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**The Kinetic Theory of Gases**

Chapter 19 The Kinetic Theory of Gases

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**Chapter-19 The Kinetic Theory of Gases**

Topics to be covered: Avogadro number Ideal gas law Internal energy of an ideal gas Distribution of speeds among the atoms in a gas Specific heat under constant volume Specific heat under constant pressure. Adiabatic expansion of an ideal gas

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**Kinetic theory of gases**

It relates the macroscopic property of gases (pressure - temperature - volume - internal energy) to the microscopic property - the motion of atoms or molecules (speed)

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**What is one mole? Avogadro’s Number NA = 6.02 X 1023**

One mole of any element contains Avogadro’s number of atoms of that element. One mole of iron contains 6.02 X 1023 iron atoms. One mole of water contains 6.02 X 1023 water molecules. From experiments: 12 g of carbon contains 6.02 X 1023 carbon atoms. Thus, 1 mole carbon = 12 g of carbon. 4 g of helium contains 6.02 X 1023 helium atoms. Thus, 1 mole helium = 4 g of helium. Avogadro’s Number NA = 6.02 X 1023 per mole = 6.02 X 1023 mol-1

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**Avogadro's Number Formula - number of moles**

n = N /NA n = number of moles N = number of molecules NA = Avogadro number M = Molar mass of a substance Msample = mass of a sample n = Msample /M

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Ideal Gas Law At low enough densities, all gases tend to obey the ideal gas law. Ideal gas law p V=n R T where R= 8.31 J/mol.K (ideal gas constant), and T temperature in Kelvin!!! p V= n R T = N k T; N is the number of molecules and K is Boltzman constant k = R/NA

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Ideal Gas Law

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**Isothermal process p =n R T/V = constant/V isotherm**

Isothermal expansion (Reverse is isothermal Compression) Quasi-static equilibrium (p,V,T are well defined) p =n R T/V = constant/V

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Checkpoint 1

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**Work done at constant temperature**

W = n R T Ln(Vf/Vi)

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**Work done at constant pressure**

isobaric process W = p (Vf-Vi)

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**Work done at constant volume**

isochoric process W = 0

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**Root Mean Square (RMS) speed vrms**

For 4 atoms having speeds v1, v2, v3 and v4 Vrms is a kind of average speed

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**Pressure, Temperature, RMS Speed**

The pressure p of the gas is related to root-mean -square speed vrms, volume V and temperature T of the gas p=(nM vrms 2)/3V but pV/n = RT Equation in the textbook Vrms = (3RT)/M

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**At room temperature (300K)**

Continue… Vrms = (3RT)/M R is the ideal gas constant T is temperature in Kelvin M is the molar mass (mass of one mole of the gas) At room temperature (300K) Gas Molar Mass (g/mol) Vrns (m/s) Hydrogen 2 1920 Nitrogen 28 517 Oxygen 32 483

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**Translational Kinetic Energy K**

Average translational kinetic energy of one molecule Kavg=(mv2/2)avg=m(vrms2)/2 Kavg=m(vrms2)/2=(m/2)[3RT/M] =(3/2)(m/M)RT=(3/2)(R/NA)T Kavg=(3/2)(R/NA)T=(3/2)kT

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Continue At a given temperature, all ideal gas molecules – no matter what their masses – have the same average translational kinetic energy.

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Checkpoint 2 A gas mixture consists of molecules of type 1, 2, and 3, with molecular masses m1>m2>m3. Rank the three types according to average kinetic energy, and rms speed, greatest first.

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**The Molar Specific Heat of an Ideal Gas**

Internal energy of an ideal gas Eint For a monatomic gas (which has individual atoms rather than molecules), the internal energy Eint is the sum of the translational kinetic energies of the atoms. Eint = N Kavg= N (3/2) k T = 3/2 (N k T) = 3/2 (n R T) Eint = 3/2 n R T The internal energy Eint of a confined ideal gas is a function of the gas temperature only, it does not depend on any other variable.

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**Change in internal energy**

Eint = 3/2 n R T, DEint = 3/2 n R DT

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**The Molar Specific Heat of an Ideal Gas**

Heat Q2 Heat Q1 Eventhough Ti and Tf is the same for both processes, but Q1 and Q2 are Different because heat depends on the path!

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**Heat gained or lost at constant volume**

For an ideal gas process at constant volume pi,Ti increases to pf,Tf and heat absorbed Q = n cv T and W=0. Then Eint = (3/2)n R T = Q = n cv T cv = 3R/2 Q = n cV T where cv is molar specific heat at constant volume

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**Eint = (3/2)n R T = Q - PDV = Q - n R DT **

Heat gained or lost at constant pressure For an ideal gas process at constant pressure Vi,Ti increases to Vf,Tf and heat absorbed Q = n cp T and W=PDV. Then Eint = (3/2)n R T = Q - PDV = Q - n R DT Q = (3/2nR+nR) DT = 5/2 n R DT cp = 5/2 R Q = n cp T where cp is molar specific heat at constant pressure

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**The Molar Specific Heats of a Monatomic Ideal Gas**

Cp = CV + R; specific heat ration = Cp/ CV For monatomic gas Cp= 5R/2, CV= 3R/2 and = Cp/ CV = 5/3 (specific heat ratio)

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Checkpoint 3

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**The Molar Specific Heat of an Ideal Gas**

monatomic diatomic polyatomic

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**(translational + rotational)**

Internal energy of monatomic, diatomic, and polyatomic gases (theoretical values) Degrees of freedom (translational + rotational) Cv Eint=n CVT Cp=Cv+R Monatomic gas (3/2) R = 12.5 (3/2) nRT 5/2 R 3 Diatomic gas (5/2) R = 20.8 (5/2) nRT 7/2 R 5 Polyatomic gas (6/2) R = 24.9 (6/2) nRT 8/2 R 6 Eint=n CV T

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**Cv of common gases in joules/mole/deg.C (at 15 C and 1 atm.)**

Symbol Cv g g (experiment) (theory) Helium He 12.5 1.666 Argon Ar Nitrogen N2 20.6 1.405 1.407 Oxygen O2 21.1 1.396 1.397 Carbon Dioxide CO2 28.2 1.302 1.298

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**Adiabatic Expansion for an Ideal Gas**

In adiabatic processes, no heat transferred to the system Q=0 Either system is well insulated, or process occurs so rapidly In this case DEint = - W

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Adiabatic Process P, V and T are related to the initial and final states with the following relations: PiVi= PfVf TiVi-1 = TfVf-1 Also T/( -1) V =constant then piTi(-1)/ = pfTf(-1)/

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**Free Expansion of an Ideal Gas**

An ideal gas expands in an adiabatic process such that no work is done on or by the gas and no change in the internal energy of the system i.e. Ti=Tf Also in this adiabatic process since ( pV=nRT), piVi=pfVf ( not PiVi= PfVf)

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