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PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics

PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics. PRESSURE IDEAL GAS EQUATION OF STATE KINETIC THEORY MODEL THERMAL PROCESSES. PHYSICS: FUN EXCITING SIMPLE. ap06/p1/thermal/ptE_gases.ppt. Overview of Thermal Physics Module:

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PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics

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  1. PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics PRESSURE IDEAL GAS EQUATION OF STATE KINETIC THEORY MODEL THERMAL PROCESSES PHYSICS: FUN EXCITING SIMPLE ap06/p1/thermal/ptE_gases.ppt

  2. Overview of Thermal Physics Module: • Thermodynamic Systems:Work, Heat, Internal Energy0th, 1st and 2nd Law of Thermodynamics • Thermal Expansion • Heat Capacity, Latent Heat • Methods of Heat Transfer:Conduction, Convection, Radiation • Ideal Gases, Kinetic Theory Model • Second Law of ThermodynamicsEntropy and Disorder • Heat Engines, Refrigerators

  3. Kinetic-Molecular Model of an Ideal Gas Thermal Processes * Ideal gas, Equation of state (§18.1 p611) * Kinetic-molecular model of an ideal (results only – not the mathematical derivations) (§18.3 p619) * Heating a gas: heat capacities, molar heat capacity (§17.5 p582 §18.4 p626 §19.6 p658 §19.7 p659) * First Law of Thermodynamics: Internal Energy, Work, Heat, Paths between thermodynamic states (§19.1 p 723-725, §19.2 p725-728, § 19.3 p728-729, §19.4 p729-735) * Thermal Processes and pV diagrams: Isothermal, Isobaric Isochoric (constant volume gas thermometer), Adiabatic Cyclic (§19.5 p735-737, §19.8 p741-744, §17.3 p644-645 References: University Physics 12th ed Young & Freedman

  4. HEAT ENGINES & GASES

  5. Phases of matter Gas - very weak intermolecular forces, rapid random motion high temp low pressure Liquid- intermolecular forces bind closest neighbours low temp high pressure Solid- strong intermolecular forces

  6. Quantity of a gas number of particles N mass of particle m molar mass M (kg.mol-1)  mass of 1 mole of a substance number of moles n ( mol)  1 mole contains NA particles Avogadro's constant NA = 6.023x1023 mol-1 1 mole is the number of atoms in a 12 g sample of carbon-12 1 mole of tennis balls would fill a volume equal to 7 Moons The mass of a carbon-12 atom is defined to be exactly 12 u u  atomic mass units, 1 u = 1.66x10-27 kg (1 u)(NA) = (1.66x10-27)(6.023x1023) = 10-3 kg = 1 g mtot = N m If N = NA mtot = NA m = M M = NA m n = N / NA = mtot / M

  7. 1.00 kg of water vapour H2O M(H2O) = M(H2) + M(O) = (1 + 1 + 16) g = 18 g = 1810-3 kg n(H2O) = mtot / M(H2O) = 1 / 1810-3 = 55.6 mol N(H2O) = nNA = (55.6)(6.0231023) = 3.351025 m(H2O) = M / NA = (1810-3) / (6.0231023) kg = 2.9910-26 kg 1 amu = 1 u = 1.6610-27 kg m(H2O) = 18 u = (18)(1.6610-27) kg = 2.9910-26 kg

  8. Pressure P Is this pressure? What pressure is applied to the ground if a person stood on one heel?

  9. Patmosphere = 1.013105 Pa ~1032 molecules strike our skin every day with an avg speed ~ 1700 km.s-1

  10. Rough estimate of atmospheric pressure air ~ 1 kg.m-3g ~ 10 m.s-2h ~ 10 km = 104 m p = F / A = mg / A =  V g / A =  A h / A =  g h Patm ~ (1)(10)(104) Pa Patm ~ 105 Pa

  11. Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg ? … and all the king's horses … What force is required to separate the hemispheres? Is this force significant?

  12. Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg p = 1x105 Pa R = 0.30 m A = 4R2 F = p A F = (105)(4)(0.3)2 N F = 105 N

  13. Gauge and absolute pressures Pressure gauges measure the pressure above and below atmospheric (or barometric) pressure. P = P = 1 atm = 101.3 kPa = 1013 hPa = 1013 millibars = atm 0 760 torr = 760 mmHg Gauge pressure P 200 g Absolute pressure P 100 300 0 400 P = P + P g atm P = 200 kPa P = 100 kPa g atm P = 300 kPa

  14. Ideal Gases – equation of state (experimental law) p V = n R T = N k T R, Universal gas constant (same value for all gases) R = 8.314 J.mol-1.K-1 Boltzmann constant k = 1.38x10-23 J.K-1 k = R / NAR = kNA must be in kelvin (K)

  15. All gases contain the same number of molecules when they occupy the same volume under the same conditions of temperature and pressure (Avogadro 1776 - 1856) p V = n R Tn = N / NA= p V / R T Ideal gas, constant mass (fixed quantity of gas)

  16. Boyle's Law (constant temperature) p = constant / V Charles Law (constant pressure) V = constant T Gay-Lussac’s Law (constant volume) p = constant T

  17. Thermodynamic system (ideal gas) work internal energy U = Q – W = n CVDT p V = n R T p V = N k T k = R / NA mtot = n M N = n NA p V T U S heat Q = n C DT CV or Cp mtot N n Q = 0 pV  = constant T V-1 = constant

  18. Provided the temperature is not too high (< 3000 K), a diatomic molecule has 5 degrees of freedom

  19. Kinetic–Molecular model for an ideal gas (p619) Large number of molecules N with mass m randomly bouncing around in a closed container with Volume V. z Experimental Law Kinetic-Molecular Model (Theory) y x For the two equations to agree, we must have:

  20. Total kinetic energy for random translational motion of all molecules, Ktr is the average translational kinetic energy of a single molecule Average translational KE of a molecule For an ideal gas, temperature is a direct measure of the average kinetic energy of its molecules.

  21. At a given temperature T, all ideal gas molecules have the same average translational kinetic energy, no matter what the mass of the molecule energy stored in each degree of freedom = ½ k T Theorem of equipartition of energy (James Clerk Maxwell): The thermal energy kT is an important factor in the natural sciences. By knowing the temperature we have a direct measure of the energy available for initiating chemical reactions, physical and biological processes.

  22. Internal energy U of an ideal gas PE = 0   Degrees of freedom (T not too high) monatomic gas, f = 3 diatomic gas, f = 5, polyatomic gas, f = 6 Only translation possible at very low temp, T rotation begins, T  oscillatory motion starts

  23. Heating a gas Molar heat capacity

  24. Heating a gas at constant volume 1st Law Thermodynamics U = n CVDT Q Constant volume process V = 0  W = 0 All the heat Q goes into changing the internal energy U hence temperature T Larger f larger CV  smaller T for a given Q

  25. Heating a gas at constant pressure W 1st Law Thermodynamics Q Constant pressure process W = p V It requires a greater heat input to raise the temp of a gas a given amount at constant pressure c.f. constant volume Q = U + WW > 0

  26. Thermal processes T1 p1 V1 U1 S1 T2 p2 V2 U2 S2 W n N mtot Q Reversible processes

  27. Isothermal change T = 0 Boyle’s Law (1627 -1691) T1= T2 p1V1 = p2 V2 U= 0pV = n R T

  28. isotherm W is the area under an isothermal curve

  29. Isochoric (V = 0)W = 0 U = Q = n CVT isochor

  30. Isobaric (p = 0) W = pV Q = n Cp T U = Q – W = n CV T T2 > T1U > 0 W > 0 Q > 0 W < Q V1/T1=V2/T2 isobar

  31. Adiabatic (Q = 0) U = - W CV = (f / 2) RCp = CV + R = (f / 2 +1)R  = Cp / CV  = (f + 2) / f pV  = constant diatomic gas f = 5 T V-1 = constant = 7 / 5 = 1.4

  32. 1 to 2: Q = 0 T1 > T2W > 0 U < 0 W adiabat An adiabat steeper on a pV diagram than the nearby isotherms since  > 1

  33. Adiabatic processes can occur when the system is well insulated or a very rapid process occurs so that there is not enough time for a significant heat to be transferred eg rapid expansion of a gas; a series of compressions and expansions as a sound wave propagates through air. Atmospheric processes which lead to changes in atmospheric pressure often adiabatic: HIGH pressure cell, falling air is compressed and warmed.LOW pressure cell, rising air expands and cooled  condensation and rain.

  34. convergence divergence divergence convergence HIGH LOW - more uniform - less uniform conditions - inhibits cloud conditions - encourages cloud formation formation sunshine Atmospheric adiabatic processes Q = 0 U = - W T V-1 = constant Burma Cyclone 5 May 2008 +50 000 killed ?

  35. Cyclic Processes: U = 0 reversible cyclic process

  36. Problem E.1 Oxygen enclosed in a cylinder with a movable piston (assume the gas is ideal) is taken from an initial state A to another state B then to state C and back to state A. How many moles of oxygen are in the cylinder? Find the values of Q, W and U for the paths A to B; B to C; C to A and the complete cycle A to B to C to A and clearly indicate the sign + or – for each process. Does this cycle represent a heat engine?

  37. Thermodynamic system (ideal gas) work internal energy U = Q – W = n CVDT p V = n R T p V = N k T k = R / NA mtot = n M N = n NA p V T U S heat Q = n C DT CV or Cp mtot N n Q = 0 pV  = constant T V-1 = constant

  38. Solution Identify / Setup oxygen diatomic f = 5 CV = (f / 2) RCp = CV + R = (f / 2 +1) R CV = 5/2 RCp = 7/2 R R = 8.315 J.mol-1.K-1 CV = 20.8 J.mol-1.K-1 Cp = 29.1 J.mol-1.K-1 p V = n R T = N k T

  39. Execute At A

  40. 1 A to B is isobaric T1= TB – TA = (400 – 100) K = 300 K pA = pB = p1 = 40 kPa = 4.00104 Pa V1 = (0.080 – 0.020) m3 = 0.060 m3 W1 > 0 since gas expands W1 = p1 V1 = (4.0104)(0.06) = 2.4103 J U1 > 0 since the temperature increases U1 = n CV T1 = (1)(20.8)(300) J = 6.2103 J Q1 > 0 since U1 > 0 and U1 = Q1 – W1 > 0 Q1 > W1 > 0 Q1 = n Cp T1= (1)(29.1)(300) J = 8.7103 J Check: First law U1 = Q1 - W1 = (8.73103 – 2.4103) J = 6.3103 J Q1

  41. 2 B to C is isochoric T2= TC – TB = (800 – 400) K = 400 K V2 = 0 m3 W2 = 0 since no change in volume U2 > 0 since the temperature increases U2 = n CV T2 = (1)(20.8)(400) J = 8.3103 J Q2 = U2 since W2 = 0 Q2 = 8.3103 J Q2

  42. 3C to A T3= TA – TC = (100 – 800) K = -700 K pA = 40 kPa = 4.0104 PapC = 80 kPa = 8.0104 PapCA = 4.0104 Pa VA = 0.02 m3VC = 0.08 m3 V3= 0.06 m3 W3 < 0 since gas is compressed W3 = area under curve = area of rectangle + area of triangle W3 = - { (0.06)(4.0104) + (½)(0.06)(8.0104- 4.0104)} J = - 3.6103 J U3 < 0 since the temperature decreases U3 = n CV T3 = (1)(20.8)(-700) J = - 14.6103 J Q3 = U3 + W3 Q3 = (- 14.5103 - 3.6103) J = - 18.2103 J Q3

  43. Complete cycle U = 0 J Refrigerator cycle: |QH| = |QC| +|W| |W| TC TH |QC| |QH|

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