Published on: **Mar 3, 2016**

Source: www.slideshare.net

- 1. Kinetic Theory of Gases
- 2. Ideal GasThe number of molecules is largeThe average separation between molecules is largeMolecules moves randomlyMolecules obeys Newton’s LawMolecules collide elastically with each other and with the wallConsists of identical molecules
- 3. The Ideal Gas Law PV = nRT in Kn: the number of moles in the ideal gas N total number n= NA of molecules Avogadro’s number: the number of atoms, molecules, etc, in a mole of a substance: NA=6.02 x 1023/mol.R: the Gas Constant: R = 8.31 J/mol · K
- 4. Pressure and Temperature Pressure: Results from collisions of molecules on the surface Force F Pressure: P= A Area dp Force: F= Rate of momentum dt given to the surfaceMomentum: momentum given by each collision times the number of collisions in time dt
- 5. Only molecules moving toward the surface hitthe surface. Assuming the surface is normal tothe x axis, half the molecules of speed vx movetoward the surface.Only those close enough to the surface hit itin time dt, those within the distance vxdtThe number of collisions hitting an area A intime dt is 1 N ⋅ A ⋅ vx ⋅ dt 2V Average densityThe momentum given by each collision tothe surface 2mvx
- 6. Momentum in time dt: 1 N dp = (2mv x )⋅ ⋅ ⋅ A ⋅v x dt 2 VForce: dp 1 N F= = (2mvx )⋅ ⋅ ⋅ A⋅ v x dt 2 V Pressure: F N 2 P = = mv x A VNot all molecules have the same v x ⇒ average v2 x N 2 P = mv x V
- 7. 2 vx 1 2 1 2 2 3 3 ( 2 = v = v x + v y + vz ) 2 1 2 1 2 vx = v = vrms 3 3 vrms is the root-mean-square speed 2 2 2 vx + vy + vz vrms = v 2 = 3 1N 2 2 N1 2Pressure: P = mv = mv 3V 3 V 2 Average Translational Kinetic Energy: 1 2 1 2 K = mv = mvrms 2 2
- 8. 2 NPressure: P = ⋅ ⋅K 3 V 2From PV = ⋅ N ⋅ K and PV = nRT 3 3 nRT 3Temperature: K= ⋅ = ⋅ k BT 2 N 2 R −23Boltzmann constant: k B = = 1.38 × 10 J/K NA
- 9. 1 2From PV = ⋅ N ⋅ mvrms 3 Nand PV = nRT = RT NA Avogadro’s number N = nN A 3RT vrms = M Molar mass M = mN A
- 10. Internal EnergyFor monatomic gas: the internal energy = sumof the kinetic energy of all molecules: 3 3 Eint = N ⋅ K = nN A ⋅ k BT = nRT 2 2 3 Eint = nRT ∝ T 2
- 11. Mean Free Path Molecules collide elastically with other molecules Mean Free Path λ: average distance between two consecutive collisions 1 λ= 2 2πd N / Vthe bigger the molecules the more molecules the more collisions the more collisions
- 12. Q = cm ⋅ ∆T Molar Specific Heat ∆Eint = Q − W 3 Eint = nRT 2Definition: For constant volume: Q = nCV ∆T For constant pressure: Q = nC p ∆TThe 1st Law of Thermodynamics: 3 ∆Eint = nR∆T = Q − W (Monatomic) 2
- 13. 3 nR∆T = Q − WConstant Volume 2 (Monatomic) Q = nCV ∆TW = ∫ PdV = 0 3 Eint = nRT3 2 nR∆T = nCV ∆T2 3CV = R 2Eint = nCV T
- 14. 3 nR∆T = Q − W 2 Constant Pressure (Monatomic) Q = nC p ∆TW = P∆V = nR∆T3 nR∆T = nC p ∆T − nR∆T2 5CV = Cp − R Cp = R 2 Cp 5γ = γ = CV 3
- 15. 1st Law dEint = dQ − dW Adiabatic Process Ideal Gas Law pV = nRT (Q=0) Eint = nCV TdEint = −dW = − pdV C p = CV + R = nCV dT Cp γ = pdV CVpdV + Vdp = nRdT = nR − nCV Divide by pV:dV dp C p − CV dV dV + = − = (1 − γ )V p CV V V
- 16. dV dp dV Ideal Gas Law + = (1 − γ ) pV = nRTV p Vdp dV +γ =0 p V γ γln p + ln V = ln( pV ) = const. γ pV = const. nRT γ( )V = const. V γ −1 TV = const.
- 17. Equipartition of EnergyThe internal energy of non-monatomicmolecules includes also vibrational androtational energies besides thetranslational energy.Each degree of freedom has associated with 1it an energy of k BT per molecules. 2
- 18. Eint = nCV T Monatomic Gases3 translational degrees of freedom: 3 3 Eint = kBT ⋅nN A = nRT 2 2 1 dEint 3 CV = ⋅ = R n dT 2
- 19. Eint = nCV T Diatomic Gases3 translational degrees of freedom2 rotational degrees of freedom2 vibrational degrees of freedomHOWEVER, different DOFs require differenttemperatures to excite. At room temperature,only the first two kinds are excited: 5 5 Eint = nRT CV = R 2 2