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IB1 Chemistry Quantitative chemistry 5. 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations. 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.
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IB1 ChemistryQuantitative chemistry 5 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations. 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. 1.4.7 Solve problems using the ideal gas equation, PV = nRT. 1.4.8 Analyse graphs relating to the ideal gas equation.
Measuring chemical quantities: gases • in volumeunits (cm3, dm3, etc.) using a gas syringe • volumedepends on temperature and pressure
Propeties of gases • Variable volume and shape • Expand to occupy volume available • Can be easily compressed • Exert pressure on whatever surrounds them • Volume, Pressure, Temperature, and the number of moles present are interrelated • Easily diffuse into one another
Mercury barometer • Defines and measures atmospheric pressure • Mercury column rises to 760 mm average at sea level • This quantity 1 atmosphere = 100 kPa
Standard temperature and pressure (STP) • Standard Temperature and Pressure (IUPAC) STP = 0oC or 273.15 K and 100kPa • Reference for comparing gas quantities • Can calculate volume at various temperatures and pressures
Charles’ Law Charles’ Law: the volume of a gas is proportional to the Kelvin temperature at constant pressure V = kT V1 = T1 V2 T2
Example: Calculate the volume at 75oC of of a gas sample that at 40 oC occupies a volume of 2.32 dm3 Convert temperatures to Kelvin. 40C = 313K 75C = 348K 2.32 dm3= 313 K V2 348K (313K)( V2) = (2.32 dm3) (348K) V2 = 2.58dm3
Gay-Lussac’s Law The pressure and temperature of a gas are directly proportional at constant volume. P = kT P1 = T1 P2 T2
Boyle’s Law Boyle’s Law: pressure and volume of a gas are inversely proportional at constant temperature. PV = constant. P1V1 = P2V2
Example: Calculate the pressure at 2.50 dm3 if the pressure of helium gas in a balloon has a volume of 4.00 dm3 at 210 kPa. P1 V1 = P2 V2 (210 kPa) (4.00 dm3) = P2(2.50 dm3) P2 = (210 kPa) (4.00 dm3) (2.50 dm3) = 340 kPa
Combined gas law P1V1 = P2V2 T1 T2
Example question Example: A gas at 110 kPa and 30 oC fills a container at 2.0 dm3. Calculate the new volume when the temperature rises to 80oC and the pressure increases to 440 kPa.
Avogadro’s Law • Equal volumes of a gas under the same temperature and pressure contain the same number of particles. • If the temperature and pressure are constant the volume of a gas is proportional to the number of moles of gas present V = constant * n where n is the number of moles of gas V/n = constant V1/n1 = constant = V2 /n2 V1/n1 = V2 /n2
Universal Gas Equation 4 factors that define the quantity of gas: Volume, Pressure, Kelvin Temperature, and the number of moles of gas present (n). PV = constant, R nT proportionality constant R is known as the universal gas constant
Universal Gas Equation The Universal gas equation is usually written as PV = nRT Where P = pressure V = volume T = Kelvin Temperature n = number of mole R = 8.314 dm3kPa mol-1 K-1