Chapter 19: The Kinec Theory of

Thermodynamics = macroscopic picture Gases  micro -> macro picture

One is the number of 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 : then So the number of moles n is given by M n=N/N sample A N = N A mmole−mass

€ Ideal Law Ideal Gases, Law It was found experimentally that if 1 mole of any gas is placed in containers that have the same V and are kept at the same T, approximately all have the same p. The small differences in pressure disappear if lower gas are used. Further experiments showed that all low- gases obey the equation pV = nRT. Here R = 8.31 K/mol ⋅ K and is known as the "." The equation itself is known as the "." The constant R can be expressed -23 as R = kNA . Here k is called the 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 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 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 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: 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 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 molecule at 20.0 0C (each N has 7 protons and 7 neutrons). At what 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 ,

1 2  1  3RT KE = m(vrms ) =  m 2  2  M

What is the relaonship between translaonal € kinec 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 )

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 of 1 mole nitrogen molecules at 1600 K? Molar Specific 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: 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 (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 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 ().

0 ≤ ΔStotal

€ Entropy: Ideal Gas Processes

1) For reversible process: dQ ΔScycle,rev = 0 = ∫ T

2) For : € 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 ΔSrev,adiabatic = 0 €

€ Checkpoint: An ideal gas has a temperature T1 at the inial 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: / processes

dQ ΔS 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 idencal 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 insulang shuer. When we li the shuer, 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