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- 1. Physics DF025 Chapter 14 CHAPTER 14: Kinetic theory of gases (3 Hours)
- 2. Physics DF025 Chapter 14 Learning Outcome: 14.1 Ideal gas equations (1 hour) At the end of this chapter, students should be able to: a. Sketch and interpret b. Use the ideal gas equation: – P-V graph at constant temperature PV nRT – V-T graph at constant pressure – P-T graph at constant volume of an ideal gas.
- 3. Physics DF025 Chapter 1414.1 Ideal gas equation • The related equations to the Boyle’s law are14.1.1 Boyle’s law• states : “The pressure PV constant of a fixed mass of gas at constant OR temperature is inversely proportional P1V1 P2V2 to its volume.” OR where P : initial pressure 1 1 P2 : final pressure P if T constant V1 : initial volume V V2 : final volume
- 4. Physics DF025 Chapter 14• Graphs of the Boyle’s law. a. P The pressure of a fixed mass of gass at constant temperature is inversely T2 propertional to its volume. T1 0 V b. P T2 T1 1 0 V
- 5. Physics DF025 Chapter 1414.1.2 Charles’s law The related equations to the Charles’ law are• states : “The volume V of a fixed mass of constant gas at constant T pressure is directly OR proportional to its V1 V2 absolute temperature.” T1 T2 OR where T1 : initial absolute temperatu re T2 : final absolute temperatu re V T If P constant V1 : initial volume V2 : final volume
- 6. Physics DF025 Chapter 14 Graphs of the Charles’s law. b. V a V . 273.15 0 T( C) 0 T(K)
- 7. Physics DF025 Chapter 1414.1.3 Gay-lussac’s The related equations to the Gay- lussac’s law are (pressure) law P states : “The pressure constant of a fixed mass of gas T OR at constant volume is directly P1 P2 proportional to its T1 T2 absolute where temperature.” T1 : initial absolute temperatu re OR T2 : final absolute temperatu re P : initial pressure 1 P T If V constant P2 : final pressure
- 8. Physics DF025 Chapter 14 Graphs of the Gay-lussac’s (pressure) law a. P b. P 273.15 0 T( C) 0 T(K)
- 9. Physics DF025 Chapter 14 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 14.1. P1 P2 P2 2nd stage V1 1st stage V V2 T1 T1 T2 Figure 14.1
- 10. Physics DF025 Chapter 14 – In 1st stage, temperature is kept at T1 , – Equating eqs. (1) and (2), thus Using Boyle’s law : PV 2 PV1 1 PV1 1 P2V2 OR PV1 T1 T2 V 1 (1) P2 – In 2nd stage, pressure is kept Initial Final constant at P2 , Using Charles’s law : PV constant (3) V V2 T T1 T2 V2T1 V (2) T2
- 11. Physics DF025 Chapter 14 • 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), 3 PVm 101.3 10 0.0224 R T 1 273.15 1 R 8.31 J K mol where R is called molar gas constant and its value is the same for all gases. – Thus PVm R T PVm RT where Vm : volume of 1 mole gas • For n mole of an ideal gas, the equation of state is written as PV nRT – where n : the number of mole gas
- 12. Physics DF025 Chapter 14 m n – If the Boltzmann constant, k is defined as M where R 23 1 m : mass of a gas k 1.38 10 JK M : molar mass of a gas NA OR then the equation of state N becomes n NA PV NkT where N : number of molecules N A : Avogadro s constant 6.02 1023 mol 1
- 13. Physics DF025 Chapter 14Example 1 : Solution : VA 3VB ; mA m; T0A 300 K;T0B 500 KThe volume of vessel A is threetimes of the volume vessel B. The Since the vessels A and B arevessels are filled with an ideal gas connected by a narrow tube thus the pressure for both vessels is same,and are at a steady state. The finally i.e.temperature of vessel A and vessel PA PB PB are at 300 K and 500 K The system is in the steady state, thusrespectively as shown in Figure14.2. T0A TA 300 K;T0B TB 500 K By applying the equation of state for an A B ideal gas, (300 K) (500 K) m Figure 14.2 PV nRT and n m MIf the mass of the gas in the vessel PV RTA is m, obtain the mass of the gas Min the vessel B in terms of m.
- 14. Physics DF025 Chapter 14 Vessel B :Therefore, VA 3VB ; mA m; T0A 300 K;T0B 500 KVessel A : mA m PVB 100 R (1) PAVA RTA M M mB PVB 500 R (2) M m P 3VB R 300 By equating the eqs. (1) and (2) hence M m mB 100 R 500 R M M m mB 5
- 15. Physics DF025 Chapter 14 Example 2 : connecting tap B A Figure 14.3 Refer to Figure 14.3. 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.
- 16. Physics DF025 Chapter 14 Solution : V0A V1A 3.00 m3 ; T0A T1A 250 K; P0A 5.00 104 Pa;V0B V1B 7.20 m3 ; T0B T1B 400 K; P0B 2.00 104 Pa After the connecting tap has been opened and the system reached equilibrium, thus P1A P1B P By using the equation of state for ideal gas, P0V0 PV1 1 T0 T1 P0AV0A P0BV0B V1A V1B P T0A T0B T1A T1B 5.00 104 3.00 2.00 104 7.20 3.00 7.20 P 250 400 250 400 4 P 3.20 10 Pa
- 17. Physics DF025 Chapter 14 Exercise 14.1 : Given R = 8.31 J mol 1 K 1 and NA = 6.0 1023 mol 1 1. A gas has a volume of 60.0 cm3 at 20 C and 900 mmHg. What would its volume be at STP? (Given the atmospheric pressure = 101.3 kPa and the density of mercury = 13600 kg m 3) ANS. : 66.2 cm3 2. 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 3. 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
- 18. Physics DF025 Chapter 14Learning Outcome: and pressure ,14.2 Kinetic theory of gases (1 hour) 1 P v2 3At the end of this chapter, students should be able in related problems. to: c. Explain and use roota. State the assumptions mean square (rms) of kinetic theory of speed, gases.b. Apply the equations of 2 kT ideal gas, v 3 m 1 of gas molecules. PV Nm v 2 3
- 19. Physics DF025 Chapter 14 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.
- 20. Physics DF025 Chapter 1414.2.2 Force exerted by an ideal • Let each molecule of the gas have the gas mass m and velocity v. • The velocity, v of each molecule can be• Consider an ideal gas of N resolved into their components i.e. vx, vy molecules are contained in a and vz. cubical container of side d as • Consider, initially a single molecule shown in Figure 14.4. 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 14.5. Figure Wa Wall B ll A 14.5 Wa ll B Wall A Wa ll A Wa Figure 14.4 ll B
- 21. Physics DF025 Chapter 14Therefore the change in the If Fx1 is the magnitude of thelinear momentum of the average force exerted by amolecule is given by molecule on the wall in the time t, thus by applying Newton’s Px mvx mvx second law of motion gives Px 2mvx Px 2mvx m 2 Fx1 Fx1 vxThe molecule has to travel a t 2d ddistance 2d (from A to B and vxback to A) before its nextcollision with wall A. The time For N molecules of the ideal gas,taken for this movement is m 2 m 2 m 2 2d Fx vx1 vx 2 ....... vxN d d d t vx Fx m 2 2 vx1 vx 2 ....... vxN 2 d
- 22. Physics DF025 Chapter 14 where vx1 is the x Thus, the x component for the component of velocity of total force exerted on the wall of molecule 1, vx2 is the x the cubical container is component of velocity of molecule 2 and so on. m 2 Fx N vx• The mean (average ) value d of the square of the velocity in the x direction for N • The magnitude of the velocity molecules is v is given by 2 2 2 2 2 2 vx1 vx 2 ....... vxN 2 vx 2 v vx vy vz N 2 2 2 v x1 vx 2 ....... vxN then 2 2 2 2 v2 vx vy vz N vx
- 23. Physics DF025 Chapter 14 • Since the velocities of the molecules in the ideal gas are completely random, there is no preference to one direction or another. Hence 2 2 2 v x v y v z 2 2 v 3 vx 2 v2 vx 3 • The total force exerted on the wall in all direction, F is given by F m N vx 2 m v2 F N d d 3 N m v2 F 3 d where v 2 : mean square velocity of the molecule
- 24. Physics DF025 Chapter 14 14.2.3 Pressure of an ideal gas • From the definition of pressure, F N m v2 P where A d2 and F A 3 d 1 Nm v 2 P 3 and d3 V 3 d 1 Nm 2 P v (14.1) 3 V 1 2 PV Nm v (14.2) 3 where Nm : mass of an ideal gas in the container
- 25. Physics DF025 Chapter 14• Since the density of the 14.2.4 Root mean square gas, is given by velocity ( vrms) Nm • is defined as vrms v2 V hence the equation • From the equation of state in terms (15.1) can be written as of Boltzmann constant, k : 1 2 PV NkT (14.4) P v (14.3) 3 • By equating the eqs. (15.4) and (14.2), thuswhere P : pressure by the gas 1 NkT Nm v 2 : density of the gas 3 v 2 : mean square velocity 2 3kT v of the gas molecules m
- 26. Physics DF025 Chapter 14• Therefore • Since 1 P v2 3 3kT 3RT vrms OR vrms m M 2 3P thus vwhere therefore the equation of rootvrms : root mean square velocity (speed) mean square velocity of the gas molecules also can be m : mass of a molecule gas written as M : molar mass of a gas 3P vrms T : absolute temperatu re
- 27. Physics DF025 Chapter 14Example 3 : Solution : a. The mean speed of the Eight gas molecules chosen molecules is given by at random are found to have speeds of 1,1,2,2,2,3,4 and N 8 5 m s 1. Determine N a. the mean speed of the vi i 1 molecules, v b. the mean square speed of N the molecules, 1 1 2 2 2 3 4 5 v c. the root mean square 8 speed of the molecules. 1 v 2.5 m s
- 28. Physics DF025 Chapter 14Solution : N 8b. The mean square speed of c. The root mean square speed the molecules is given by of the molecules is N 2 vi v 2 i 1 vrms v2 N v 2 1 2 1 2 2 2 2 2 2 2 3 2 4 2 5 2 vrms 8 8 1 v 2 8m s 2 2 vrms 2.83 m s
- 29. Physics DF025 Chapter 14 Example 4 : (Given R = 8.31 J mol 1 K 1, k = 1.38 10 23 J K 1, molar mass of oxygen, M = 32 A cylinder of volume 0.08 m3 g mol 1, NA = 6.02 1023 mol 1) contains oxygen gas at a Solution : temperature of 280 K and pressure of 90 kPa. V 0.08 m3 ; T 280 K; Determine P 90 103 Pa a. the mass of oxygen in the cylinder, a. By using the equation of b. the number of oxygen state, thus molecules in the cylinder, m c. the root mean square PV nRT and n M speed of the oxygen m molecules in the cylinder. PV RT M
- 30. Physics DF025 Chapter 14 Solution : V 0.08 m3 ; T 280 K; P 90 103 Pa a. 3 m 90 10 0.08 8.31 280 0.032 m 9.90 10 2 kg b. The number of oxygen molecules in the cylinder is m N m n N NA M NA M 9.90 10 2 23 N 6.02 10 0.032 24 N 1.86 10 molecules c. The root mean square speed of the oxygen molecules is 3RT 3 8.31 280 1 vrms vrms 467 m s M 0.032
- 31. Physics DF025 Chapter 14 Exercise 15.2 : Given R = 8.31 J mol 1 K 1, Boltzmann constant, k = 1.38 10 23K1 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
- 32. Physics DF025 Chapter 14 Learning Outcome: 14.3 Molecular kinetic energy and internal energy At the end of this chapter, students should be able to:a. Explain and use c. Define degree of freedom translational kinetic energy of gases, d. State the number of 3 R 3 degree of freedom for K tr T kT monoatomic, diatomic, 2 NA 2 and polyatomic gas molecules.b. State the principle of equipartition of energy.
- 33. Physics DF025 Chapter 14 Learning Outcome: 14.3 Molecular kinetic energy and internal energy At the end of this chapter, students should be able to:e. Explain internal energy f. Explain and use of gas and relate the internal energy of an internal energy to the ideal gas number of degree of freedom,
- 34. Physics DF025 Chapter 14 14.3 Molecular kinetic energy and internal energy 14.3.1 Translational kinetic energy of molecule • From equation (14.1), thus 1 Nm 2 P v 3 V 2 N 1 P m v2 (14.5) 3 V 2 This equation shows that N increases V P increases ( ) When 1 m v2 increases 2
- 35. Physics DF025 Chapter 14 14.3.1 Translational kinetic energy of molecule This equation shows that N increases P increases ( ) When V 1 m v2 increases 2 Rearrange equation (14.5), thus 2 1 PV N m v2 and PV NkT 3 2
- 36. Physics DF025 Chapter 14 2 1 2 3 3 R NkT N m v K tr kT T 3 2 2 2 NA 1 2 3 where m v kT 2 2 K tr : average translati onal kinetic energy of a and molecule 1 T : absolute temperatu re m v2 K tr k : Boltzmann constant 2 R : molar gas constant N A : Avogadro constant
- 37. Physics DF025 Chapter 14 E NKtr• For N molecules of 3 an ideal gas in the E N kT cubical container, 2 the total average 3 (mean) translational E NkT 2 kinetic energy, E is given by OR 3 E nRT 2
- 38. Physics DF025 Chapter 14 14.3.2 Principle of equipartition of energy • States : “the mean (average) kinetic energy of every degrees of freedom of a molecule is 1 kT . Therefore 2 f Mean (average) kinetic K kT energy per molecule 2 OR f Mean (average) kinetic K RT energy per mole 2 where f : degrees of freedom T : absolute temperatu re
- 39. Physics DF025 Chapter 14Example 5 : Solution : NA vessel contains hydrogen gas of 7.50 1017 ; vrms 2.50 103 m s 1 ;7.50 1017 molecules per unit Vvolume and the root mean T 303.15 Ksquare speed of the molecules is2.50 km s 1at a temperature of 30 a. The average translational C. Determine kinetic energy of a moleculea. the average translational kinetic isenergy of a molecule for 3 hydrogen gas, K tr kTb. the pressure of hydrogen gas. 2(Given the molar mass of hydrogengas = 2 g mol 1, NA= 6.02 1023 3 23 K tr 1.38 10 303.15mol 1and k = 1.38 10 23 J K 1) 2 21 K tr 6.28 10 J
- 40. Physics DF025 Chapter 14 N Solution : 7.50 1017 ; vrms 2.50 103 m s 1 ; T 303.15 K V b. The pressure of hydrogen gas is given by 1 N 2 M 2 2 P m v where m and v vrms 3 V NA 1 N M 2 P vrms 3 V NA 1 17 0.002 3 2 P 7.50 10 23 2.50 10 3 6.02 10 3 P 5.19 10 Pa
- 41. Physics DF025 Chapter 1414.3.3 Degree of freedom ( f )• is defined as a number of independent ways in which y an atom or molecule can absorb or release or store  the energy. vy He  xMonatomic gas (e.g. He,Ne,Ar)  vx vz• The number of degrees of freedom z Figure 14.6 is 3 i.e. three direction of translational motion where contribute translational kinetic energy as shown in Figure 14.6.
- 42. Physics DF025 Chapter 14Diatomic gas (e.g. H2, O2, N2) y• The number of degrees of freedom is  Translational kinetic energy 3 vy H x Rotational kinetic energy 2  H  5 vz v x z Figure 14.7 yPolyatomic gas (e.g. H2O, CO2, NH3)• The number of degrees of freedom is  Translational kinetic energy 3 vy Rotational kinetic energy 3 O 6  x H  vx vz H z Figure 14.8
- 43. Physics DF025 Chapter 14 • Table 14.1 shows the degrees of freedom for various molecules. Degrees of Freedom ( f ) Average kinetic Molecule Example energy per Translational Rotational Total molecule,<K> 3Monatomic He 3 0 3 2 kT 5Diatomic H2 3 2 5 2 kT 6Polyatomic H 2O 3 3 6 2 kT 3kT (At temperature of 300 K) Table 14.1
- 44. Physics DF025 Chapter 14• Degrees of freedom depend on the absolute temperature of the gases. H H – For example : Diatomic gas (H2) vibration Figure 14.9 – Hydrogen gas have the vibrational kinetic energy (as shown in Figure 14.9) where contribute 2 degrees of freedom which when the temperature, correspond to the kinetic energy and the potential At 250 K f 3 energy associated with vibrations along the bond At 250 – 750 K f 5 between the atoms. At >750 K f 7
- 45. Physics DF025 Chapter 14 Solution :Example 6 : T1 293.15 K; E1 3.00 10 6 J;A vessel contains an ideal E2 9.00 10 6 J ; m2 2m1diatomic gas at temperature of By applying the equation of the total20 C. The total translational translational kinetic energy, thuskinetic energy of the gas 3molecules is 3.00 10 6 J. E NkT 2The mass of the gas is then m and k Rdoubled and the total where N NAtranslational kinetic energy M NAof the molecules becomes 3 m R9.00 10 6 J. Determine the E NA T 2 M NAnew temperature of the gas. 3 m E RT 2 M
- 46. Physics DF025 Chapter 14 Solution : T1293.15 K; E1 3.00 10 6 J; E2 9.00 10 6 J m22m1 3 m1 For temperature T1 : E RT (1) 1 1 2 M 3 m2 For temperature T2 : E2 RT2 2 M 3 2m1 E2 RT2 (2) E2 2T2 2 M (2) (1) : E1 T1 6 9.00 10 2T2 6 3.00 10 293.15 T2 440 K OR 167  C
- 47. Physics DF025 Chapter 14 Learning Outcome: At the end of this chapter, students should be able to:• Explain internal energy • Define molar specific heat of gas and relate the at constant pressure and internal energy to the volume. number of degree of • Use equations, freedom.• Explain and use C P CV R internal energy of an ideal gas and f CP U NkT 2 CV
- 48. Physics DF025 Chapter 1414.3.4 Internal energy of gas Thus for N molecules, and relate the internal energy to the number U N K of degree of freedom. f R• is defined as the sum of total U NkT and k 2 NA kinetic energy and total potential energy of the gas OR molecules. f• But in ideal gas, the U nRT intermolecular forces are 2 assumed to be negligible thus the potential energy of the where U : internal energy molecules can be neglected. of the gas
- 49. Physics DF025 Chapter 14 Table 14.2 shows the properties for 1 mole of an ideal gas. Table 14.2 Monatomic Diatomic Polyatomic Degrees of freedom, f 3 5 6 Average kinetic energy per 3 5 6 molecule, <K> kT kT kT 3kT 2 2 2 3 5 6 Internal energy, U RT RT RT 3RT 2 2 2
- 50. Physics DF025 Chapter 14 Exercise 14.3 : Given R = 8.31 J mol 1 K 1, Boltzmann constant, k = 1.38 10 23 K 1 1. One mole of oxygen has a mass of 32 g. Assuming oxygen behaves as an ideal gas, calculate a. the volume occupied by one mole of oxygen gas b. the density of oxygen gas c. the r.m.s. speed of its molecules d. the average translational kinetic energy of a molecule at 273 K and pressure of 1.01 105 Pa. ANS. : 2.25 10 2 m3; 1.42 kg m 3; 461 m s 1; 5.65 10 21 J
- 51. Physics DF025 Chapter 14 THE END… Next Chapter… CHAPTER 15 : Thermodynamics

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