Lecture 16. The van der Waals Gas (Ch. 5)

The simplest model of a liquid-gas phase transition - the van der Waals model of “real” gases – grasps some essential features of this phase transformation. (Note that there is no such transformation in the ideal gas model). This will be our attempt to take intermolecular interactions into account.

Nobel 1910

Outline

• Thermodynamics of vdW gas

• critical point

• liquid-gas phase transition

van der Waals gas = ideal gas + interaction

The main reason for the transformation of gas into liquid at decreasing T and (or) increasing P - interaction between the molecules.

TNkNbVV

aNP B

2

2

NbVVeff

2 2

vdW IG 2eff

N a N aU U P P

V V

the strong short-range repulsion: the molecules are rigid: P as soon as the molecules “touch” each other.

The vdW equationof state

U(r)

short-distancerepulsion

long-distanceattraction

-3

-2

-1

0

1

2

3

4

1.5 2.0 2.5 3.0 3.5 4.0

distance

Ene

rgy

r

the weak long-range attraction:

Two ingredients of the model:

- the constant a is a measure of the long-range attraction

- the constant b (~ 43/3) is a measure of the short-range repulsion, the “excluded volume” per particle

12 6

U rr r

r =

2

2

V

aN

NbV

TNkP B

2N N aU N a

V V

Lennard-Jones

Substance a’(J. m3/mol2)

b’(x10-5 m3/mol)

Pc

(MPa)

Tc

(K)Air .1358 3.64 3.77 133 K

Carbon Dioxide (CO2) .3643 4.27 7.39 304.2 K

Nitrogen (N2) .1361 3.85 3.39 126.2 K

Hydrogen (H2) .0247 2.65 1.30 33.2 K

Water (H2O) .5507 3.04 22.09 647.3 K

Ammonia (NH3) .4233 3.73 11.28 406 K

Helium (He) .00341 2.34 0.23 5.2 K

Freon (CCl2F2) 1.078 9.98 4.12 385 K

2

3

23

22

' '

J m'

molJ m

molecule molecules

mole

A A

A

N a a N b b

a

a

N

TNkNbVV

aNP B

2

2

When can TNkPV Bbe reduced to

V

NaTkB

VNb

- high temperatures (kinetic energy >> interaction energy)

?

V

NaTkNPV B

b – roughly the volume of a molecule, (3.5·10-29 – 1.7 ·10-28) m3 ~(few Å)3

a – varies a lot [~ (8·10-51 – 3 ·10-48) J · m3] depending on the intermolecular

interactions (strongest – between polar molecules, weakest – for inert gases).

The van der Waals Parameters

- low densities

Problem

The vdW constants for N2: NA2a = 0.136 Pa·m6 ·mol-2, NAb = 3.85·10-5 m3 ·mol-1.

How accurate is the assumption that Nitrogen can be considered as an ideal gas at normal P and T?

TNkNbVV

aNP B

2

2

1 mole of N2 at T = 300K occupies V1 mol RT/P 2.5 ·10-2 m3 ·mol-1

NAb = 3.9·10-5 m3 ·mol-1 NAb / V1 mol ~1.6%

NA2a / V2 = 0.135 Pa·m6 ·mol-2 /(2.5 ·10-2 m3 ·mol-1) 2 = 216 Pa NA

2a / V2P = 0.2%

Entropy of monatomic van der Waals gas

Multiplicity of ideal gas:

3 / 2

IG 2

1 2, , 2

! 3 / 2 !

NNV mUN V U p

N N h

NbVVeff van der Waals gas: 2

vdW IG

N aU U

V

Multiplicity of monatomic van der Waals gas:

2 3 / 2

vdW 2

21, , 2

! 3 / 2 !

NN N a

Vm UV NbN V U p

N N h

Entropy of monatomic van der Waals gas:

2 3/ 2

vdW 2

4 5( , , ) ln

3 2

N aV

B

UV Nb mS N V U Nk

N h N

Is it correct?

Equations of state (vdW)

Equations of state:

2 3/ 2

vdW 2

4 5( , , ) ln

3 2

N aV

B

UV Nb mS N V U Nk

N h N

,

1

V N

S

T U

2

vdW

3

2 B

N aU Nk T

V

,U N

SP T

V

2

2

V

aN

NbV

TNkP B

,U V

ST

N

3/ 2

2

2 2ln B

B

mk TV Nb Na Nbk T

N h V V Nb

(take home exercise: derive these results)

G

S, F, and G for the monatomic van der Waals gas

F

V

aNTNkTk

h

m

N

NbVTkN

Tkh

m

N

NbVTkN

V

aNTNkTSUF

BBB

BBBvdW

22/3

2

2/3

2

2

2ln

2

52ln

2

3

2

2

V

aN

NbV

TNk

V

FP B

TvdW

S dVNbV

Nk

T

dTNk

fdV

T

PdT

T

CdV

V

SdT

T

SdS B

BV

V

TV

2

2

52ln1ln

2/3

2Tk

h

m

N

NbVkN

V

NbNkSS BBBidealvdW

- the same “volume” in the momentum space, smaller accessible volume in the coordinate space.

(see Pr. 5.12) constNbVNkTNkf

S BBvdW lnln2

V

aN

V

NbTkNFF BidealvdW

2

1ln

3/ 2 2 2

2

2 2lnvdW B B B

V Nb m aN N bG F PV Nk T k T k T

N h V V Nb

(take home exercise: derive these results)

N

Quasi-static Processes in a vdW Gas (low n, high T)

isochoric ( V = const )

isobaric ( P = const )

021 W

TCTTNkQ VB 02

31221

0),( 12

2

1

21 VVPdVTVPW

1 2 PQ C T

21QdU

2121 QWdU

V

P

V1,2

T1

T21

2

V

P

V1

T1

T212

V2

U Q W dU TdS PdV

same as ideal gas

PTVP dT

dVP

V

UCC

complicate

Isothermal Process in a vdW Gas (low n, high T)

2

1

2

1

1 2

2

2

2 22

1 1 2

( , )

ln

V

V

V

B

V

B

W P V T dV

Nk T N adV

V Nb V

V Nb N a N aNk T

V Nb V V

Wi-f > 0 if Vi >Vf (compression)

Wi-f < 0 if Vi <Vf (expansion)

isothermal ( T = const ) :

V

P

PV= NkBT

V1V2

W

21 2 1 2

1

lnB

V NbQ U W Nk T

V Nb

2 2

1 2T

N a N aU

V V

kJkJkJ

mollmoll

mollmollK

molK

Jmol

llmol

lkJmolW TvdW

7.2464.3194.6

/0385.034.3

/0385.0317.0ln3103.83

17.0

1

4.3

1138.09

22

kJl

lK

Kmol

Jmol

V

VTNkW BTideal 12.23

4.3

17.0ln3103.83ln

1

2

Depending on the interplay between the 1st and 2nd terms, it’s either harder or easier to compress the vdW gas in comparison with an ideal gas. If both V1 and V2 >> Nb, the interactions between molecules are attractive, and WvdW < Wideal However, as in this problem, if the final volume is comparable to Nb , the work against repulsive forces at short distances overweighs that of the attractive forces at large distances. Under these conditions, it is harder to compress the vdW gas rather than an ideal gas.

r

Upot

For N2, the vdW coefficients are N2a = 0.138 kJ·liter/mol2 and Nb = 0.0385 liter/mol. Evaluate the work of isothermal and reversible compression of N2 (assuming it is a vdW gas) for n=3 mol, T=310 K, V1 =3.4 liter, V2 =0.17 liter. Compare this value to that calculated for an ideal gas. Comment on why it is easier (or harder, depending on your result) to compress a vdW gas relative to an ideal gas under these conditions.

Isothermal Process in a vdW Gas (low n, high T)

Adiabatic Process in a vdW Gas

adiabatic (thermally isolated system)

3/ 2 const.V Nb T

2

2const

N aP V Nb

V

1 2 0 0Q S 21WdU

2

1

),(21

V

V

dVTVPW

V

P

V1

T1

T21

2

V2

3/ 2

vdW 2

2 5( , , ) ln

2B

B

mk TV NbS N V T Nk

N h

1const.T V Nb

2 2

2 11 2

3

2 B

N a N aU Nk T T

V V

2

vdW

3

2 B

N aU Nk T

V

recall:

take home exercise: Carnot engine efficiency using vdW gas

Problem (vdW + heat engine)The working substance in a heat engine is the vdW gas with a known constant b and a temperature-independent heat capacity cV (the same as for an ideal gas). The gas goes through the cycle that consists of two isochors (V1 and V2) and two adiabats. Find the efficiency of the heat engine.

P

C

D

A

B

V V1 V2

A-D DAVH TTcQ

B-C CBVC TTcQ

2 / 2 /

1 12 / 1

2 2 1 1

2 2

1 1 1 1 1

f f

fA DB CC

H A D A D

V Nb V NbT T

T T V Nb V NbQ V Nb V Nbe

Q T T T T V Nb V Nb

H

C

Q

Qe

1

DA

CB

TT

TTe

1

constNbVNkTNkf

S BBvdW lnln2 NbV

NbVNk

T

TNk

fS

i

fB

i

fBvdW

lnln

2

0lnln2

NbV

NbVNk

T

TNk

f

i

fB

i

fB constNbVT

f

2 adiabatic process for the vdW gas

The relationship between TA, TB, TC, TD – from the adiabatic processes B-C and D-A

The van der Waals Isotherms

2

2

V

aN

NbV

TNkP B

0

3223

P

abNV

P

aNV

P

TNkNbV B

0

N·b

unstable

The Critical Point

- the critical coefficient

67.23

8

CC

CC VP

RTK

substance H2 He N2 CO2 H20

KC 3.0 3.1 3.4 3.5 4.5

TC (K) 33.2 5.2 126 304 647

PC (MPa) 1.3 0.23 3.4 7.4 22.1

The critical pointcritical point is the unique point where both (dP/dV)T = 0 and (d2P/dV 2)T = 0 (see Pr. 5.48)

- in terms of P,T,V normalized by the critical parameters:

- the materials parameters vanish if we introduce the proper scales.

Critical parameters:

2

1 83

27 27C C B C

a aV Nb P k T

b b

ˆ ˆ ˆC C C

P V TP V T

P V T

2

ˆ83 1ˆ ˆˆ 3 3

Bk TP V

V

Problems For Argon, the critical point occurs at a pressure PC = 4.83

MPa and temperature TC = 151 K. Determine values for

the vdW constants a and b for Ar and estimate the diameter of an Ar atom.

C

CB

C

CBCBC P

Tka

P

Tkb

b

aTk

b

aP

2

2 64

27

827

8

27

1

3296

23

m 104.5Pa 104.838

K 151J/K 1038.1

8

C

CB

P

Tkb Am 86.31086.3 103/1 b

/PaJ 108.3

Pa 1083.4

K 151J/K 1038.1

64

27

64

27 2496

2232

C

CB

P

Tka

Per mole: a=0.138 Pa m6 mol-2; b=3.25x10-5 m3 mol-1

Problem

3/

lnln2

5lnln

2

5

0000 C

ffff

VV

VR

T

TR

NbV

VR

T

TRS

J/K 5.25261024.51004.0101

102.2ln

273

2.266ln

2

5 133

2

RRS

One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume V0=1liter. The critical parameters for N2 are: VC = 3Nb = 0.12 liter/mol, TC = (8/27)(a/kBb) = 126K. The gas goes through the free expansion process (Q = 0, W = 0), in which the pressure drops down to the atmospheric pressure Patm=1 bar. Assume that the gas obeys the van der Waals equation of state in the compressed state, and that it behaves as an ideal gas at the atmospheric pressure. Find the change in the gas entropy.

vdW

ideal

ln ln ( , )2

ln ln ( , )2

A

fS R T R V N b g N m

fS R T R V g N m

22 3

0 00 0

2 9266.2 K 2.2 10 m

5 20fC CA

f fatm

RTT VaNT T T V

RV V P

Phase Separation in the vdW ModelThe phase transformation in the vdW model is easier to analyze by minimizing F(V) rather than G(P) (dramatic changes in the term PV makes the dependence G(P) very complicated, see below).

At T< TC, there is a region on the F(V) curve in which F makes a concave protrusion (2F/V

2<0) – unlike its ideal gas counterpart. Due to this protrusion, it is possible to draw a common tangent line so that it touches the bottom of the left dip at V = V1 and the right dip at V = V2. Since the common tangent line lies below the free energy curve, molecules can minimize their free energy by refusing to be in a single homogeneous phase in the region between V1 and V2, and by preferring to be in two coexisting phases, gas and liquid:

21

12

21

21

21

VV

VVF

VV

VVFN

N

FN

N

FF gasliquid

As usual, the minimum free energy principle controls the way molecules are assembled together. V < V1 V1 < V < V2 V2 < V

V

P

V

F

V1 V2

F1

(liquid)

F2

(gas)

T = const (< TC)

12

2

12

1

VV

VV

N

N

VV

VV

N

NNNN liquidgas

liquidgas

- we recognize this as the common tangent line.

Phase Separation in the vdW Model (cont.)

P

VVgasVliq

Pvap(T1)

P

TT1

triplepoint

criticalpoint

Since the tangent line F(V) maintains the same slope between V1 and V2, the pressure remains constant between V1 and V2: P

V

F

NT

,

In other words, the line connecting points on the PV plot is horizontal and the two coexisting phases are in a mechanical equilibrium. For each temperature below TC, the phase transformation occurs at a well-defined pressure Pvap, the so-called vapor pressure.

Two stable branches 1-2-3 and 5-6-7 correspond to different phases. Along branch 1-2-3 V is large, P is small, the density is also small – gas. Along branch 5-6-7 V is small, P is large, the density is large – liquid. Between the branches – the gas-liquid phase transformation, which starts even before we reach 3 moving along branch 1-2-3.

P

V

1

2

3

4

5

7

6Pvap

T < TC

V

F

Fliq

Fgas

Cnn / CPP /

C

CBC

C

Q

CB

VT n

nTk

nnn

n

n

nTk

N

F

4

9

1/3

11

3ln

3ln

,

CPP /

Cnn /

CTT /

idea

l gas

@ T

/TC=

1.2

For T<TC, there are three values of n with the same . The outer two values of n correspond to two stable phases which are in equilibrium with each other.

The kink on the G(V) curve is a signature of the 1st order transition. When we move along the gas-liquid coexistence curve towards the critical point, the transition becomes less and less abrupt, and at the critical point, the abruptness disappears.

Phase Separation in the vdW Model (cont.)

The Maxwell Construction

G

P

1

2,6

34

57

the areas 2-3-4-2 and 4-5-6-4 must be equal !

-Maxwell construction

the lowest branch represents the stable

phase, the other branches are unstable

[finding the position of line 2-6 without analyzing F(V)]

On the one hand, using the dashed line on the F-V plot:

On the other hand, the area under the vdW isoterm 2-6 on the P-V plot:

2 6 2 6,

gas liquid vapT N

FF F V V P V V

V

liquidgas

V

V

V

V NT

FFdVV

FPdV

2

6

2

6 ,

62

2

6

VVPdVVP vap

V

V

vdW Thus,

P

V

1

2

3

4

5

7

6Pvap

T < TC

V

F

Fliq

Fgas

Problem

P

V

The total mass of water and its saturated vapor (gas) mtotal = mliq+ mgas = 12 kg. What are the masses of water, mliq, and the gas, mgas, in the state of the system shown in the Figure?

Vliq increases from 0 to V1 while the total volume decreases from V2 to V1. Vgas decreases from V2 to 0 while the total volume decreases from V2 to V1. When V = V1, mtotal = mliq. Thus, in the state shown in the Figure, mliq 2 kg and mgas 10 kg.

V

Vliq

Vgas

V1 V2

Phase Diagram in T-V Plane

Single Phase

V

T

TC

VC

At T >TC, the N molecules can exist in a single phase in any volume V, with any density n = N/V. Below TC, they can exist in a homogeneous phase either in volume V < V1 or in volume V > V2. There is a gap in the density allowed for a homogeneous phase.

There are two regions within the two-phase “dome”: metastable (P/V< 0) and unstable (P/V>0). In the unstable region with negative compressibility, nothing can prevent phase separation. In two metastable regions, though the system would decrease the free energy by phase separation, it should overcome the potential barrier first. Indeed, when small droplets with radius R are initially formed, an associated with the surface energy term tends to increase F. The F loss (gain) per droplet:

metastab

le

met

asta

ble unstable

V1(T) V2(T)

0,02

2

V

F

V

P

condensation:

1

31 /3

4VR

N

NF

interface:24 R

R F

1

312 /3

44 VR

N

NFRF

Total balance: RC

P

V

Pvap

T < TC

V

F

Fliq

Fgas

Joule-Thomson Process for the vdW Gas

0

PP

HT

T

HH

TP

The JT process corresponds to an isenthalpic expansion:

PP

HTCC

T

H

TPP

P

VP

ST

P

HPVSTH

TT

(see Pr. 5.12)P

P

P

T

H C

VT

VT

C

P

H

P

T

We’ll consider the vdW gas at low densities: NbVV

aNP

2

2

PBB TTNkPNb

V

aNPVTNkNbV

V

aNP ...

2

2

2

2

22

2

VaN

P

Nk

T

VNk

T

V

V

aN

T

VP B

PB

PP

PT T

V

P

S

P

B

H C

NbTk

Na

P

T

2

This is a pretty general (model-independent) result. By applying this result to the vdW equation, one can qualitatively describe the shape of the inversion curve (requires solving cubic equations...).

cooling

heating

Joule-Thomson Process for the vdW Gas (cont.)

cooling

heating 02

bTk

a

INVB

The upper inversion temperature:(at low densities)

CB

INV Tbk

aT

4

272

SubstanceTINV

(P=1 bar)

CO2 (2050)

CH4 (1290)

O2 893

N2 621

H2 205

4He 51

3He (23)

(TC – the critical temperature of the vdW gas, see below)

Cooling: 02

bTk

a

B

If b = 0, T always decreases in the JT process: an increase of Upot at the expense of K. If a = 0, T always increases in the JT process (despite the work of molecular forces is 0):

Heating: 02

bTk

a

B

PNbTCPNbTRCPVUHPNbRTPVRTNbVP PV

Thus, the vdW gas can be liquefied by compression only if its T < 27/4TC.

Problem

The vdW gas undergoes an isothermal expansion from volume V1 to volume V2.

Calculate the change in the Helmholtz free energy.

In the isothermal process, the change of the Helmholtz free energy is

PdVdNPdVSdTdF NTNT ,,

NbV

NbVRT

VVaNdV

V

aN

NbV

RTPdVF

V

V

V

V 1

2

21

22

2

ln112

1

2

1

21

2 11

VVaNU TvdW

compare with TSUF TvdWST