Chemistry

1. Chemistry

2. States of matter – Session 2

3. Session Objectives

4. Session Objectives • Graham’s law of diffusion/effusion • Postulates of kinetic theory of gases • Kinetic gas equation and kinetic energy of gases • Velocity of gas molecules • Maxwell-Boltzmann velocity distribution • Explanation of gas laws on the basis of kinetic theory of gases

5. Questions

6. Illustrative example 7 Which of the following compounds issteam volatile?

7. Solution The compound which forms intramolecular H-bonding has lower boiling point and hence, steam volatile. Hence, the answer is (b).

8. Illustrative example 1 A mixture CO and CO2 is found to have a density of 2gL–1 at 250C and 740 torr. Find the compositionof the mixture Solution:

9. Illustrative example 2 A 1500 ml flask contains 400 mg of O2and 60 mg H2 at 1000Cand they are allowed to react to form water vapour.What will be the partial pressure of the substances present at that temperature? Solution:

10. Animation for diffusion

11. Graham’s Law of Diffusion (i) For same volume of two gases (ii) For two gases with different volumes and same diffusion time

12. Effusion It happens under pressure through a Small aperture. Rate of effusion A=area of aperture Normally considered against vacuum

13. Question

14. Illustrative example 3 4:1 molar mixture of He and CH4 iseffusing through a pinhole at aconstant temperature.What is thecomposition of the mixture effusingout initially? Solution:

15. Solution Let x mole of CH4 is effusing in t sec. Then, 8x mole of He is effusing in same time Composition of He in the mixture = 89% Composition of CH4 in the mixture = 11%

16. Question

17. Illustrative example 4 A balloon filled with ethylene (C2H4) ispricked with a sharp point & quicklydropped in a tank full of hydrogen at the same pressure. After a while theballoon will have (a) Shrunk (b) Enlarged (c) Completely collapsed (d) Remain unchanged in size Solution Since, molar mass of H2 is much less than C2H4. H2 will diffuse into the balloon. Hence, answer is (b).

18. Kinetic molecular theory • All gases are made up of very large number of extremely small particles called molecules. • The actual volume of the molecules is negligible as compared to the total volume of the gas. • The distances of separation between the molecules are so large that the forces of attraction or repulsion between them are negligible.

19. Kinetic molecular theory • The molecules are in a constant state of motion in all directions. During their motion, they collide with one another and also the walls of the container. • The molecular collisions are perfectly elastic. • The pressure exerted by the gas is due to bombardment of the gas molecules on the walls of the container.

20. Kinetic Theory of Gases The kinetic gas equation, Where, m = mass of each gas molecule n’ = number of gas molecules c = velocity of gas molecule KE of one molecule =

21. (where n = number of moles,n’ = number of molecules) (for one mole of a gas, n =1 and n’ = NA ) Kinetic Theory of Gases = Average kinetic energy per mole

22. Ask yourself What will be the average kinetic energy of one molecule? Average kinetic energy of one molecule = k = Boltzmann constant = 1.38 × 10–16 ergs k–1 molecule–1

23. Questions

24. Average KE per molecule of the gas = Illustrative example 5 Calculate the average kinetic energyper molecule and total kinetic energyof 2 moles of an ideal gas at 25oC. Solution: = 6.17 × 10–21 J

25. Average KE per mole of the gas = RT = x 8.314 x 298 = 3.72 kJ/mole Total KE of 2 moles of the gas = RT x 2 = 7.44 kJ Solution

26. Molecular velocity Average velocity Most probable velocity (CMP)

27. Molecular velocity Root mean square velocity (CRMS) Interrelation of molecular velocities CAvg : CRMS = 0.9213 CMP : CRMS = 0.8165 CAvg: : CMP = 1.1286

28. Questions

29. The gas molecules have root mean square velocity of 1000 m/s. What is its average velocity? (a) 1000 m/s (b) 921.58 m/s (c) 546 m/s (d) 960 m/s Illustrative example 6 Solution: From (1) and (2)

30. Illustrative example 7 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,

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

32. Maxwell-Boltzmann velocity distribution

33. Characteristic features of Maxwell’s distribution curve • A very small fraction of molecules has very low or very high speeds. • The fraction of molecules possessing higher and higher speed goes on increasing till it reaches a peak. The fraction with still higher speed then goes on decreasing. • The peak represents maximum fraction of molecules at that speed. This speed, corresponding to the peak in the curve,is known as the most probable speed. • On increasing temperature, the value of the most probable speed also increases.

34. Explanation of Boyle’s law on the basis of kinetic theory According to kinetic gas equation

35. Explanation of Boyle’s law on the basis of kinetic theory

36. Explanation of charl’s law on the basis of kinetic theory As deduced from the kinetic gas equation, we have Hence, if P is kept constant, = constant which is Charles’ law.

37. Explanation of Dalton’s law on the basis of kinetic theory Let us consider only two gases. According to kinetic gas equation, Now, if only the first gas is enclosed in the vessel of volume V, the pressure exerted would be Again, if only the second gas is enclosed in the same vessel (so that V is constant), then the pressure exerted would be

38. Explanation of Dalton’s law on the basis of kinetic theory Lastly, if both the gases are enclosed together in the same vessel then since the gases do not react with each other, their molecules behave independent of each other. Hence, the total pressure exerted would be = P1 + P2 Similarly, if more than two gases are present, then it can be proved that P = P1 + P2 + P3 + ...

39. Explanation of Avogadro’s hypothesis on the basis of kinetic theory Let us assume equal volume of two gases at the same temperature and pressure, then from kinetic gas equation. Since average kinetic energy per molecule depends on temperature,

40. Dividing equation (i) by (ii), we have • This is Avogadro’s hypothesis. Explanation of Avogadro’s hypothesis on the basis of kinetic theory

41. Explanation of Graham’s law on the basis of kinetic theory From kinetic gas equation, at constant pressure This is in accordance with Graham’s law.

42. Illustrative example 8 One mole of N2 at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound. (Xe = 131, F = 19)

43. Solution

44. Class Exercise

45. Class exercise 1 20 dm3 of SO2 diffuses through a porous partition in 60 s. What volume of O2 will diffuse under similar conditions in 30 s? Solution: For diffusion, 1 for O2, 2 for SO2 Ans. 14.14 dm3

46. Class exercise 2 180 cm3 of an organic compound diffuses through a pinhole in vacuum in 15 minutes, while 120 cm3 of SO2 under identical condition diffuses in 20 minutes. What is the molecular weight of the organic compound? Solution:

47. Class exercise 3 The ratio of rates of diffusion of gases A and B is 1 : 4, if the ratio of their masses present in the mixture is 2 : 3, calculate the ratio of their mole fractions. Solution: Let WA and WB are the weights of two gases in the mixture WA : WB = 2 : 3

48. Solution Similarly, mole fraction of B,

49. Solution \ xA : xB = 1 : 24 Ans. 1 : 24

50. Class exercise 4 Calculate the root mean square velocity of (i) O2 if its density is 0.0081 g ml–1 at 1 atm. (ii) ethane at 27° C and 720 mm of Hg Solution: P = 1 × 76 × 13.6 × 981 dyne cm–2