Chapter 19: The Kine c 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
€ Adiaba c 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 par cles, we then know the kine c energy,
1 2 1 3RT KE = m(vrms ) = m 2 2 M
What is the rela onship between transla onal € kine c energy & internal energy? Transla onal Kine c 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 ini al 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?
T From i to a: 2 ΔS = nCV ln T1
From i to b: T2 Vb ΔS = nCV ln + nRln T1 Va Entropy: Liquid/solid processes
dQ ΔS phase change = ∫ 1) For phase changes: T Temperature = constant Q = phase change T mL = T
2) For temperature changes: € dQ ΔSliquid / solid = S f − Si = ∫ T mcdT = ∫ T
Tf = mcln Ti
€ Sample Problem #20‐2: Two iden cal copper 0 blocks of mass m=1.5kg: Block L is at TiL= 60 C and 0 block R is at TiR=20 C. The blocks are in a thermally insulated box and are separated by an insula ng shu er. When we li the shu er, the blocks come 0 to equilibrium with Tf=40 C. 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.
T f dQ f mcdT T dQ = mcdT ΔS = = = mcln f = −35.86J / K L ∫ T ∫ T T i Til iL 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 le 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 V Q = nRT ln f Vi
€ Put in numbers nRT ln V / V Q ( f i) ) Vf ΔS = = = nRln T T V i ΔS = +5.76J / K