Ideal GasesExperimental results on gases:Boyle’s LawPV = constant (fixed amount of gas, constant T)Charles’ LawV/T = constant (fixed amount of gas, constant P)(T is in Kelvins)Combining these two laws:THE IDEAL GAS LAW PV = nRTwhere n is the number of moles of gasand R is the gas constant, R = 8.31 J/(mol • K)Alternate form: PV = NkTwhere N is the number of molecules inthe gasand k is the Boltzmann’s constant, k = 1.38x10-23 J/K(Comparing the two forms gives R=NAk.)All real gases approach the “ideal gas” inthe limit of very low density.Work done at constant T (ideal gas)Isothermal expansion:PV = nRT = constant P=nRT/V• Isotherm: a curve along which T isconstant.Work done by gas: DW = ∫ P dVDW = ∫ (nRT/V) dV = nRT ln(Vf/Vi)Note: if Vf > Vi (expansion), then DW is + if Vf < Vi (compression), then DW is -PVifisothermfifiWork done in otherprocesses DW = ∫ P dVConstant Volume:DW = 0Constant Pressure:DW = P (Vf-Vi)fiPVifP Vi fKinetic Theory of GasesConsider N molecules (n moles withn=N/NA) in a cubical box of side L, i.e.Volume = L3.Change in momentumat the x-wall isDpx = 2 m vxTime between collisions with thex-wall isDt = 2 L / vxyxzLAverage rate of change of momentum inx-direction:Dpx/Dt = (2mvx)/(2L/vx) = m vx2/ LThis is force exerted by the molecule.Total Force = ∑ (m vx2)i / LPressure P = Force/Area = F/L2 = (m/L3) ∑ (vx2)ifi P = (m/L3) N vx2mN = nM is the total mass.where n = # of moles M = molar massfi P = (nM/V) vx2i=1NaverageFor any molecule: v2 = vx2 + vy2 + vz2fi vx2 = (1/3) v2fi P = (nM/3V) v2Define root-mean-square speed vrms: vrms = √ v2fi PV = (nM/3) v2rmsFrom ideal gas law: PV = nRTfi (nM/3) v2rms = nRTfi vrms = √3RT/Mmicroscopic vrms ¤ macroscopic TKinetic Energy Average (translational) kinetic energy per molecule = (1/2) m v2 = (1/2) m (3RT/M)Using M/m = NA, K = 3RT/(2NA) = (3/2) k Tfi K = (3/2) k TA measurement of the Temperature of agas is equivalent to a measure of theaverage kinetic energy of its molecules.Molecular Speeds at Room Temp (T=300 K) Gas molar mass(g) vrms(m/s)Hydrogen, H2 2 1920Helium, He 4 1370Water vapor, H2O 18 645Nitrogen, N2 28 517Oxygen, O232 483Carbon Dioxide, CO244 412Sulfer Dioxide, SO264 342Table 20.1Maxwell’s Speed DistributionThe distribution of molecular speeds wasfirst written down by Maxwell (1852):P(v) dv is the probability that a moleculehas speed between v and v + dv.It is normalized to 1. ∫ P(v) dv = 1.v (m/s)† P(v) = 4pM2pRTÊ Ë Á ˆ ¯ ˜ 3 / 2v2e-Mv22RT P(v)(10-3 s/m)0•vpvvrmsCurve for O2 At T = 300 KAverage speed: v = ∫ v P(v) dv = √8RT/(pM)Root-mean-square speed: v2 = v2rms = ∫ v2 P(v) dv = 3RT/MMost probable speed (maximum ofdistribution curve): vp = √2RT/M0•0•Internal Energy (due only to kinetic energy of atoms)• Monatomic gas - Single atoms:U = N (3/2) kT = (3/2) nNAkT = (3/2) nRTEach atom has 3 Degrees of Freedom.(K. E. in x, y, or z directions).• Diatomic molecule:Rotates (in two planes)fi 5 degrees of freedom.U = (5/2) nRT• Polyatomic molecule:Rotates in all 3 planesfi 6 degrees of freedom.(3 translational + 3 rotational).U = (6/2) nRT = 3 nRT“Equipartion of Energy”Molar Specific Heats(of ideal gas)Recall: Specific heat tells how T changesas Q is added.This depends on the conditions: Constant V or Constant P.Constant Volume: DQ = n CVDTwhere CV is specific heat at constant V.1st Law of TD: DQ = DU + DWAt constant V, DW = 0.fi DQ = DU = (3/2) nRDT (monatomic gas)Comparing with definition of CV gives:CV = (3/2) R = 12.5 J/(mol˙K)Constant Pressure: DQ = n CPDTwhere CP is specific heat at constant P.1st Law of TD: DQ = DU + DWAt constant P, DW = P DV.Ideal gas law: P DV = nR DTfi DQ = (3/2) nR DT + nR DT = (5/2) nR DTfi CP = (5/2) R (again, for monatomic gas)In General:CP = CV + RQuantum Mechanics andEquipartition of EnergyQuantum Theory predicts: rotational energies are quantized(only have certain discrete values).• Rotational degrees of freedom only“turn on” above some minimum temperature (roughly when kTis larger than the lowest rotationalenergy level of the molecule). (Fig. 20-12 of HRW)c´V/R5,0002,0001,000500200100205010DissociationVibration1234725232RotationTranslationTemperature (K)01210,000©2005 by Prentice Hall, Inc.A Pearson CompanyFigure Number: 19 11 Fishbane/Gasiorowicz/ThorntonPhysics for Scientists and Engineers, withModern Physics 3EAADYWKO0Adiabatic Expansion (DQ = 0)Occurs if:• change is made sufficiently quickly• and/or with good thermal isolation.Governing formula:PVg = constantwhere g = CP/CVBecause PV/T is constant (ideal gas): Vg-1 T = constant (for adiabatic)PVAdiabatIsothermsProof of PVg=constant (for adiabatic process)1) Adiabatic: dQ = 0 = dU + dW = dU + PdV2) U only depends on T:dU = n CV dT (derived for constant volume, but true in general)3) Ideal gas: T = PV/(nR) dT = [(dP)V + P(dV)]/(nR)Plug into 2): dU = (CV/R)[VdP + PdV]Plug into 1): 0 = (CV/R)[VdP + PdV] + PdVPVIsotherms(constant T)Same DURearrange: (dP/P) = - (CV+R)/CV (dV/V) = - g (dV/V)where g = (CV+R)/CV = CP/CVIntegrate both sides:ln(P) = - g ln(V) + constantorln(PVg) = constantorPVg = constantQEDIrreversible ProcessesExamples:• Block sliding on table comes to restdue to friction: KE converted to heat.• Heat flows from hot object to coldobject.• Air flows into an evacuated chamber.Reverse process allowed by energyconservation, yet it does not occur. arrow of timeWhy?2nd Law of Thermodynamics (entropy)Heat EnginesHeat engine: a cyclic device designed toconvert heat into work.2nd Law of TD (Kelvin form):It is impossible for a cyclic process toremove thermal energy from a system at asingle temperature and convert it tomechanical work without changing thesystem or surroundings in some other way.Hot Reservoir, THCold Reservoir, TCQHQCWork,