectly and has no attractive/repulsive forces Works as a simple prediction breaks down at high P/T 5C5I = 5[5E5G R = 0.08206 5?5N5a5Z 5Z55Y5> *STP = standard temperature and pressure = 273K and 1.00atm Calculate the mol NH3 contained in a 3.0L vessel at 1.50atm and 300K. 5[ = 5C5I/ 5E5G = (1.505N5a5Z)(3.05?)/ (0.08206 5?5N5a5Z 5Z55Y5>)(3005>) = 0.25Z55Y 5A5;3 Made with nCreator - tiplanet.org
>>
Compatible OS 3.0 et ultérieurs.
<<
U10 Gases.pptx Units of Note for Gas Laws Pressure " atmosphere (atm), Torricelli (torr), or millimeters of Hg (mmHg) " 1atm = 760torr = 760mmHg Volume " liter (L) or milliliter (mL) " 1L = 1000mL Temperature " Kelvin (K) or Celsius (C) " K = C + 273 *STP = standard temperature and pressure = 273K and 1.00atm Daltons law of partial pressures Total pressure exerted by all gases in a mixture is the sum of all partial pressures of individual gases in the mixture p= pressure exerted by a given gas n= dedsignation for a specific gas p1= pressure of first gas Partial Pressure Example A sample of gas X evaporates over water in a closed system. What is the pressure of gas X if the total pressure is 780.0 torr and water vapor pressure is 1.00 atm? 780.05a55_5_ 15N5a5Z/ 7605a55_5_ = 1.035N5a5Z 5a55a5N5Y 1.035N5a5Z 5a55a5N5Y 1.005N5a5Z 55S 5d5N5a5R5_ = 0.03005N5a5Z 55S 5T5N5` Mole Ratio of Gases The mole ratio of a specific gas is described by& 5K5V = 5[5V/ 5[5a55a5N5Y = 5C5V /5C5a55a5N5Y = 5I5V /5I5a55a5N5Y 5K5V = mole ratio of a given gas 5[5V = moles of a given gas 5C5V = pressure of a given gas 5I5V = volume of a given gas *Mole ratios are used as conversion factors for mixtures of gases Boyles Law Robert Boyle (1627-1691) Volume of a gas inversely proportional to the Pressure exerted by the gas. 5C15I1 = 5C25I2 As volume and pressure are inversely proportional, as volume increases, pressure must decrease by the same factor. Therefore, since volume doubled, pressure must be reduced by half 5C2 = 5C15I1/ 5I2 = (0.5005N5a5Z)(1.005?)/ (5.005?) = 0.1005N5a Charles Law Jaques Charles (1746-1823) Volume of a gas is directly proportional to the Temperature of a gas 5I1/ 5G1 = 5I2/ 5G2 Since volume and temperature are directly proportional, if temperature increases the volume of the gas must also increase by the same factor. In this case, the temperature was doubled, therefore the volume must double as well 5I2 = 5I15G2/ 5G1 = (2.325?)(3485>)/ (3135>) = 2.58 Guy-Lussacs Law Guy-Lussacs Law Joseph Guy-Lussac (1778-1850) Pressure of a gas is directly proportional to the Temperature of a gas 5C1/ 5G1 = 5C2/ 5G2 Since pressure and temperature are directly proportional, any increase in temperature would cause a corresponding increase in pressure. As the temperature doubles, the pressure must also double 5C2 = 5C15G2/ 5G1 = (5.005N5a5Z)(2635>)/ (2985>) = 4.415N5a Avogadros Law Amedeo Avogadro (1776-1856) Volume of gas is directly proportional to the number of Moles of a gas 5I1/ 5[1 = 5I2/ 5[2 As volume and the number of moles of a gas are directly proportional, reducing the number of moles by half would also reduce the volume by half. The vessel would contract to half its current volume, or 2.5L A balloon is filled to a volume of 1.90L with 0.0920mol He gas. If 0.0210mol of additional He is added to the balloon, while temperature and pressure and held constant, what is the new volume of the balloon? 5I2 = 5I15[2/ 5[1 = (1.905?)(0.09205Z55Y 5;5R + 0.02105Z55Y 5;5R)/ 0.09205Z55Y 5;5R = 2.33 Combined Gas Law Combine the Laws of Boyle, Charles, and Guy-Lussac P is inversely proportional to V. P and V are directly proportional to T. 5C15I1/ 5G1 = 5C25I2/ 5G2 A gas at 1.10atm and 303K fills a flexible container with an internal volume of 2.00L. If the temperature is raised to 353K and the pressure increases to 4.40atm, what is the new volume? 5I2 = 5C15I15G2 5G15C2 = (1.105N5a5Z)(2.005?)(3535>)/ (3035>)(4.405N5a5Z) = 0.583 Ideal Gas Law Gas that behaves perfectly and has no attractive/repulsive forces Works as a simple prediction breaks down at high P/T 5C5I = 5[5E5G R = 0.08206 5?5N5a5Z 5Z55Y5> *STP = standard temperature and pressure = 273K and 1.00atm Calculate the mol NH3 contained in a 3.0L vessel at 1.50atm and 300K. 5[ = 5C5I/ 5E5G = (1.505N5a5Z)(3.05?)/ (0.08206 5?5N5a5Z 5Z55Y5>)(3005>) = 0.25Z55Y 5A5;3 Made with nCreator - tiplanet.org
>>
