1. Number of lone electron pairs 0123 3 trigonal planarbent 4 tetrahedraltrigonal pyramidalbent 5 trigonal bipyramidalsee-sawT-shapedlinear 6 octahedralsquare pyramidalsquare planarT-shaped Total number of electron pairs Recap VSEPR Theory

2. 2 Gases Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids

3. Boyle’s Law - the volume of a gas is inversely proportional to the pressure it exerts: V = k B (1/P) or PV = k B Empirical Gas Laws Charles’ Law - the volume of a gas is directly proportional to its absolute temperature V = k C T Avogadro’s Law – equal volumes of gas at the same pressure and temperature contain equal number of molecules V = k A n 3

4. Charles’ Law predicts that the volume of a gas keeps decreasing as the temperature is lowered The temperature at which the V is zero and the gas occupies no space is called absolute zero: it is the lowest possible temperature Charles’ Law and Absolute Temperature V = k 2 T 4 On the Kelvin scale, this temperature is set as O K: 0 K is the same as -273 o C 100. K is the same as (-273 + 100.) = -173 o C 0 o C is the same as 273 K

5. Ideal Gas Equation pressure volume number of moles absolute temperature gas constant PV = nRT 5 PV = k B V = k C T V = k A n

6. Ideal Gas Equation PV = nRT if n and T are fixed then (Boyle’s Law) PV = nRT if n and P are fixed then (Charles’ Law) PV = nRT if T and P are fixed then V = nRT / P V = (nR/P ) × T (Avogadro’s Law) V = (RT/P ) × n 6

7. unitatmospheric pressure pascal (Pa) kilopascal (kPa) 1.01325  10 5 Pa 101.325 kPa atmosphere1.00 atm mm mercury = torr760 torr pounds/sq inch (psi)14.7 lb in –2 bar1.01325 bar Most convenient unit is the atmosphere (atm) Common Units of Pressure SI unit is the Pascal (Pa) 7

8. R = 8.314 J mol –1 K –1 or Pa m 3 mol –1 K –1 The Gas Constant R Use if you are using SI units e.g. pressure in Pa R = 0.08206 atm L mol –1 K –1 Use if you are using atmospheres and litres PV = nRT 8

9. Q:The lung capacity of an average human born & living at sea level is ~6 L. How many moles of air will be present in full lungs at 25  C and 1.00 atm? A:Volume and pressure has been provided in L and atm so use R = 0.08206 atm L mol –1 K –1 Example: Lung Capacity RT PV n = = (0.08206 atm L mol –1 K –1 x 298 K) (1.00 atm) x (6 L) = 0.3 mol PV = nRT 9

10. Effect of Changing Volume PV = nRT If the size of a sealed container is changed then n, R and T do not change If the initial pressure and volume are P i and V i and the new pressure and volume are P f and V f P i V i = P f V f 10

11. Gas Mixtures: Partial Pressures Dalton’s Law: the total pressure (P T ) is the sum of the pressures due to the individual gases, P i : Air is a mixture of gases with ~79% N 2 and ~21% O 2 : At atmospheric pressure, P T = 1.0 atm so: P N 2 = 0.79 x 1.0 atm = 0.79 atm P O 2 = 0.21 x 1.0 atm = 0.21 atm At a depth of 50 m, P T = 6.0 atm so: P N 2 = 0.79 x 6.0 atm = 4.7 atm P O 2 = 0.21 x 6.0 atm = 1.3 atm 11

12. Gas Stoichiometry V is proportional to the number of moles: 2 mol of H 2 (g) requires 1 mol of O 2 (g) 2 volumes H 2 (g) requires 1 volumes of O 2 (g) 2H 2 (g) + O 2 (g)  2H 2 O(l) V = (RT/P ) × n If we react 2 L of H 2 with 1 L of O 2, then all of the reactants will react If we react 1 L of H 2 with 1 L of O 2, then H 2 is the limiting reagent and O 2 will be left over at the end of the reaction If we react 2 L of H 2 with 0.5 L of O 2, then O 2 is the limiting reagent and H 2 will be left over at the end of the reaction 12

13. Gas Stoichiometry Q:What is the volume of CO 2 produced at body temperature (37 o C) and 1.00 atm when 5.60 g of glucose is burnt in air? C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l) A: (i) Convert mass into moles: number of moles = m / M = 5.60 g / 180.2 g mol -1 = 0.0311 mol of glucose (ii) Use chemical equation for stoichometry: number of moles of CO 2 (g) = 6 x 0.0311 = 0.187 mol (iii) Use ideal gas law to calculate volume: V = nRT / P = (0.187)(0.082606)(37 + 273)/(1.00) = 4.76 L don’t forget to convert to Kelvin!

14. By the end of this lecture, you should: −be able to describe how the volume of a gas varies with pressure, temperature and the number of molecules present −be able to use the ideal gas equation and choose the appropriate value of R to use −be able to work out and use partial pressures for gas mixtures −be able to complete the worksheet (if you haven’t already done so…) 14 Learning Outcomes:

15. 15 Questions to complete for next lecture: 1.One mole of gas occupies 22.414 L at STP (1 atm and 0 o C). What is the volume in m 3 occupied by one mole of gas at STP? (Hint : 1 m 3 = 1000 L) 2.Use the ideal gas law and your answer to Q1 to work out the value of the gas constant, R, and its units when volume and pressure are given in S.I. units (m 3 and Pa respectively). (Hint: from the data sheet, 1 atm = 1.013  10 5 Pa.)

16. 16 Questions to complete for next lecture: 3.A 2.0 L pressure cooker is sealed at atmospheric pressure and 25 o C. What will the pressure in the cooker be at 120 o C? 4.A 12 L air cylinder is filled to a pressure of 200. atm in an air conditioned diving shop at 22 °C. What will be the pressure inside the tank once it has been left in the sun at 35 °C?