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Entropy: Ideal Gas Processes Chapter 19: The Kinec Theory of Gases Thermodynamics = macroscopic picture Gases micro -> macro picture One mole is the number of atoms in 12 g sample Avogadro’s Number of carbon-12 23 -1 C(12)—6 protrons, 6 neutrons and 6 electrons NA=6.02 x 10 mol 12 atomic units of mass assuming mP=mn Another way to do this is to know the mass of one molecule: then So the number of moles n is given by M n=N/N sample A N = N A mmole−mass € Ideal Gas Law Ideal Gases, Ideal Gas Law It was found experimentally that if 1 mole of any gas is placed in containers that have the same volume V and are kept at the same temperature T, approximately all have the same pressure p. The small differences in pressure disappear if lower gas densities are used. Further experiments showed that all low-density gases obey the equation pV = nRT. Here R = 8.31 K/mol ⋅ K and is known as the "gas constant." The equation itself is known as the "ideal gas law." The constant R can be expressed -23 as R = kNA . Here k is called the Boltzmann constant and is equal to 1.38 × 10 J/K. N If we substitute R as well as n = in the ideal gas law we get the equivalent form: NA pV = NkT. Here N is the number of molecules in the gas. The behavior of all real gases approaches that of an ideal gas at low enough densities. Low densitiens m= enumberans tha oft t hemoles gas molecul es are fa Nr e=nough number apa ofr tparticles that the y do not interact with one another, but only with the walls of the gas container. Boltzmann Constant gas constant -23 R = 8.315 J/(mol⋅K) pV = nRT kB = 1.38×10 J/K = 0.0821 (L⋅atm)/(mol⋅K) pV = Nk T = 1.99 calories/(mol⋅K) B R = kBNA Work Done by Isothermal (ΔT = 0) Expansion/Compression of Ideal Gas On p-V graph, the green lines are isotherms… … each green line corresponds to a system at a constant temperature. From ideal gas law, this means that for a given isotherm: 1 pV = constant ⇒ p = (nRT) Relates p and V V Vf The work done by the gas€ is then: ⇒ Wby,isothermal = nRT ln T =0 Δ Vi V f V f V f nRT dV Wby = ∫ pdV = ∫ dV = nRT ∫ pi Vi Vi V Vi V = nRT ln p f For expansion, we have: Vf > Vi Wby > 0 € For compression, we have: V < V W < 0 f i € by Checkpoint 1: An ideal gas has an initial pressure of 3 pressure units and an initial volume of 4 volume units. The table gives the final pressure and volume of the gas in five processes. Which processes start and end on the same isotherm? – a b c d e • p 12 6 5 4 1 • V 1 2 7 3 12 pV=nRT in this case T is a constant so pV=C=12 Sample problem #19-2 One mole of oxygen expands at a constant temperature T of 310 K from an initial volume Vi of 12 L to a final volume Vf of 19 L. How much work is done by the gas during the expansion. We calculated W for isothermal W=nRT ln (Vf /VI) W= (1 mole)(8.31J/mole K)(310K) ln(19/12) W=1180 J Work Done by Isobaric (Δp = 0) Expansion of an Ideal Gas V f Wby = ∫ pdV Vi ⇒ Wby,isobaric = pΔV ΔP=0 € = nRΔT € Adiabac Expansion of an ideal gas (Q = 0) Because a gas is thermally insulated, or expansion/compression happens suddenly ⇒ adiabatic st Remember “Adiabatic means Q = 0” or, by 1 Law of Thermo ⇒ ΔEint = -Wby γ In this case pV = constant where γ = cp/cV = (R+ cV )/cV example: monatomic gas γ = 5/3 γ γ p1V1 = p2V2 γ −1 γ −1 T1V1 = T2V2 Compare with Isothermal Expansion (ΔT = 0) € T1 = T2 ⇔ [p1V1 = p2V2 ]isothermal € Pressure, Temperature, & RMS speed Assume the collision of the gas molecule with the wall is elastic then the : - Δpx = (−mvx ) − (mvx ) = 2mvx The molecule travels to the back wall, collides and comes back. The time it takes is 2L/vx. Δp 2mv mv2 = x = x The is F/A Δt 2l / vx L If we calculated the average velocity F 1 n mv2 m n p = x = xi = v2 and use the fact that the number in the sum 2 ∑ 3 ∑ xi is nN then: n A L i=1 L L i=1 A 2 2 vx = vxi nNA 2 ( )avg ∑ nmN A 2 nM (vx )avg i=1 p = 3 (vx )avg = L V 2 2 2 2 2 2 v = vx + vy + vz 2 nM (v )avg nMv € rms 2 (v )avg = vrms = = v 3V 3V v2 = x 3 RMS = Root-Mean-Square RMS Speeds We have nMv 2 pressure p = rms 3V For ideal gas pV = nRT € 1/2 1/2 3pV 3nRT v = = rms nM nM 3RT v = rms M The RMS velocity depends on: Molar mass & Temperature Problem #19-3: Here are five numbers: 5, 11, 32, 67, and 89. (a) What is the average value navg? (b) What is the rms value nrms of the numbers? (a) 5 + 11+ 32 + 67 + 89 n = = 40.8 avg 5 (b) n 2 2 2 2 2 2 1 2 5 + 11 + 32 + 67 + 89 (n )avg = ∑ni = = 2714.41 n i=1 5 2 (n )avg = 52.1 2 navg ≠ n = nrms ( )avg € Problem #19-21: (a) Compute the rms speed of a nitrogen molecule at 20.0 0C (each N atom has 7 protons and 7 neutrons). At what temperatures will the rms speed be (b) half that value and (c) twice that value? 3RT v = rms M Remember to use: T = 20 °C + 273 = 293 K € If we know the “average” speed of parcles, we then know the kinec energy, 1 2 1 3RT KE = m(vrms ) = m 2 2 M What is the relaonship between translaonal € kinec energy & internal energy? Translaonal Kinec Energy & Internal energy 1 2 1 3RT M = mN A KE = m(vrms ) = m 2 2 M kB= R N A 3 The KE of all ideal gas molecules KE = k T € 2 B depends only on the temperature € € (not mass!) € Monoatomic ideal gas : He, Ar, Ne, Kr… (no potential energies) 3 3 The internal energy Eint,monotonic = NA kBT = nRT ( 2 ) 2 of an ideal gas depends only on the temperature 3 ΔE int,monotonic = 2 nR(ΔT) € € Problem #19-26: What is the average translational kinetic energy of 1 mole nitrogen molecules at 1600 K? Molar Specific Heat at Constant Volume: Monatomic Ideal Gas Molar Specific Heat at Constant Volume (Wby=0) Q = ncV ΔT & Wby = 0 ΔEint = Q = ncV ΔT € 3 ΔE int = n( 2 R)ΔT = n(CV )ΔT = Q € 3 cV ,monatomic = R =12.5 ⋅ J/mol⋅ K € 2 € Molar Specific Heat: Monatomic ideal gas Molar Specific Heat at Constant Volume (Wby=0) ΔEint = Q = ncV ΔT 3 cV ,monatomic = R =12.5 ⋅ J/mol⋅ K € 2 Molar Specific Heat at Constant Pressure (W =pΔV) € by ΔEint = Q − Wby = nc p ΔT − pΔV ncV ΔT = nc p ΔT − nRΔT c = c − R or c = c + R € V p p V 5 € c = R = 20.8 ⋅ J/mol⋅ K p,monatomic 2 € € Remember special cases… Adiabatic expansion/contraction - NO TRANSFER OF ENERGY AS HEAT Q = 0 [ΔE int = −W]adiabatic Constant-volume processes (isochoric)- Wby = ∫ pdV V f =Vi NO WORK IS DONE = area under p −V graph W = 0 Wby = ∫ pdV = 0 ΔE = Q Vi € int QΔV = 0 = nCV ΔT € Constant-pressure processes V f =Vi € Wby = pdV = pΔV ΔE int = Q − pΔV ∫ € Vi CV = CP − R QΔP= 0 = nCP ΔT 3)Cyclical process (closed cycle) ΔEint,closed cycle =0 ΔE = 0 ⇒ Q = W €net area in p-V curve is Q int by € € Chapter 20: Entropy and the Second law of thermodynamics 0th law Thermal Equilibrium: A = B & B = C then A = C Q = ncΔT Q → 0 as ΔT → 0 st 1 law Conservation of energy: ΔEint = Q - Wby= Q + Won Change in Internal energy = heat added minus work done by Today: 2nd law HEAT FLOWS NATURALLY FROM HOT OBJECT TO A COLD OBJECT Heat will NOT flow spontaneously from cold to hot 0 ≤ ΔStotal € Hall of fame Ludwig Boltzmann (1844-1906) -23 Boltzmann constant kB =1.38×10 J/K W = number of states Irreversible Processes How to understand this: Entropy How to describe a system: Entropy, S, like T,V, P, Eint, and n is P, T, V, Eint, and n state variable How to define entropy? Easier to define Change of entropy during a process. f dQ ΔS = S − S = Temp in Kelvin part f i ∫ Units: [J/K] i T where Q is energy transferred to or from a system during a process [ Note: since T > 0, if Q is positive (negative) the ΔS is positive (negative) ] What€ is a process? expansion, compression, temperature rise, add mass Reversible process {processes can be done infinitely slowly to ensure thermal equilibrium} The Second Law of Thermodynamics The entropy of a closed system (no energy and no mass comes in and out) never decreases. It either stays constant (reversible process) or increases (irreversible process). 0 ≤ ΔStotal € Entropy: Ideal Gas Processes 1) For reversible process: dQ ΔScycle,rev = 0 = ∫ T 2) For isothermal process: € ΔE = 0 ⇒ Q = W Q int S V Δ isothermal = Vf f T W = nRT ln ⇒ ΔSisothermal = nR ln V Vi i 3) In general for gas, using 1st law: dE int = dQ − dW nC €dT dQ pdV nRT V € dQ nRT dV nCV dT = − & p€= ⇒ ∫ = ∫ + ∫ T T T V T V T T Now integrate: V f T f S S S nR ln nC ln pV nRT € Δ general,gas = f − i = + V ⇐ [ = ] € Vi Ti 4) For adiabatic (reversible) adiabatic compression/expansion (Q=0): Entropy is a State Function ΔSrev,adiabatic = 0 € € Checkpoint: An ideal gas has a temperature T1 at the inial state i shown in the p‐V diagram.
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