Presentation on theme: "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."— Presentation transcript:
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)
What is one mole? Avogadro’s Number N A = 6.02 X One mole of any element contains Avogadro’s number of atoms of that element. One mole of iron contains 6.02 X iron atoms. One mole of water contains 6.02 X water molecules. From experiments: 12 g of carbon contains 6.02 X carbon atoms. Thus, 1 mole carbon = 12 g of carbon. 4 g of helium contains 6.02 X helium atoms. Thus, 1 mole helium = 4 g of helium. Avogadro’s Number N A = 6.02 X per mole = 6.02 X mol -1
Avogadro's Number Formula - number of moles n = N /N A n = number of moles N = number of molecules N A = Avogadro number M = Molar mass of a substance M sample = mass of a sample n = M sample /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/N A
Ideal Gas Law
Isothermal process Isothermal expansion (Reverse is isothermal Compression) isotherm 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(V f /V i )
Work done at constant pressure isobaric process W = p (V f -V i )
Work done at constant volume isochoric process W = 0
Root Mean Square (RMS) speed v rms For 4 atoms having speeds v 1, v 2, v 3 and v 4 V rms is a kind of average speed
Pressure, Temperature, RMS Speed The pressure p of the gas is related to root-mean -square speed v rms, volume V and temperature T of the gas p=(nM v rms 2 )/3V Equation in the textbook V rms = (3RT)/M but pV/n = RT
Continue… V rms = (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) V rns (m/s) Hydrogen21920 Nitrogen28517 Oxygen32483
K avg =(3/2)(R/N A )T=(3/2)kT Translational Kinetic Energy K Average translational kinetic energy of one molecule K avg =(mv 2 /2) avg =m(v rms 2 )/2 K avg =m(v rms 2 )/2=(m/2)[3RT/M] =(3/2)(m/M)RT=(3/2)(R/N A )T
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 m 1 >m 2 >m 3. Rank the three types according to average kinetic energy, and rms speed, greatest first.
The Molar Specific Heat of an Ideal Gas For a monatomic gas (which has individual atoms rather than molecules), the internal energy E int is the sum of the translational kinetic energies of the atoms. Internal energy of an ideal gas E int E int = N K avg = N (3/2) k T = 3/2 (N k T) = 3/2 (n R T) E int = 3/2 n R T The internal energy E int 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 E int = 3/2 n R T, E int = 3/2 n R T
The Molar Specific Heat of an Ideal Gas Heat Q 1 Heat Q 2 Eventhough T i and T f is the same for both processes, but Q 1 and Q 2 are Different because heat depends on the path!
For an ideal gas process at constant volume p i,T i increases to p f,T f and heat absorbed Q = n c v T and W=0. Then E int = (3/2)n R T = Q = n c v T c v = 3R/2 Q = n c V T Heat gained or lost at constant volume where c v is molar specific heat at constant volume
For an ideal gas process at constant pressure V i,T i increases to V f,T f and heat absorbed Q = n c p T and W=P V. Then E int = (3/2)n R T = Q P V = Q n R T Q = (3/2nR+nR) T = 5/2 n R T c p = 5/2 R Q = n c p T Heat gained or lost at constant pressure where c p is molar specific heat at constant pressure
The Molar Specific Heats of a Monatomic Ideal Gas C p = C V + R; specific heat ration = C p / C V For monatomic gas C p= 5R/2, C V = 3R/2 and = C p / C V = 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) (3/2) R = 12.5 (3/2) nRT5/2 R3 Diatomic gas (5/2) R = 20.8 (5/2) nRT7/2 R5 (6/2) R = 24.9 (6/2) nRT8/2 R6 E int =n C V T Monatomic gas Polyatomic gas CvCv E int =n C V T C p =C v +R Degrees of freedom (translational + rotational)
C v of common gases in joules/mole/deg.C (at 15 C and 1 atm.) GasSymbol C v (experiment) (theory) HeliumHe ArgonAr NitrogenN2N OxygenO Carbon DioxideCO
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 E int = - W In this case
Adiabatic Process P, V and T are related to the initial and final states with the following relations: P i V i = P f V f T i V i -1 = T f V f -1 Also T /( -1) V =constant then p i T i ( -1)/ = p f T f ( -1)/
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. T i =T f Also in this adiabatic process since ( pV=nRT), p i V i =p f V f ( not P i V i = P f V f ) Free Expansion of an Ideal Gas