Lecture 15. 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.

In particular, van der Waals was able to explain the existence of a critical point for the vapor-liquid transition and to derive a Law of Corresponding States (1880).

In his Nobel prize acceptance speech, van der Waals saw the qualitative agreement of his theory with experiment as a major victory for the atomistic theory of matter – stressing that this view had still remained controversial at the turn of the 20th century!

Nobel 1910

The van der Waals ModelThe 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

VaNPPeff +=

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 4.0di tr

the weak long-range attraction: the long-range attractive forces between the molecules tend to keep them closer together; these forces have the same effect as an additional compression of the gas.

Two ingredients of the model:

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

( )

- the constant b (~ 4πσ3/3) is a measure of the short-rangerepulsion, the “excluded volume” per particle

612

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛∝

rrrU σσ

r = σ

( ) 2

2

VaN

NbVTNkP B −

−=

Substance a’(J. m3/mol2)

b’(x10-5 m3/mol)

Pc(MPa)

Tc(K)

Air .1358 3.64 3.77 133 KCarbon 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

2

6

2

6

molemolecules

molm J

moleculem J

⎟⎠⎞

⎜⎝⎛

⎥⎦

⎤⎢⎣

⎡ ⋅

=⎥⎦

⎤⎢⎣

⎡ ⋅

⇔⇔

A

AA

N

aa

bbNaaN

'

''2

( ) TNkNbVV

aNP B=−⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

When can TNkPV B=be reduced to

VNaTkB >>

VNb <<

- high temperatures(kinetic energy >> interaction energy)

?

⎟⎠⎞

⎜⎝⎛ −≈

VNaTkNPV 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%

The vdW Isotherms( ) 2

2

VaN

NbVTNkP B −

−=

032

23 =−+⎟⎠⎞

⎜⎝⎛ +−

PabNV

PaNV

PTNkNbV B

0

N·b

CPP /

Cnn /

CTT /

idea

l gas

@ T

/TC=1

.2

2

313

8⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

CC

C

C nn

TT

nnP

P

The vdW isotherms in terms of the gas density n:

A real isotherm cannot have a negative slope dP/dn (or a positive slope dP/dV), since this corresponds to a hegative compressibility. The black isotherm at T=TC(the top panel) corresponds to the critical behavior. Below TC, the system becomes unstable against the phase separation (gas ↔ liquid) within a certain range V(P,T). The horizontal straight lines show the true isotherms in the liquid-gas coexistence region, the filled circles indicating the limits of this region.

The critical isotherm represents a boundary between those isotherms along which no such phase transition occurs and those that exhibit phase transitions. The point at which the isotherm is flat and has zero curvature (∂P/∂V= ∂2P/∂V2=0) is called a critical point.

The Critical Point

baTk

baPNbV CBCC 27

82713 2 ===

- the critical coefficient

67.238==≡

CC

CC VP

RTKsubstance H2 He N2 CO2 H20RTC/PCVC 3.0 3.1 3.4 3.5 4.5

TC (K) 33.2 5.2 126 304 647PC (MPa) 1.3 0.23 3.4 7.4 22.1

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

032

23 =−+⎟⎠⎞

⎜⎝⎛ +−

PabNV

PaNV

PTNkNbV B

Three solutions of the equation should merge into one at T = TC:

( ) 033 32233 =−+−=− CCCC VVVVVVVV

- in terms of P,T,V normalized by the critical parameters:3

831ˆ

ˆ3ˆ

2

TkVV

P B=⎟⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛ +

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

Critical parameters:

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

TkaPTkb

baTk

baP

2

2 6427

8278

271

====

3296

23

m 104.5Pa 104.838

K 151J/K 1038.18

−−

⋅=⋅×

×⋅==

C

CB

PTkb Am 86.31086.3 103/1 =⋅= −b

( ) ( ) /PaJ 108.3Pa 1083.4

K 151J/K 1038.16427

6427 249

6

2232−

−

⋅=⋅

×⋅==

C

CB

PTka

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

Energy of the vdW Gas

(low n, high T)

Let’s start with an ideal gas (the dilute limit, interactions can be neglected) and compress the gas to the final volume @ T,N = const:

[ ] PTPT

TP

VSP

VSTPdVTdSdU

VU

VVTTT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=−⎟⎠⎞

⎜⎝⎛∂∂

=−==⎟⎠⎞

⎜⎝⎛∂∂

From the vdW equation: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

−=⎟

⎠⎞

⎜⎝⎛∂∂

2

21V

aNPTNbV

NkTP B

V2

2

VaN

VU

T

=⎟⎠⎞

⎜⎝⎛∂∂

VaNTkNfdV

VUUU B

constT TidealvdW

2

2−=⎟

⎠⎞

⎜⎝⎛∂∂

+= ∫=

This derivation assumes that the system is homogeneous – it does not work for the two-phase (P, V) region (see below). The same equation for U can be obtained in the model of colliding rigid spheres. According to the equipartition theorem, K = (1/2) kBT for each degree of freedom regardless of V. Upot – due to attraction between the molecules (repulsion does not contribute to Upot in the model of colliding rigid spheres). The attraction forces result in additional pressure aN2/V2 . The work against these forces at T = const provides an increase of U in the process of anisothermal expansion of the vdW gas:

(see Pr. 5.12)

VNaNUU idealvdW −=

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=Δ

fiTvdW VV

aNU 112

( ) 2

2

VaN

NbVTNkP B −

−=

dVVUUU

constT TidealvdW ∫

=

⎟⎠⎞

⎜⎝⎛∂∂

+=

Isothermal Process for the vdW gas at low n and/or high T ( ) ⎟

⎟⎠

⎞⎜⎜⎝

⎛−=Δ

fiTvdW VV

aNU 112

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−=−= ∫∫

fii

fB

V

V

BV

V VVaN

NbVNbV

TNkdVV

aNNbVTNkPdVW

f

i

f

i

11ln 22

2

δ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=− NbVNbVTNkQ HB

1

221 lnδWQU δδ +=Δ

( )

kJkJkJmollmollmollmollK

molKJmol

llmollkJmolW TvdW

7.2464.3194.6/0385.034.3/0385.0317.0ln3103.83

17.01

4.31138.09 2

2

=+−=

⎟⎠⎞

⎜⎝⎛

⋅−⋅−

⋅⋅

⋅−⎟⎠⎞

⎜⎝⎛ −

⋅⋅=Δ

( ) kJllK

KmolJmol

VVTNkW BTideal 12.23

4.317.0ln3103.83ln

1

2 =⎟⎠⎞

⎜⎝⎛⋅⋅

⋅⋅−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=Δ

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 ΔWvdW < ΔWidealHowever, 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.

VaNTC

VaNTkNfU

idealV

BvdW

2

2

2

−=

−=

T

U Uideal

UvdW

VaN 2

−

V = const

V

UUideal

UvdW

T = const

idealUaN 2

idealVvdWV CC =

Problem: One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume V0=1liter. The gas goes through the free expansion process (δQ = 0, δ 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 temperature.

The internal energy of the gas is conserved in this process (δQ = 0, δ W = 0), and, thus, ΔU = 0:

( )0

2

00

2

00 25,

VaNRT

VaNUVTU idealvdW −=−=

fRTV

aNRT25

25

0

2

0 =−

Tf is the temperatureafter expansion

0

2

0 52

RVaNTTf −=

For 1 mole of the “ideal” Nitrogen (after expansion): fideal RTU25

=

For 1 mole of the “vdW” Nitrogen (diatomic gas)

The final temperature is lower than the initial temperature: the gas molecules work against the attraction forces, and this work comes at the expense of their kinetic energy.

( )NbVTVNk

VaNTNkTk

hm

NNbVTkNPVFG B

BBBvdW −+−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

−=+=22/3

2

22ln πG

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

F ( )

( )V

aNTNkTkh

mN

NbVTkN

Tkh

mN

NbVTkNV

aNTNkTSUF

BBB

BBBvdW

22/3

2

2/3

2

2

2ln

252ln

23

−−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

−−=−=

π

π

2

2

VaN

NbVTNk

VFP B

TvdW −

−=⎟

⎠⎞

⎜⎝⎛∂∂

−=

S dVNbV

NkTdTNkfdV

TPdT

TCdV

VSdT

TSdS B

BV

V

TV −+=⎟

⎠⎞

⎜⎝⎛∂∂

+=⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=2

( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

=⎟⎠⎞

⎜⎝⎛ −+=

252ln1ln

2/3

2 Tkh

mNNbVkN

VNbNkSS BBBidealvdW

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

(see Pr. 5.12) ( ) constNbVNkTNkfS BBvdW +−+= lnln2

VaN

VNbTkNFF BidealvdW

2

1ln −⎟⎠⎞

⎜⎝⎛ −−=

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 −=Δ

B-C ( )CBVC TTcQ −=Δ

( )( )

VcRf

DA

f

D

f

A

DA

CB

H

C

NbVNbV

NbVNbV

TTNbVNbVT

NbVNbVT

TTTT

QQe

/

2

1

/2

2

1

/2

2

1

/2

2

1

11111 ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=−−

−=ΔΔ

−=

H

C

QQe

δδ

−=1

( )( )DA

CB

TTTTe

−−

−=1

( ) constNbVNkTNkfS BBvdW +−+= lnln2 NbV

NbVNk

TT

NkfSi

fB

i

fBvdW −

−+=Δ lnln

2

0lnln2

=−

−+

NbVNbV

NkTT

Nkf

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

Problem

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=Δ3/

lnln25lnln

25

0000 C

ffff

VVV

RTT

RNbV

VR

TT

RS

32

00 m 102.2K 2.266

209 −⋅===−=

atm

ff

CCf P

RTV

VVTTT

( ) ),(lnln2

mNfNbVRTRfSvdW +−+=

J/K 5.25261024.51004.0101

102.2ln273

2.266ln25 1

33

2

≈+⋅−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅−⋅

⋅+=Δ −

−−

−

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 (δQ = 0, δW = 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.

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

VVVVF

VVVVFN

NFN

NFF gasliquid −

−+

−−

=+=

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

V

P

VF

V1 V2

F1(liquid)

F2(gas)

T = const (< TC)

12

2

12

1

VVVV

NN

VVVV

NN

NNN liquidgasliquidgas −

−=

−−

=+=

- 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:

PVF

NT

−=⎟⎠⎞

⎜⎝⎛∂∂

,

In other words, the line connecting points on the PVplot 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 Vis 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

VF

Fliq

Fgas

μ μ

Cnn / CPP /

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+⎟⎠⎞

⎜⎝⎛ −−⎟

⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛∂∂

=C

CBC

C

Q

CB

VT nnTk

nnnn

nnTk

NF

49

1/3113ln3ln

,

μ

CPP /

Cnn /

CTT /

idea

l gas

@ T

/TC=1

.2

For T<TC, there are three values of n with the same μ . The outer two values of ncorrespond 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

P1

2,6

34

57

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

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-Vplot:

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

( )

( )62

62,,

2

6

VVP

VVVFdV

VFFF

vap

NT

V

V NTliquidgas

−−=

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=− ∫

( )liquidgas

V

V

V

V NT

FFdVVFPdV −−=⎟⎠⎞

⎜⎝⎛∂∂

−=∫ ∫2

6

2

6 ,

( ) ( )62

2

6

VVPdVVP vap

V

VvdW −=∫Thus,

P

V

1

2

3

4

5

7

6Pvap

T < TC

VF

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 V2to 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 < V1or 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:

metastable

met

asta

ble unstable

V1(T) V2(T)

0,0 2

2

=∂∂

=∂∂

VF

VP

condensation: ⎟⎠⎞

⎜⎝⎛Δ− 1

31 /34 VR

NNF π

interface: 24 Rπγ

Rδ F

⎟⎠⎞

⎜⎝⎛Δ−= 1

312 /344 VR

NNFRF ππγδ

Total balance: RC

P

V

Pvap

T < TC

VF

Fliq

Fgas

Joule-Thomson Process for the vdW Gas

0=Δ⎟⎠⎞

⎜⎝⎛∂∂

+Δ⎟⎠⎞

⎜⎝⎛∂∂

=Δ PPHT

THH

TP

The JT process corresponds to an isenthalpic expansion:

PPHTCC

TH

TPP

P

Δ⎟⎠⎞

⎜⎝⎛∂∂

−=Δ=⎟⎠⎞

⎜⎝⎛∂∂

VPST

PHPVSTH

TT

+⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

Δ+Δ=Δ

(see Pr. 5.12)P

P

P

T

H C

VTVT

CPH

PT

−⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

−=⎟

⎠⎞

⎜⎝⎛ΔΔ

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

aNP >>>> 2

2

( ) ( )PBB TTNkPNb

VaNPVTNkNbV

VaNP ...

2

2

2

∂∂

=−+=−⎟⎟⎠

⎞⎜⎜⎝

⎛+

2

22

2

VaNP

NkTVNk

TV

VaN

TVP B

PB

PP −=⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

PT TV

PS

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

P

B

H C

NbTk

Na

PT

−≈⎟

⎠⎞

⎜⎝⎛ΔΔ

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=− b

Tka

INVB

The upper inversion temperature:(at low densities)

CB

INV TbkaT

4272

==

SubstanceTINV

(P=1 bar)

CO2 (2050)

CH4 (1290)

O2 893

N2 621

H2 2054He 513He (23)

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

Cooling: 02>− b

Tka

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<− b

Tka

B

( ) ( ) PNbTCPNbTRCPVUHPNbRTPVRTNbVP PV +=++=+=−==−

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

ProblemThe 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 −=+−−= ,, μ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−=−=Δ ∫∫ NbV

NbVRTVV

aNdVV

aNNbV

RTPdVFV

V

V

V 1

2

21

22

2

ln112

1

2

1

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=Δ

21

2 11VV

aNU TvdW

compare with TSUF −=( )TvdWSTΔ