Chemistry

1. Chemistry

2. States of Matter - 3

3. Order of diffusion would be CO2 > SO2 > SO3 > PCl3 Session Opener The rates of diffusion of SO2, CO2, PCl3 and SO3 are in the following order Solution: Rate of diffusion Hence, answer is (d)

4. The root mean square velocity of an ideal gas at constant pressure varies with density (d) as (a) d2 (b) d \ Vrms Illustrative example - 1 Solution: Hence, answer is (d)

5. Illustrative example - 2 Calculate the temperature at which root mean square velocity of SO2 molecules is same as that of O2 molecules at 27° C. Solution: For O2 at 27° C, For SO2 at t° C,

6. Solution Since both these velocities are equal, • or 600 = 273 + t • or t = 600 – 273 = 327° C

7. Session Objectives

8. Session Objectives • concept of real gas • Explanation for deviation from idealbehaviour. • van der Waals’ equation for real gases • Liquefaction of gases and critical phenomena • The liquid state and its characteristics:Compressibility, diffusion, evaporation.

9. These gases are real gases Deviation: Ideal behaviour Gases behave ideally at low pressures and high temperatures. • For an ideal gas, PV = ZnRT and Z = 1, Z is compressibility factor, measures extent of ideal nature

10. Deviation: Ideal behaviour

11. Reason for deviation • Near the liquefaction pointthere is greater deviation from ideal behaviour. 1. The volume of the gas molecule is negligible as compared to the total volume of the gas. 2. The forces of attraction between gas molecules are negligible.

12. van der Waals’ equation

13. Molecule equally attracted in all directions Molecule strikes wall with lesser velocity van der Waals’ equation

14. van der Waals’ equation p = Correction of pressure Intermolecular forces of attraction proportional to (i) the number of molecules per unit volume in the bulk of the gas. (ii) the number of molecules colliding per unit volume. a is van der Waals’ constant

15. Excluded volume for a pair of molecules = Excluded volume per molecule = van der Waals’ equation Vm = Actual volume of a gas molecule. Excluded volume for one mole b = 4 × Vm × NA. observed volume = (V – nb) (For n mole of gas) Van der waals’ equation for real gas

16. Questions

17. Illustrative example - 3 • Calculate the pressure exerted by 5 mol of CO2 in one litre vessel at 47°C using van der Waals equation.Also report the pressure of gas if it behaves ideally in nature. Given that a = 3.592 atm litre2 mol–2, b = 0.0427 litre mol–1.

18. Solution • Given n = 5, v = 1 litre, T = 320 K • a = 3.592, b = 0.0427 • Using van der Waals equation • P = 77.218 atm • Also if gas behaves ideally, then • PV = nRT • P × 1 = 5 × .0821 × 320 • P = 131.36 atm

19. Illustrative example - 4 Using van der Waals equation, calculate the constant ‘a’ when two mol of a gas confined in a four litre flask exerts a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 litre/mol. Solution: van der Waals equation a = 6.48 atm litre2 mol–2

20. Ask yourself What is the significance of a and b? Larger value of ‘a’ indicatehigher intermolecular forcesof attraction,i.e. liquefaction of gases become easier. While value of ‘b’ indicates theeffective size of the gas molecules.

21. At very low pressure, volume will be too high. and b can be neglected. Application For extremely low pressure Hence, PV = RT (Ideal gas equation)

22. Hence negligible value of and b. Application For very high temperature The volume will be extremely high \ PV = RT

23. PV = RT + Pb – Small size and less molecular mass makesnegligible i.e. Z = Application For hydrogen gas PV = RT + Pb

24. Critical states Critical temperature Critical pressure VC = 3b Critical volume Boyle temperature(TB) At TB Z=1 I.e., gases behave ideally Every real gas has a characteristic value of TB.

25. Ask yourself What is triple point? Triple point is the point at which solid, liquid and vapour are in equilibrium with eachother.

26. Liquid State – Characteristic Properties • The liquid state of matter hasintermediate properties. • Less orderly thansolid state but more orderly than thegaseous state.

27. Liquid State Liquid state is explained on the basisof kinetic theory model as follows: • Liquids are made up of molecules. • Molecules in a liquid are quite close to each other. • Force of attraction between the molecules in a liquid is quite large. • The molecules in a liquid are in a state of constant random motion. • The average kinetic energy of the molecules in a liquid is to their absolutetemperature.

28. Properties of Liquid Diffusion • There is diffusion in liquids but it is slower than in gases. • Diffusion involves movement of molecules from higher concentration to lower concentration.. Compressibility Liquids are relatively less compressible than gases.

29. Properties of Liquid Evaporation • The process of conversion of a liquid into its vapours at room temperature is known as evaporation. • Evaporation can be easily explained on the basis of kinetic theory. • At a given temperature, the average kinetic energy of the liquid molecules is constant but all the molecules do not have the same kinetic energy.

30. Properties of Liquid • Consequently, a certain fraction ofmolecules will have kinetic energies large enough to overcome the attractive forces of their neighbours and to escape into space above the liquid surface. • Evaporation causes cooling. • Rate of evaporation is influenced by the surface area of the liquid, the temperature and the strength ofintermolecular forces of attraction.

31. Class exercise

32. When an ideal gas undergoes unrestricted expansion, no cooling occurs because the molecules • exert no attractive forces on each other • (b) do not work • (c) collide without loss of energy • (d) are above the inversion temperature Class exercise - 1

33. Solution When a gas undergoes unrestricted expansion, there will be no intermolecular forces of attraction. Hence, no cooling is possible. Hence, the answer is (a)

34. The speed possessed by most of the gas molecules is (a) most probable speed (b) average speed (c) root mean square speed (d) None of these Class exercise - 2

35. Solution Most probable velocity is the velocity possessed by most of the gas molecules. Hence, the answer is (a)

36. Class exercise - 3 • The value of critical coefficient is • 2.66 (b) 3.55 • (c) 1.55 (d) 4.33 Solution: Critical coefficient Hence, the answer is (a)

37. Class exercise - 4 The value of van der Waals’ constant ‘a’ for the gases O2, N2, NH3, CH4 are 1.36, 1.39, 4.17 and 2.25 litre2 atm mol–2. The gas which can be most easily liquefied is (a) O2 (b) N2 (c) NH3 (d) CH4 Solution: The gas with the highest value of ‘a’ is having highest intermolecular forces of attraction. Hence, it can be liquefied most easily. Hence, the answer is (c).

38. Class exercise - 5 At the same temperature and pressure, which of the following gases will have the highest rate of diffusion? (a) Hydrogen (b) Oxygen (c) Methane (d) All will have the same rate of diffusion Solution: Hence, the answer is (a).

39. The ratio between the rms velocityof H2 at 50 K to that of O2 at 800 K is (a) 4 (b) 2 (c) 1 (d) Class exercise - 6

40. Solution Hence, the answer is (c).

41. Class exercise - 7 One litre of one mole of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors are 1.072 and 1.375 respectively at the initial and final steps. Find the volume. Solution: • We know that P1V1 = Z1nRT1 ... (i) • P2V2 = Z2nRT2 ... (ii)

42. Solution Dividing (ii) by (i) = 370.15 ml Ans. 370.15

43. Class exercise - 8 Three moles of CO2 (a van der Waals’ gas) occupy 10 L at 15 atm. Find out the temperature. (Given a = 3.59 L2 atm mol–2, b = 0.043 L mol–1)

44. Solution According to van der Waals’ gas equation, • T = 614.1 K Ans. T = 614.1 K

45. Thank you