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Thermal Conduction … Ideal Gas Law… Kinetic Molecular Theory… Thermodynamics…

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Presentation on theme: "Thermal Conduction … Ideal Gas Law… Kinetic Molecular Theory… Thermodynamics…"— Presentation transcript:

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2 Thermal Conduction … Ideal Gas Law… Kinetic Molecular Theory… Thermodynamics…

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5 Heat As Energy Transfer
(c) Specific heat of the material (Q) Thermal energy added (m) Mass of the object (ΔT) Change in temperature m

6 Calorimetry Conservation of thermal energy: Final Temperature: m1 m2
TL Final Temperature:

7 Latent Heat Energy is required for a material to change phase, even though its temperature is not changing. (m) Mass of the object (L) Latent heat of the fusion or vaporization (Q) Thermal energy added

8 TL A TH Dx Thermal conductivity Surface constant area Heat transferred
Change in temperature Time for transfer Thickness

9 Hint ~ ratios

10 kB = Boltzmann’s Constant = 1.38 x 10-23 J/K = R/NA
The Ideal Gas Law PV=nRT P = Pressure (Pa) V = Volume (m3) n = # of moles (mol) R = 8.31 J/mol K = (L.atm)/(mol.K) If N = # of particles T = Temp. (K) B PV=NkBT PV = nRT = NkBT kB = Boltzmann’s Constant = 1.38 x J/K = R/NA

11 Consider the way pressure changes with temperature in a gas:
Gay-Lussac’s Law P α T Pressure Temperature Gas 1 Gas 2 Pressure must always be greater than zero So temperature can never go below this point: T = ° C = Absolute Zero

12 Fluids?

13 PV=nRT PV=nRT Ideal Gas at Constant Temperature (Isothermal)
PV=constant Boyle’s Law constant PiVi= PfVf or Ideal Gas at Constant Pressure Charles’ Law PV=nRT constant or

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15 What about the speeds of the molecules in an ideal gas……
This molecule has an average speed This one has a different average speed What about the Kinetic Energy of a gas molecule? m v KE = ½mv2 What about the average Kinetic Energy of a group of gas molecules? KE = ½mvrms2

16 Kinetic Theory B Average KE of an ideal gas molecule
Temperature is a measure of the average molecular kinetic energy (k = Boltzmann’s constant = 1.38 x J/K) Kelvin temperature So now you can relate the mass, speed, and temperature of molecules in an ideal gas: B Temperature is simply a measure of the average Kinetic Energy of the molecules of a substance. A single molecule has no temperature since temperature is an averaging effect.

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18 Maxwell Distribution Curves

19 Why?

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21 U W Q Walls can be either: Diathermal - heat can flow through
Surroundings System U W Q Walls can be either: Diathermal - heat can flow through or Adiabatic – heat can’t flow through Car engine Q W Rub hands together W Q

22 The internal energy of an ideal gas depends only on its temperature
The internal energy of an ideal gas depends only on its temperature. Therefore, the change in internal energy of a fixed amount of ideal gas (n moles) is given by: Units: Joules (J) R = 8.31 J/mol K

23 1st Law of Thermodynamics
The total increase in the internal energy of a system is equal to the ‘sum’ of the work done on the system or by the system and the heat added to or removed from the system. (Conservation of Energy!) +ve Q = system GAINS heat -ve Q = system LOSES heat

24 Pressure, Volume ,Temperature & Work
In Thermodynamics we talk about the state of the system: Pressure, Volume ,Temperature & Work Work Done on a Gas F W = F Dx DV = A Dx W= PA Dx W = -P DV Dx Area +ve W=work done ON system COMPRESSION OF GAS DV -ve W=work done BY system EXPANSION OF GAS

25 Pressure - Volume Graph
T4 T3 Isotherms (lines of constant temperature) T2 P T1 Area under curve represents work Pressure Internal energy is proportional to temperature Volume T4 >T1 V

26 Isobaric (Constant Pressure)
Expansion W = -PDV Po W = -ve DU increases T4 T3 T2 1st Law: DU = Q + W T1 DV V ΔV = +ve T4 >T1

27 Isovolumetric or Isochoric (Constant Volume)
P W = 0 Decrease in Pressure Po Pf DU decreases 1st Law: Q = DU T3 T2 T1 DU = Q + W V

28 Isothermal (Constant Temperature)
Expansion P Po so W is -ve (ie gas is expanding) T3 Pf DU = 0 T2 1st Law: DU = Q+W 0 = Q+W Q = -W = (-)(-) = +ve! i.e. the work done is provided by Q added T1 Vo Vf V

29 Adiabatic (No Heat Exchange)
P Po Pf Vo Vf W = DU Compression W = +ve DU = increases T2 1st Law: Q = 0 DU = W T1 V

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33 2nd Law of Thermodynamics
Higher Temperature Lower Temperature Heat Heat will not flow spontaneously from a lower temperature object to a higher temperature object.

34 Heat Engine – any device that uses heat to do work
QH (input heat) Conservation of Energy: QH = W + QC W (Work) QC (rejected heat)

35 2nd Law of Thermodynamics (Entropy form)
Reversible Process – when the system and its surroundings can return to the exact states they were in before the process. P1 , V1 , T1 P2 , V2 , T2 P1 , V1 , T1 For a Reversible Process S = 0 For an Irreversible Process S > 0 2nd Law of Thermodynamics (Entropy form) Entropy (S) a measure of disorder

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