1. Gases and moles

2. Gas volumes It is easier to measure the volume of a gas than its mass. The volume of a gas depends on; The temperature. The pressure. The number of moles present.

3. Avogadro’s Law Gases are not very dense, so the size of gas particles is negligible compared to the distance between them. A gas consisting of single atoms, eg Argon, will behave the same as a gas composed of molecules of two atoms such as chlorine. Hence Avogadro’s Law; Equal volumes of gases contain the same number of particles under the same temperature and pressure.

4. The Kelvin Temperature Scale One degree of the Kelvin scale represents the same change in temperature as one degree on the Celsius scale. But it starts at -273 o C. IE K = C + 273 Eg; A room is at 20 o C, what is its temperature in K? T = 20 + 273 = 293K. NB units are K, not oK!

5. Standard conditions (STP) Standard conditions are defined as; A temperature of 273K (0 o C) A pressure of 101kPa (Kilo Pascals) = 1 atmosphere. One mole of any gas under standard conditions will occupy a volume of 22.4 dm 3.

6. A mole of gas!

7. Equation for stp N o moles = Volume in dm 3 22.4

8. Number of moles Volume (dm 3 ) n V 22.4 Think of the equation as a triangle At stp;

9. Number of moles = Volume (dm 3 ) 22.4 n V n = v/22.4 Rearranging;

10. Number of moles Volume n V 22.4 V = n22.4 X 22.4 Rearranging;

11. Calculating the number of moles at stp. Number of moles = volume (dm 3 ) / 22.4 Eg; How many moles are there in 150 cm 3 of oxygen at stp? First convert the volume to decimetres. V = 150/1000 = 0.15 dm 3 Then divide by 22.4. N = 0.15 / 22.4 = 0.0067 moles.

12. Calculating the volume of a gas at stp. Eg; What is the volume, at stp, of 2.5 moles of carbon dioxide? First rearrange the basic equation; N = V /22.4 So V = N x 22.4 V = 2.5 x 22.4 = 56 dm 3.

13. Room temperature and pressure. Room temperature is taken as being 298K (25 o C). Room pressure is assumed to be 101 kPa (1 atm). Under these conditions one mole of any gas occupies a volume of 24 dm 3.

14. Equation for rtp N o moles = Volume in dm 3 24

15. Number of moles Volume (dm 3 ) n V 24 Think of the equation as a triangle At rtp

16. Number of moles = Volume (dm 3 ) 24 n V n = v/24 Rearranging;

17. Number of moles Volume n V 24 V = n24 X 24 Rearranging;

18. Calculating the number of moles at rtp. Number of moles = volume (in dm 3 ) / 24 Eg; How many moles are there in 325 cm 3 of hydrogen at rtp? First convert the volume to decimetres. V = 325 / 1000 = 0.325 dm 3 Then divide by 24 N = 0.325 / 24 = 0.0135 moles.

19. Calculating the volume of a gas at rtp. Eg; What is the volume, at rtp, of 0.75 moles of nitrogen dioxide? First rearrange the basic equation; N = V / 24 So V = N x 24 V = 0.75 x 24 = 18 dm 3.

20. Ideal gas equation PV = nRT P is pressure in Pascals (Pa) aka Newton/m 2 V is volume in m 3. NB 1m 3 = 1,000 dm 3. n is the number of moles. R is the gas constant = 8.31 T is the temperature in Kelvin.

21. Limitations of the ideal gas equation The ideal gas equation makes two assumptions; 1) Gas particles have mass, but zero volume. 2) There are no internal forces between particles. Gases conform to these assumptions unless the pressures are very high and/or the temperatures very low.

22. Using the ideal gas equation to calculate relative molecular mass. n = m/M Where m = mass present M = molecular mass M = mRT/PV

23. Calculations on the ideal gas law 512cm 3 of a gas has a mass of 1.236g at 30 o C and 1atm pressure. What is its molecular mass? First write out what you know; m = 1.236g V = 512cm 3 P =1 atm T = 20 o C Then convert the units. V = 512 x 1x 10 -6 = 5.12x10 -4 m 3 T = 20 + 273 = 293K. P = 1 x 101375 = 101375 Nm -2

24. Substitute into the ideal gas equation; M = mRT/PV 1.236x8.31 x293 / 101325x 5.12x10 -4 = 58 NB Molecular mass is relative so should have no units.