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CHAPTER 14: Kinetic Theory of Gases (3 Hours)

CHAPTER 14: Kinetic Theory of Gases (3 Hours). Learning Outcome:. 14.1 Ideal gas equation ( 1 hour). At the end of this chapter, students should be able to: Sketch P-V graph at constant temperature V-T graph at constant pressure P-T graph at constant volume of an ideal gas.

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CHAPTER 14: Kinetic Theory of Gases (3 Hours)

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  1. CHAPTER 14: Kinetic Theory of Gases(3 Hours)

  2. Learning Outcome: 14.1 Ideal gas equation (1 hour) At the end of this chapter, students should be able to: • Sketch • P-V graph at constant temperature • V-T graph at constant pressure • P-T graph at constant volume of an ideal gas. • Use the ideal gas equation:

  3. 14.1 Ideal Gas Equation 14.1.1 Boyle’s law • states : “The pressure of a fixed mass of gas at constant temperature is inversely proportionalto its volume.” if OR where

  4. PV P T2 T1 T2 T1 0 0 V P PV P T2 T1 T2 T1 V 0 0 a. b. T2 > T1 • Graphs of the Boyle’s law. d. c.

  5. 14.1.2 Charles’s law • states : “The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature.” If where

  6. V V 0 0 273.15 T(K) T(C) • Graphs of the Charles’s law. a. b.

  7. 14.1.3 Gay-lussac’s (pressure) law • states : “The pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature.” If where

  8. P P 0 273.15 0 T(C) T(K) • Graphs of the Gay-lussac’s (pressure) law. a. b.

  9. 14.1.4 Equation of state for an ideal gas • An ideal gas is defined as a perfect gas which obeys the three gas laws (Boyle’s, Charles’s and Gay-Lussac’s) exactly. • Consider an ideal gas in a container changes its pressure P, volume V and temperature T as shown in Figure 15.1. 2nd stage 1st stage Figure 14.1

  10. (3) (1) (2) • In 1st stage, temperature is kept at T1 , Using Boyle’s law : • In 2nd stage, pressure is kept constant at P2 , Using Charles’s law : • Equating eqs. (1) and (2), thus OR Final Initial

  11. Consider 1 mole of gas at standard temperature and pressure (S.T.P.), T= 273.15 K, P = 101.3 kPa and Vm = 0.0224 m3 • From equation (3), where R is called molar gas constant and its value is the same for all gases. • Thus • For n mole of an ideal gas, the equation of state is written as • where n : the number of mole gas where

  12. where OR • If the Boltzmann constant, k is defined as then the equation of state becomes where

  13. A B (300 K) (500 K) Example 14.1 : The volume of vessel A is three times of the volume vessel B. The vessels are filled with an ideal gas and are at a steady state. The temperature of vessel A and vessel B are at 300 K and 500 K respectively as shown in Figure 15.2. If the mass of the gas in the vessel A is m, obtain the mass of the gas in the vessel B in terms of m. Figure 14.2

  14. Solution :

  15. Solution :

  16. connecting tap B  A Example 14.2 : Refer to Figure 15.3. Initially A contains 3.00 m3 of an ideal gas at a temperature of 250 K and a pressure of 5.00  104 Pa, while B contains 7.20 m3 of the same gas at 400 K and 2.00  104 Pa. Calculate the pressure after the connecting tap has been opened and the system reached equilibrium, assuming that A is kept at 250 K and B is kept at 400 K. Figure 14.3

  17. Solution :

  18. Exercise 14.1 : Given R = 8.31 J mol1 K1 and NA = 6.0  1023 mol1 1. Estimate the number of molecules in a flask of volume 5.0  104 m3 which contains oxygen gas at a pressure of 2.0  105 Pa and temperature of 300 K. ANS. : 2.41  1022 molecules 2. A cylinder contains a hydrogen gas of volume 2.40  103 m3 at 17 C and 2.32  106 Pa. Calculate a. the number of molecules of hydrogen in the cylinder, b. the mass of the hydrogen, c. the density of hydrogen under these conditions. (Given the molar mass of hydrogen = 2 g mol1) ANS. : 1.39  1024 molecules; 4.62 g; 1.93 kg m3

  19. Learning Outcome: 14.2 Kinetic theory of gases (1 hour) At the end of this chapter, students should be able to: • State the assumptions of kinetic theory of gases. • Apply the equations of ideal gas, and pressure , in related problems. • Explain and use root mean square (rms) speed, ofgas molecules.

  20. 14.2 Kinetic theory of gases • The macroscopic behaviour of an ideal gas can be describe by using the equation of state but the microscopic behaviour only can be describe by kinetic theory of gases. 14.2.1 Assumption of kinetic theory of gases • All gases are made up of identical atoms or molecules. • All atoms or molecules move randomly and haphazardly. • The volume of the atoms or molecules is negligible when compared with the volume occupied by the gas. • The intermolecular forces are negligible except during collisions. • Inter-atomic or molecular collisions are elastic. • The duration of a collision is negligible compared with the time spent travelling between collisions. • Atoms and molecules move with constant speed between collisions. Gravity has no effect on molecular motion.

  21. WallB WallA 14.2.2 Force exerted by an ideal gas • Consider an ideal gas of N molecules are contained in a cubical container of side d as shown in Figure 15.4. • Let each molecule of the gas have the mass m and velocity v. • The velocity, v of each molecule can be resolved into their components i.e. vx, vy and vz. Figure 14.4

  22. WallA WallB WallA WallB • Consider, initially a single molecule moving with a velocity vx towards wall A and after colliding elastically , it moves in the opposite direction with a velocity vx as shown in Figure 15.5. • Therefore the change in the linear momentum of the molecule is given by Figure 14.5

  23. The molecule has to travel a distance 2d (from A to B and back to A) before its next collision with wall A. The time taken for this movement is • If Fx1 is the magnitude of the average force exerted by a molecule on the wall in the time t, thus by applying Newton’s second law of motion gives • For N molecules of the ideal gas,

  24. where vx1 is the x component of velocity of molecule 1, vx2 is the x component of velocity of molecule 2 and so on. • The mean (average ) value of the square of the velocity in the x direction for N molecules is • Thus, the x component for the total force exerted on the wall of the cubical container is • The magnitude of the velocity v is given by then

  25. Since the velocities of the molecules in the ideal gas are completely random, there is no preference to one direction or another. Hence • The total force exerted on the wall in all direction, F is given by where

  26. (14.1) (14.2) 14.2.3 Pressure of an ideal gas • From the definition of pressure, where and and where

  27. (14.3) • Since the density of the gas,  is given by • hence the equation (14.1) can be written as where

  28. (14.4) 14.2.4 Root mean square velocity ( vrms) • is defined as • From the equation of state in terms of Boltzmann constant, k : • By equating the eqs. (14.4) and (14.2), thus • Therefore OR

  29. where • Since therefore the equation of root mean square velocity of the gas molecules also can be written as thus

  30. Example 14.3 : Eight gas molecules chosen at random are found to have speeds of 1,1,2,2,2,3,4 and 5 m s1. Determine a. the mean speed of the molecules, b. the mean square speed of the molecules, c. the root mean square speed of the molecules. Solution : a.

  31. Solution : b.

  32. Example 14.4 : A cylinder of volume 0.08 m3 contains oxygen gas at a temperature of 280 K and pressure of 90 kPa. Determine a. the mass of oxygen in the cylinder, b. the number of oxygen molecules in the cylinder, c. the root mean square speed of the oxygen molecules in the cylinder. (Given R = 8.31 J mol1 K1, k = 1.38  1023 J K1, molar mass of oxygen, M = 32 g mol1, NA = 6.02  1023 mol1) Solution : a.

  33. Solution : .

  34. Exercise 14.2 : Given R = 8.31 J mol1 K1, Boltzmann constant, k = 1.381023 K1 1. In a period of 1.00 s, 5.00  1023 nitrogen molecules strike a wall with an area of 8.00 cm2. If the molecules move with a speed of 300 m s1 and strike the wall head-on in the elastic collisions, determine the pressure exerted on the wall. (The mass of one N2 molecule is 4.68  1026 kg) ANS. : 17.6 kPa 2. Initially, the r.m.s. speed of an atom of a monatomic ideal gas is 250 m s1. The pressure and volume of the gas are each doubled while the number of moles of the gas is kept constant. Calculate the final translational r.m.s. speed of the atoms. ANS. : 500 m s1 3. Given that the r.m.s. of a helium atom at a certain temperature is 1350 m s1, determine the r.m.s. speed of an oxygen (O2) molecule at this temperature. (The molar mass of O2 is 32.0 g mol1 and the molar mass of He is 4.00 g mol1) ANS. : 477 m s1

  35. Learning Outcome: 14.3 Molecular kinetic energy and internal energy At the end of this chapter, students should be able to: • Explain and use translational kinetic energy of gases, • State the principle of equipartition of energy. • Define degree of freedom. • State the number of degree of freedom for monoatomic, diatomic and polyatomic gas molecules. • Explain internal energy of gas and relate the internal energy to the number of degree of freedom. • Explain and use internal energy of an ideal gas

  36. (14.5) 14.3.1 Molecular kinetic energy • From equation (14.1), thus This equation shows that • Rearrange equation (14.5), thus P increases () When and

  37. and where

  38. For Nmolecules of an ideal gas in the cubical container, the total average (mean) translational kinetic energy, E is given by OR

  39. Principle of equipartition of energy • States : “the mean (average) kinetic energy of every degrees of freedom of a molecule is Therefore Mean (average) kinetic energy per molecule OR Mean (average) kinetic energy per mole where

  40. He Degree of freedom ( f ) • is defined as a number of independent ways in which an atom or molecule can absorb or release or store the energy. Monoatomic gas (e.g. He, Ne, Ar) • The number of degrees of freedom is 3 i.e. three direction of translational motion where contribute translational kinetic energy as shown in Figure 15.6. Figure 15.6

  41. H H O H H Translational kinetic energy Diatomic gas (e.g. H2, O2, N2) • The number of degrees of freedom is Polyatomic gas (e.g. H2O, CO2, NH3) • The number of degrees of freedom is Rotational kinetic energy Figure 14.7 Figure 14.8 Translational kinetic energy Rotational kinetic energy

  42. Table 14.1 shows the degrees of freedom for various molecules. Monatomic He Diatomic H2 Polyatomic H2O (At temperature of 300 K) Table 14.1

  43. H H vibration • Degrees of freedom depend on the absolute temperature of the gases. • For example : Diatomic gas (H2) • Hydrogen gas have the vibrational kinetic energy (as shown in Figure 15.9) where contribute 2 degrees of freedom which correspond to the kinetic energy and the potential energy associated with vibrations along the bond between the atoms. when the temperature, At 250 K At 250 – 750 K At >750 K Figure 15.9

  44. Example 14.5 : A vessel contains hydrogen gas of 2.20  1018 molecules per unit volume and the mean square speed of the molecules is 4.50 km s1 at a temperature of 50 C. Determine a. the average translational kinetic energy of a molecule for hydrogen gas, b. the pressure of hydrogen gas. (Given the molar mass of hydrogen gas = 2 g mol1, NA= 6.02  1023 mol1 and k = 1.38  1023 J K1) Solution :

  45. Solution :

  46. 14.3.2 Internal energy, U • is defined as the sum of total kinetic energy and total potential energy of the gas molecules. • But in ideal gas, the intermolecular forces are assumed to be negligible thus the potential energy of the molecules can be neglected. Thus for N molecules, • For N molecules of monoatomic gas , and OR OR where

  47. Example 14.7 : Neon is a monoatomic gas. Determine the internal energy of 1 kg of neon gas at temperature of 293 K. Molar mass of neon is 20 g. Solution :

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