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## Chapter 19

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**Chapter 19**The Kinetic Theory of Gases**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**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) http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm**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 1023water 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**Avogadro's NumberFormula - 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**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**Isothermal process**isotherm Isothermal expansion (Reverse is isothermal Compression) Quasi-static equilibrium (p,V,T are well defined) p =n R T/V = constant/V**Work done at constant temperature**W = n R T Ln(Vf/Vi)**Work done at constant pressure**• isobaric process W = p (Vf-Vi)**Work done at constant volume**• isochoric process W = 0**Root Mean Square (RMS) speed vrms**• For 4 atoms having speeds v1, v2, v3 and v4 Vrms is a kind of average speed http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm**p=(nM vrms2)/3V**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 but pV/n = RT Equation 19-21 in the textbook Vrms = (3RT)/M**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)**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**Continue**At a given temperature, all ideal gas molecules – no matter what their masses – have the same average translational kinetic energy.**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.**Eint = 3/2 n R T**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) 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.**Change in internal energy**Eint = 3/2 n R T, DEint = 3/2 n R DT**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!**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 cvT and W=0. Then Eint = (3/2)n R T = Q = n cvT cv = 3R/2 Q = n cVT where cv is molar specific heat at constant volume**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 cpT 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 cpT where cp is molar specific heat at constant pressure**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)**The Molar Specific Heat of an Ideal Gas**monatomic diatomic polyatomic**Internal energy of monatomic, diatomic, and polyatomic gases**(theoretical values) Degrees of freedom (translational + rotational) Eint=n CVT Cv 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 (6/2) R = 24.9 (6/2) nRT 8/2 R 6 Polyatomic gas Eint=n CV T**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**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)/**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 ( notPiVi= PfVf)