Presentation on theme: "The Kinetic Theory of Gases"— Presentation transcript:
1The Kinetic Theory of Gases Chapter 19The Kinetic Theory of Gases
2Chapter-19 The Kinetic Theory of Gases Topics to be covered:Avogadro numberIdeal gas lawInternal energy of an ideal gasDistribution of speeds among the atoms in a gasSpecific heat under constant volumeSpecific heat under constant pressure.Adiabatic expansion of an ideal gas
3Kinetic theory of gases It relatesthe macroscopic property of gases(pressure - temperature - volume - internal energy)tothe microscopic property - the motion of atoms or molecules(speed)
4What 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
5Avogadro's Number Formula - number of moles n = N /NAn = number of molesN = number of moleculesNA = Avogadro numberM = Molar mass of a substanceMsample = mass of a samplen = Msample /M
6Ideal Gas LawAt low enough densities, all gases tend to obey the ideal gas law.Ideal gas lawp V=n R Twhere 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 moleculesand K is Boltzman constant k = R/NA
10Work done at constant temperature W = n R T Ln(Vf/Vi)
11Work done at constant pressure isobaric processW = p (Vf-Vi)
12Work done at constant volume isochoric processW = 0
13Root Mean Square (RMS) speed vrms For 4 atoms having speeds v1, v2, v3 and v4Vrms is a kind of average speed
14Pressure, Temperature, RMS Speed The pressure p of the gas is related to root-mean -square speed vrms, volume V and temperature T of the gasp=(nM vrms 2)/3Vbut pV/n = RTEquation in the textbookVrms = (3RT)/M
15At room temperature (300K) Continue…Vrms = (3RT)/MR is the ideal gas constantT is temperature in KelvinM is the molar mass (mass of one mole of the gas)At room temperature (300K)GasMolar Mass(g/mol)Vrns(m/s)Hydrogen21920Nitrogen28517Oxygen32483
16Translational Kinetic Energy K Average translational kinetic energy of one moleculeKavg=(mv2/2)avg=m(vrms2)/2Kavg=m(vrms2)/2=(m/2)[3RT/M]=(3/2)(m/M)RT=(3/2)(R/NA)TKavg=(3/2)(R/NA)T=(3/2)kT
17ContinueAt a given temperature, all ideal gas molecules – no matter what their masses – have the same average translational kinetic energy.
18Checkpoint 2A 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.
19The Molar Specific Heat of an Ideal Gas Internal energy of an ideal gas EintFor 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 TThe internal energy Eint of a confined ideal gas is a function of the gas temperature only, it does not depend on any other variable.
20Change in internal energy Eint = 3/2 n R T, DEint = 3/2 n R DT
21The Molar Specific Heat of an Ideal Gas Heat Q2Heat Q1Eventhough Ti and Tf is the same for both processes, but Q1 and Q2 are Different because heat depends on the path!
22Heat gained or lost at constant volume For an ideal gas process at constant volume pi,Ti increases to pf,Tf and heat absorbedQ = n cv T and W=0. ThenEint = (3/2)n R T = Q = n cv Tcv = 3R/2Q = n cV Twhere cv is molar specific heat at constant volume
23Eint = (3/2)n R T = Q - PDV = Q - n R DT Heat gained or lost at constant pressureFor an ideal gas process at constant pressure Vi,Ti increases to Vf,Tf and heat absorbedQ = n cp T and W=PDV. ThenEint = (3/2)n R T = Q - PDV = Q - n R DTQ = (3/2nR+nR) DT = 5/2 n R DTcp = 5/2 RQ = n cp Twhere cp is molar specific heat at constant pressure
24The Molar Specific Heats of a Monatomic Ideal Gas Cp = CV + R;specific heat ration = Cp/ CVFor monatomic gas Cp= 5R/2, CV= 3R/2and = Cp/ CV = 5/3 (specific heat ratio)
26The Molar Specific Heat of an Ideal Gas monatomicdiatomicpolyatomic
27(translational + rotational) Internal energy of monatomic, diatomic, and polyatomic gases (theoretical values)Degrees of freedom(translational + rotational)CvEint=n CVTCp=Cv+RMonatomic gas(3/2) R= 12.5(3/2) nRT5/2 R3Diatomic gas(5/2) R= 20.8(5/2) nRT7/2 R5Polyatomic gas(6/2) R= 24.9(6/2) nRT8/2 R6Eint=n CV T
28Cv of common gases in joules/mole/deg.C (at 15 C and 1 atm.) Symbol Cv g g (experiment)(theory)HeliumHe12.51.666ArgonArNitrogenN220.61.4051.407OxygenO2 21.11.3961.397Carbon DioxideCO2 28.21.3021.298
29Adiabatic Expansion for an Ideal Gas In adiabatic processes, no heat transferred to the system Q=0Either system is well insulated, or process occurs so rapidlyIn this caseDEint = - W
30Adiabatic ProcessP, V and T are related to the initial and final states with the following relations:PiVi= PfVfTiVi-1 = TfVf-1Also T/( -1) V =constant thenpiTi(-1)/ = pfTf(-1)/
31Free 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=TfAlso in this adiabatic process since ( pV=nRT), piVi=pfVf ( not PiVi= PfVf)