1. ERT 108/3 PHYSICAL CHEMISTRY INTRODUCTION Prepared by: Pn. Hairul Nazirah Abdul Halim

2. Chemistry is the study of matter and the changes it undergoes

3. What is Physical Chemistry? is the study of macroscopic, atomic, subatomic, and particulate phenomena in chemical systems in terms of physical laws and concepts. It applies the principles, practices and concepts of physics such as motion, energy, force, time, thermodynamics, quantum chemistry, statistical chemistry and dynamic.

4. Subtopic The properties of gases a) The perfect gas b) Real Gases

5. THE PERFECT GAS The general form of an equation of state: If we know the values of T, V and n for a particular substance, then the pressure has a fixed value. Equation of state of a ‘perfect gas’:

6. a)PRESSURE Definition: A force divided by the area to which the force is applied. p = F/A SI unit of pressure: Pascal (Pa) = Nm -2 = kgm -1 s -2 Pressure is measured with a barometer

7. Units of Pressure Pascal (Pa)1Pa = 1 Nm -2 Bar1 bar = 10 5 Pa Atmosphere1 atm = 1.01325 bar Torr760 Torr = 1 atm mmHg1mmHg = 133.322 Pa Psi1 psi = 6.894757 kPa

8. When a region of high pressure is separated from a low pressure by a movable wall, the wall will be pushed into one region.

9. If the two pressure is identical, the wall will not move.

10. Example 1.1Calculating pressure Suppose Isaac Newton weighed 65 kg. Calculate the pressure he exerted on the ground when wearing boots with soles of total area 250 cm 2 in contact with the ground. Solution

11. b)Temperature Is the property that indicates the direction of the flow of energy through a thermally conducting, rigid wall. Two types of boundary: a) Diathermic – if a change of state is observed when two objects at different temp. are bought into contact. Example: metal container b) Adiabatic – if no change occurs even though the two objects have different temperatures.

12. Thermal equilibrium: no change of state occurs when two objects are in contact through a diathermic boundary. Zeroth Law Thermodynamics: If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A.

13. a)Boyle’s law b)Charles’s Law c)Avogadro’s Principle d)The perfect gas law e)Mixture of gases f)Mole fractions and partial pressures The Gas Laws

14. a)Boyle’s Law At constant temp., the pressure of a sample gas is inversely proportional to its volume. The volume it occupies is inversely proportional to its pressure:

15. Fig 1.5 The pressure-volume dependence of a fixed amount of perfect gas at different temperatures. Each curve is a hyperbola (pV = constant) and is called an isotherm.

16. Straight lines are obtained when the pressure is plotted against 1/V at constant temp.

17. b) Charles’s Law at constant pressure at constant volume

20. c) Avogadro’s principle At constant pressure and temp.

21. d) The perfect gas law perfect gas = ideal gas R = gas constant

23. Molar volume (V m ) of a perfect gas under Standard Ambient Temperature Pressure (SATP): Temp. at 298.15 K Pressure at 1 bar (10 5 Pa)

24. To calculate the change in conditions when constant amount of gas is subjected to different temp., pressures and volume: Combined gas equation:

25. Example 1.3Using combined gas equation In an industrial process, nitrogen is heated to 500K in a vessel of constant volume. If it enters the vessel at 100 atm and 300K, what pressure would it exert at the working temp., if it behaved as a perfect gas?

26. Solution

27. e) Mixture of Gases Dalton’s Law: the pressure exerted by a mixture of gases is the sum of the partial pressures of the gases. Where p = total pressure p A = partial pressure of perfect gas A p B = partial pressure of perfect gas B

28. Example 1.4Using Dalton’s Law A container of volume 10.0L holds 1.00 mol N 2 and 3.0 mol H 2 at 298 K. What is the total pressure in atmospheres if each component behaves as a perfect gas?

29. Solution

30. f) Mole fractions and partial pressures Where; X J = mole fraction; amount of J expressed as a fraction of the total amount of molecules, n in the sample.

31. ExampleMole fractions. A mixture of 1.0 mol N 2 and 3.0 mol H 2 consist of: Mol fraction of N 2 = 1.0 mol / (1.0 + 3.0 mol) = 0.25 Mol fraction of H 2 = 3.0 mol / (1.0 + 3.0 mol) = 0.75

32. Partial Pressure, p J of a gas J in a mixture Where p = total pressure The sum of partial pressures is equal to the total pressure:

33. Example 1.5 Calculating Partial Pressures The mass percentage composition of dry air at sea level is approximately N 2 = 75.5; O 2 = 23.2; Ar = 1.3. What is the partial pressure of each component when the total pressure is 1.00 atm? Solution Assume; total mass of the sample = 100g.

34. N2N2 O2O2 Ar Mole fraction0.7800.2100.0096 Partial pressure (atm)0.7800.2100.0096

35. Real Gases Real gases do not obey the perfect gas law exactly. 1.3Molecular Interactions Real gases show deviations from the perfect gas law because molecules interact with each other. Repulsive forces between molecules assist expansion. Attractive forces assist compression.

36. a)The compression factor Compression factor, Z is the ratio of its molar volume, V m to the molar volume of a perfect gas, at the same pressure and temp. For perfect gas, Z = 1 Deviation of Z from 1 is a measure of departure from perfect behavior.

37. Fig 1.13 The variation of compression factor, Z, with pressure for several gases at 0 0 C.

38. At very low pressures, all gases have Z≈1 and behave nearly perfectly. At high pressure, all the gases have Z > 1, signifying they have a larger molar volume than a perfect gas. Repulsive forces are dominant. At intermediate pressure, most gases have Z < 1. Attractive forces are reducing the molar volume.

39. b)Virial Coefficients Fig 1.14 Experimental isotherms of carbon dioxide at several temperatures.

40. b)Virial Coefficients Refer to Figure 1.14 At large molar volumes and high temp., the real- gas isotherms do not differ greatly from perfect gas isotherms. The small different suggest that the perfect gas law is in fact the first term in an expression of the form: Known as ‘Virial equation of state’

41. A more convenient expansion for many applications is: The term in parentheses can be identified with the compression factor, Z. Known as ‘Virial equation of state’ First virial coefficient = 1 Second virial coefficient = B Third virial coefficient = C

42. 1.4The van der Waals equation Eq. 1.25a The equation is often written in terms of the molar volume V m = V/n. Eq. 1.25b Constant a and b = van der Waals coefficients.

44. Example 1.6Using the van der Waals equation to estimate a molar volume Estimate the molar volume of CO 2 at 500K and 100 atm by treating it as a van der Waals gas. Solution 1. Arrange equation 1.25b to become:

46. Refer to Table 1.5, For CO 2, a = 3.610 L 2 atm mol -2 b = 4.29 x 10 -2 L mol -1

47. Thefore, on writing x = V m, the equation to solve is: The acceptable root is x = 0.366 Hence V m = 0.366 L mol -1.