1 Compiled by MAH 100’s of free ppt’s from librarywww.pptpoint.com

2 Avogadro’s Hypothesis At a constant temperature and pressure, a given volume of gas always has the same number of particles. The coefficients of a balanced reaction is the same ratio as the volumes of reactants and products

3 2CO (g) + O 2 (g)  2CO 2 (g) For the above example, it is understood that half the volume of oxygen is needed to react with a given volume of carbon monoxide. This can be used to carry out calculations about volume of gaseous product and the volume of any excess reagents.

4 Example 10cm 3 of ethyne (C 2 H 2 ) is reacted with 50cm 3 of hydrogen to produce ethane (C 2 H 6 ), calculate the total volume and composition of the remaining gas mixture, assuming constant T& P. 1 st get balanced equation: C 2 H 2 (g) + 2H 2 (g)  C 2 H 6 (g) 2 nd look at the volume ratios: 1 mol ethyne to 2 mol of hydrogen, therefore 1 vol to 2 vol 3 rd analyse: If all 10cm 3 of ethyne is used, it needs only 20cm 3 of hydrogen, therefore hydrogen is in excess by 50cm 3 -20cm 3 = 30 cm 3. In the end you’ll have 10 cm 3 Ethane and the leftover 30 cm 3 hydrogen

5 Molar volume The temperature and pressure are specified and used to calculate the volume of one mole of gas. Standard temperature and pressure (STP) is at sea level 1 atm = kPa and 0 o C = 273 K this volume is 22.4 dm 3 (or 22.4 L) Molar gas volume, V m. It contains 6.02 x molecules of gas

6 Example Calculate how many moles of oxygen molecules are there in 5.00 dm 3 at STP n= V STP = 5.00 = mol 22.4 dm dm 3

7 Boyle’s LawBoyle’s Law (1659) Boyle noticed that the product of the volume of air times the pressure exerted on it was very nearly a constant, or PV=constant. If V increases, P decreases proportionately and vice versa. (Inverse proportions) Temperature must be constant. Example: A balloon under normal pressure is blown up (1 atm), if we put it under water and exert more pressure on it (2 atm), the volume of the balloon will be smaller (1/2 its original size) P 1 V 1 =P 2 V 2

8 Boyle’s Law

9 Plotting Boyle’s Law data

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13 Charles’ Law (1787) Gas expands (volume increases) when heated and contracts (volume decreases) when cooled. The volume of a fixed mass of gas varies directly with the Kelvin temperature provided the pressure is constant. V= constant x T V 1 = V 2 T 1 T 2

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16 Gay-Lussac’s Law The pressure of a gas increases as its temperature increases. As a gas is heated, its molecules move more quickly, hitting up against the walls of the container more often, causing increased pressure. P 1 = P 2 T 1 T 2

17 Laws combined… P 1 V 1 = P 2 V 2 T 1 T 2 T must be in Kelvins, but P and V can be any proper unit provided they are consistently used throughout the calculation

18 The Combined Gas Law

19 Constant Volume

20 Constant Pressure

21 All variables considered

22 Practice If a given mass of gas occupies a volume of 8.50 L at a pressure of 95.0 kPa and 35 o C, what volume will it occupy at a pressure of 75.0 kPa and a temperature of 150 o C? 1 st convert o C to K: = 308 K = 423 K 2 nd rearrange equation and solve problem: V 2 = V 1 x P 1 x T 2 = 8.50 x 95.0 x 423 = 14.8 L P 2 x T x 308

23 Temperature Kelvin temperature is proportional to the average kinetic energy of the gas molecules. It is a measure of random motion of the gas molecules More motion = higher temperature

24 Ideal gas behaviour Ideal behaviour is when a gas obeys Boyle’s, Charle’s and Gay-Lussac’s laws well At ordinary temperature and pressures, but there is deviation at low temperature and high pressures

25 Ideal gas where all collisions between molecules are perfectly elastic and in which there are no intermolecular attractive forces. Its like hard spheres bouncing around, but NO interaction.

26 Ideal gas law PV = nRT P= pressure (kPa) Volume = (dm 3 ) n= number of moles R=universal gas constant = J mol -1 K -1 T= temperature (K)

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28 Notice N 2 becomes nearer to ideal at Higher temperatures

29 Ideal vs Real Gases

30 Example g of a gas occupies dm 3 at 17.6 o C and a pressure of kPa, determine its molar mass. PV= nRT rearrange equation for n n= PV/RT = (96.73 x 2.368) / (8.314 x 290.6) = mol Molar mass = mass/ mole = g / mol = g/mol

31 Postulates of the kinetic molecular theory 1.The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero) 2.The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3.The particles are assumed to exert no forces on each other; they are assumed neither to attract nor repel each other 4.The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas

32 What happens to the energy distribution of the particles as temperature increases ?

33 Gases, moles and equations Gas to gas are easy N 2 (g) + 3H 2 (g) → 2NH 3 (g) 200cm 3 600cm 3 400cm 3 100cm 3 400cm 3 Limiting excess200cm 3 100cm 3 Can do volume to volume without having to change to moles

34 What about solid to gas? What volume of carbon dioxide at STP can be obtained from 5.0g of copper(II)carbonate and excess hydrochloric acid? CuCO 3 (s) + 2HCl(aq) →CuCl 2 (aq) + CO 2 (g) + H 2 O(l) 5.0g excess 5.0 mole mole x 22.4 dm If the conditions were not STP work out as above then apply gas law to convert to new conditions