1. 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. 1.4.7 Solve problems using the ideal gas equation, PV = nRT. 1.4.8 Analyse graphs relating to the ideal gas equation.

2. Measuring chemical quantities: gases  in volume units (cm 3, dm 3, etc.) using a gas syringe  volume depends on temperature and pressure

3. 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

4. Mercury barometer  Defines and measures atmospheric pressure  Mercury column rises to 760 mm average at sea level  This quantity 1 atmosphere = 100 kPa

5. Pressure

6. Standard temperature and pressure (STP)  Standard Temperature and Pressure (IUPAC) STP = 0 o C or 273.15 K and 100kPa  Reference for comparing gas quantities  Can calculate volume at various temperatures and pressures

7. Charles’ Law Charles’ Law: the volume of a gas is proportional to the Kelvin temperature at constant pressure V = kT V1 = T1 V2 T2

8. 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 dm 3 = 313 K V 2 348K (313K)( V 2 ) = (2.32 dm 3 ) (348K) V 2 = 2.58dm 3

9. Gay-Lussac’s Law The pressure and temperature of a gas are directly proportional at constant volume. P = kT P1 = T1 P2 T2

10. Boyle’s Law Boyle’s Law: pressure and volume of a gas are inversely proportional at constant temperature. PV = constant. P 1 V 1 = P 2 V 2

11. Boyle’s Law

12. Example: Calculate the pressure at 2.50 dm 3 if the pressure of helium gas in a balloon has a volume of 4.00 dm 3 at 210 kPa. P 1 V 1 = P 2 V 2 (210 kPa) (4.00 dm 3 ) = P 2 (2.50 dm 3 ) P 2 = (210 kPa) (4.00 dm 3 ) (2.50 dm 3 ) = 340 kPa

13. Combined gas law P1V1 = P2V2 T1 T2

14. Example question Example: A gas at 110 kPa and 30 o C fills a container at 2.0 dm 3. Calculate the new volume when the temperature rises to 80 o C and the pressure increases to 440 kPa.

15. 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 V 1 /n 1 = constant = V 2 /n 2 V 1 /n 1 = V 2 /n 2

16. 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

17. 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 dm 3 kPa mol -1 K -1