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Chapter 19: The Kine0c Theory of Gases Thermodynamics = macroscopic picture Gases micro -> macro picture Avogadro’s Number One mole is the number of atoms in 12 g sample of carbon-12 NA=6.02 x 1023 mol-1 So the number of moles n is given by n=N/NA C(12)—6 protrons, 6 neutrons and 6 electrons 12 atomic units of mass assuming mP=mn Another way to do this is to know the mass of one molecule: then N= M sample m mole−mass NA 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 as R = kN A . Here k is called the Boltzmann constant and is equal to 1.38 × 10-23 J/K. If we substitute R as well as n = 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. = number moles Nenough = number of particles Low densitiesn means thatofthe gas molecules are far apart that they do not interact with one another, but only with the walls of the gasgas container. constant Boltzmann Constant pV = nRT pV = NkBT kB = 1.38×10-23 J/K R = kBNA R = 8.315 J/(mol⋅K) = 0.0821 (L⋅atm)/(mol⋅K) = 1.99 calories/(mol⋅K) 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: ⇒ p = (nRT ) pV = constant Wby = ∫ Vi Vf f i i V nRT dV pdV = ∫ dV = nRT ∫ V V V V For expansion, we have: Vf > Vi € For compression, we have: Vf < Vi Wby > 0 € Relates p and V Vf ⇒ Wby,isothermal = nRT ln ΔT =0 Vi p = nRT ln i pf € is then: The work done by the gas Vf 1 V Wby < 0 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 • p 12 • V 1 b c d e 6 2 5 7 4 3 1 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 Vf Wby = ∫ pdV Vi ⇒ Wby,isobaric = pΔV ΔP=0 = nRΔT € € Adiaba0c Expansion of an ideal gas (Q = 0) Because a gas is thermally insulated, or expansion/compression happens suddenly ⇒ adiabatic Remember “Adiabatic means Q = 0” or, by 1st Law of Thermo ⇒ ΔEint = -Wby In this case pVγ = constant where γ = cp/cV = (R+ cV )/cV example: monatomic gas γ = 5/3 γ 1 1 γ 2 p V = p2V T1V1γ −1 = T2V2γ −1 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. 2 The is F/A n p= Δp 2mvx mvx = = Δt 2l / vx L 2 xi n Fx 1 mv m = 2∑ = 3 ∑ vxi2 A L i =1 L L i =1 2 x avg If we calculated the average velocity and use the fact that the number in the sum n is nNA then: 2 2 nM (v ) nmN A 2 p= (vx )avg = 3 L V 2 2 2 2 v = v + v + v x y z 2 nM (v 2 )avg nMvrms € 2 = = v 3V 3V vx2 = 3 (v ) x avg = ∑v xi nN A i=1 2 (v )avg = vrms RMS = Root-Mean-Square RMS Speeds We have pressure p= For ideal gas € vrms 3pV = nM vrms = 2 nMv rms 3V pV = nRT 1/2 3nRT = nM 1/2 3RT 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) navg 5 + 11 + 32 + 67 + 89 = = 40.8 5 (b) (n 2 )avg 1 n 2 5 2 + 112 + 32 2 + 67 2 + 89 2 = ∑ ni = = 2714.41 n i =1 5 (n 2 )avg = 52.1 n avg ≠ (n 2 )avg = n rms 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 par0cles, we then know the kine0c energy, 1 3RT 1 2 KE = m (vrms ) = m 2 M 2 What is the rela0onship between transla0onal € kine0c energy & internal energy? Transla0onal Kine0c Energy & Internal energy 1 3RT 1 2 KE = m (vrms ) = m 2 M 2 3 KE = kB T 2 € M = mN A kB= R N A The € KE of all ideal gas molecules depends only on the temperature €(not mass!) € Monoatomic ideal gas : He, Ar, Ne, Kr… (no potential energies) Eint, monotonic = N A ( 3 2 ) kB T = 32 nRT The internal energy ΔE int,monotonic = 32 nR(ΔT ) € of an ideal gas depends only on the temperature 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 € ΔE int = n( 32 R)ΔT = n(CV )ΔT = Q € cV , monatomic = € € 3 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 € cV , monatomic = 3 R =12.5 ⋅ J/mol ⋅ K 2 Molar Specific Heat at Constant Pressure (Wby=pΔV) € ΔEint = Q − Wby = nc p ΔT − pΔV ncV ΔT = nc p ΔT − nRΔT cV = c p − R € € € € c p, monatomic = or c p = cV + R 5 R = 20.8 ⋅ J/mol ⋅ K 2 Remember special cases… Adiabatic expansion/contraction - NO TRANSFER OF ENERGY AS HEAT Q = 0 [ΔE int = −W ] adiabatic Constant-volume processes (isochoric)- NO WORK IS DONE ΔE int = Q W = 0 Wby = ∫ pdV = area under p − V graph ∫ pdV = 0 Vi QΔV = 0 = nCV ΔT Constant-pressure processes € Wby = V f =Vi ΔE int = Q − pΔV € CV = C P − R 3)Cyclical process (closed cycle) € net area in p-V curve is Q € € V f =Vi Wby = ∫ pdV = pΔV Vi QΔP= 0 = nC P ΔT ΔEint,closed cycle =0 ΔE int = 0 ⇒ Q = Wby Chapter 20: Entropy and the Second law of thermodynamics 0th law Thermal Equilibrium: A = B & B = C then A = C 1st law Q = ncΔT Q → 0 as ΔT → 0 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 ≤ ΔS total Hall of fame Ludwig Boltzmann (1844-1906) Boltzmann constant kB =1.38×10-23 J/K W = number of states Irreversible Processes How to understand this: Entropy How to describe a system: P, T, V, Eint, and n Entropy, S, like T,V, P, Eint, and n is state variable How to define entropy? Easier to define Change of entropy during a process. f ΔS part = S f − S i = ∫ i dQ T Temp in Kelvin Units: [J/K] 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 ≤ ΔS total € Entropy: Ideal Gas Processes 1) For reversible process: ΔS cycle,rev = 0 = ∫ dQ T 2) For isothermal process: € ΔS isothermal Q = T 3) In general for gas, using 1st law: dE int = dQ − dW nCV € dT dQ pdV = − T T T nRT & p€= V ΔE int = 0 ⇒ Q=W Vf W = nRT ln ⇒ Vi ⇒ ΔS isothermal nRT dV € dQ = ∫ ∫ + T V T ∫ Vf = nR ln Vi nCV dT T Now integrate: V f T f ΔS general, gas = S f − S i = nR ln + nC V ln Ti € Vi € 4) For adiabatic (reversible) adiabatic compression/expansion (Q=0): € ΔS rev,adiabatic = 0 ⇐ [ pV = nRT ] Entropy is a State Function Checkpoint: An ideal gas has a temperature T1 at the iniDal state i shown in the p‐V diagram. The gas has a higher temperature T2 at the final states a and b, which can reach along the paths shown. Is the entropy change along the path to state a larger than, smaller than, or the same as that along path to state b? From i to a: From i to b: T2 ΔS = nCV ln T1 T2 Vb ΔS = nCV ln + nR ln T1 Va Entropy: Liquid/solid processes 1) ΔS phase change = For phase changes: Temperature = constant ∫ dQ T Qphase change T mL = T = 2) For temperature changes: € ΔS liquid / solid = S f − S i = ∫ mcdT ∫ T Tf = mc ln Ti = € dQ T Sample Problem #20‐2: Two idenDcal copper blocks of mass m=1.5kg: Block L is at TiL= 600C and block R is at TiR=200C. The blocks are in a thermally insulated box and are separated by an insulaDng shuTer. When we liV the shuTer, the blocks come to equilibrium with Tf=400C. What is the entropy of this irreversible process? Specific heat of Cu is 386 J/kg‐K. The left block is initially at 600C, and comes to equilibrium with final temperature 400C. Heat is transferred from the left block to the right. f dQ = mcdT Tf Tf dQ mcdT ΔSL = ∫ =∫ = mc ln = −35.86J / K T T TiL i Til Now set reservoir at 200C and put in contact with R. Raise the temperature slowly to 400C. ΔSR = 38.23J / K ΔSrev = ΔSR + ΔSL = 2.4J / K Sample problem 20‐1: Suppose 1.0 mol of nitrogen gas is confined to the leV side of the container in the figure. You open the stop‐cock and the volume of the gas doubles. What is the entropy change of the gas for this irreversible process? Free Expansion so ΔT = 0 Q = nRT ln ( ) € Vf Q nRT ln V f / Vi ) ΔS = = = nR ln T T Vi Vf Vi Put in numbers ΔS = +5.76J / K