The van der Waals gas model takes into account intermolecular interactions that the ideal gas model neglects. It explains the liquid-gas phase transition through a critical point, where the vapor and liquid phases become indistinguishable. The model approximates molecules as rigid spheres that experience short-range repulsion and long-range attraction. It derives an equation of state relating pressure, volume, and temperature. This equation reduces to the ideal gas law under conditions of high temperature or low density.

1. 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

2. The van der Waals Model
The main reason for the transformation of
gas into liquid at decreasing T and (or)
increasing P - interaction between the
molecules.
  TNkNbV
V
aN
P B





 2
2
NbVVeff 
2
2
V
aN
PPeff 
the strong short-range repulsion: the molecules are rigid: P   as soon as the
molecules “touch” each other.
The vdW equation
of state
U(r)
short-distance
repulsion
long-distance
attraction
-3
-2
-1
0
1
2
3
4
1.5 2.0 2.5 3.0 3.5 4.0
distance
Energy
r
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 (~ 43/3) is a measure of the short-range
repulsion, the “excluded volume” per particle
 
612













rr
rU

r = 
  2
2
V
aN
NbV
TNk
P B




3. 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
2
6
2
6
mole
molecules
mol
mJ
molecule
mJ











 





 

A
AA
N
a
a
bbNaaN
'
''
2
  TNkNbV
V
aN
P B





 2
2
When can TNkPV Bbe reduced to
V
Na
TkB 
VNb 
- high temperatures
(kinetic energy >> interaction energy)
?







V
Na
TkNPV 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

4. Problem
The vdW constants for N2: NA
2a = 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?
  TNkNbV
V
aN
P 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%
NA
2a / V2 = 0.135 Pa·m6 ·mol-2 /(2.5 ·10-2 m3 ·mol-1) 2 = 216 Pa NA
2a / V2P = 0.2%

5. The vdW Isotherms
  2
2
V
aN
NbV
TNk
P B



0
32
23







P
abN
V
P
aN
V
P
TNk
NbV B
0
N·b
CPP /
Cnn /
CTT /
2
3
13
8














CC
C
C n
n
T
T
n
nP
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.

6. The Critical Point
b
a
Tk
b
a
PNbV CBCC
27
8
27
1
3 2

- the critical coefficient
67.2
3
8

CC
C
C
VP
RT
K
substance H2 He N2 CO2 H20
RTC/PCVC 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 point is the unique point where both (dP/dV)T = 0
and (d2P/dV 2)T = 0 (see Pr. 5.48)
0
32
23







P
abN
V
P
aN
V
P
TNk
NbV 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
8
3
1ˆ
ˆ
3ˆ
2
Tk
V
V
P B













- the materials parameters vanish if we introduce the proper scales.
Critical parameters:

7. 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
CB
CBC
P
Tk
a
P
Tk
b
b
a
Tk
b
a
P
2
2
64
27
827
8
27
1

329
6
23
m104.5
Pa104.838
K151J/K1038.1
8






C
CB
P
Tk
b Am 86.31086.3 103/1
 
b
    /PaJ108.3
Pa1083.4
K151J/K1038.1
64
27
64
27 249
6
2232






C
CB
P
Tk
a
Per mole: a=0.138 Pa m6 mol-2; b=3.25x10-5 m3 mol-1

8. 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:
  P
T
P
T
T
P
V
S
P
V
S
TPdVTdSdU
V
U
VVTTT














































From
the vdW
equation:
















2
2
1
V
aN
P
TNbV
Nk
T
P B
V
2
2
V
aN
V
U
T








V
aN
TkN
f
dV
V
U
UU B
constT T
idealvdW
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 an
isothermal expansion of the vdW gas:
(see Pr. 5.12)
V
Na
NUU idealvdW 
  








fi
TvdW
VV
aNU
112
  2
2
V
aN
NbV
TNk
P B



dV
V
U
UU
constT T
idealvdW 










9. Isothermal Process for the vdW
gas at low n and/or high T   








fi
TvdW
VV
aNU
112
























  fii
f
B
V
V
B
V
V
VV
aN
NbV
NbV
TNkdV
V
aN
NbV
TNk
PdVW
f
i
f
i
11
ln 2
2
2










NbV
NbV
TNkQ HB
1
2
21 lnWQU  
 
kJkJkJ
mollmoll
mollmoll
K
molK
J
mol
llmol
lkJ
molW TvdW
7.2464.3194.6
/0385.034.3
/0385.0317.0
ln3103.83
17.0
1
4.3
1
138.09 2
2




















  kJ
l
l
K
Kmol
J
mol
V
V
TNkW BTideal 12.23
4.3
17.0
ln3103.83ln
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.

10. V
aN
TC
V
aN
TkN
f
U
idealV
BvdW
2
2
2


T
U Uideal
UvdW
V
aN 2

V = const
V
U
Uideal
UvdW
T = const
idealU
aN 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 (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 temperature.
The internal energy of the gas is conserved in this
process (Q = 0,  W = 0), and, thus, U = 0:
 
0
2
0
0
2
00
2
5
,
V
aN
RT
V
aN
UVTU idealvdW 
fRT
V
aN
RT
2
5
2
5
0
2
0 
Tf is the temperature
after expansion
0
2
0
5
2
RV
aN
TTf 
For 1 mole of the “ideal” Nitrogen (after expansion): fideal RTU
2
5

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.

11.  
NbV
TVNk
V
aN
TNkTk
h
m
N
NbV
TkNPVFG B
BBBvdW

















22/3
2
22
ln

G
S, F, and G for the monatomic van der Waals gas
F
 
 
V
aN
TNkTk
h
m
N
NbV
TkN
Tk
h
m
N
NbV
TkN
V
aN
TNkTSUF
BBB
BBBvdW
22/3
2
2/3
2
2
2
ln
2
52
ln
2
3










































2
2
V
aN
NbV
TNk
V
F
P B
T
vdW 










S dV
NbV
Nk
T
dT
Nk
f
dV
T
P
dT
T
C
dV
V
S
dT
T
S
dS B
B
V
V
TV 

























2
 






























2
52
ln1ln
2/3
2
Tk
h
m
N
NbV
kN
V
Nb
NkSS BBBidealvdW

- the same “volume” in the
momentum space, smaller
accessible volume in the
coordinate space.
(see Pr. 5.12)   constNbVNkTNk
f
S BBvdW  lnln
2
V
aN
V
Nb
TkNFF BidealvdW
2
1ln 







12. 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
VV1
V2
A-D  DAVH TTcQ 
B-C  CBVC TTcQ 
 
 
VcRf
DA
f
D
f
A
DA
CB
H
C
NbV
NbV
NbV
NbV
TT
NbV
NbV
T
NbV
NbV
T
TT
TT
Q
Q
e
/
2
1
/2
2
1
/2
2
1
/2
2
1
11111 








































H
C
Q
Q
e


1
 
 DA
CB
TT
TT
e


1
  constNbVNkTNk
f
S BBvdW  lnln
2 NbV
NbV
Nk
T
T
Nk
f
S
i
f
B
i
f
BvdW


 lnln
2
0lnln
2




NbV
NbV
Nk
T
T
Nk
f
i
f
B
i
f
B   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

13. Problem















3/
lnln
2
5
lnln
2
5
0000 C
ffff
VV
V
R
T
T
R
NbV
V
R
T
T
RS
32
0
0 m102.2K2.266
20
9 

atm
f
f
CC
f
P
RT
V
V
VT
TT
  ),(lnln
2
mNfNbVRTR
f
SvdW 
J/K5.25261024.5
1004.0101
102.2
ln
273
2.266
ln
2
5 1
33
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.

14. Phase Separation in the vdW Model
The 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
1
2
21
2
1
21
VV
VV
F
VV
VV
FN
N
F
N
N
F
F 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
N
NNN
liquidgas
liquidgas






- we recognize this as the common tangent line.

15. Phase Separation in the vdW Model (cont.)
P
VVgasVliq
Pvap(T1)
P
TT1
triple
point
critical
point
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
6
Pvap
T < TC
V
F
Fliq
Fgas

16.  
Cnn / CPP /








































C
CB
C
C
Q
C
B
VT n
n
Tk
nnn
n
n
n
Tk
N
F
4
9
1/3
1
1
3
ln
3
ln
,

CPP /
Cnn /
CTT /
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.)

17. The Maxwell Construction
G
P
1
2,6
3
4
5
7
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-V
plot:
On the other hand, the area under the vdW isoterm
2-6 on the P-V plot:
 
 62
62
,,
2
6
VVP
VV
V
F
dV
V
F
FF
vap
NT
V
V NT
liquidgas

















 
 liquidgas
V
V
V
V NT
FFdV
V
F
PdV 







 
2
6
2
6 ,
   62
2
6
VVPdVVP vap
V
V
vdW Thus,
P
V
1
2
3
4
5
7
6
Pvap
T < TC
V
F
Fliq
Fgas

18. 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

19. 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:
unstable
V1(T) V2(T)
0,0 2
2






V
F
V
P
condensation: 





 1
31
/
3
4
VR
N
N
F 
interface:
2
4 R
R F






 1
312
/
3
4
4 VR
N
N
FRF 
Total balance: RC
P
V
Pvap
T < TC
V
F
Fliq
Fgas

20. Joule-Thomson Process for the vdW Gas
0















 P
P
H
T
T
H
H
TP
The JT process corresponds to an isenthalpic expansion:
P
P
H
TCC
T
H
T
PP
P
















V
P
S
T
P
H
PVSTH
TT

















(see Pr. 5.12)P
P
P
T
H C
V
T
V
T
C
P
H
P
T


























We’ll consider the vdW gas at low densities: NbV
V
aN
P  2
2
   PBB
T
TNkPNb
V
aN
PVTNkNbV
V
aN
P ...
2
2
2









2
22
2
V
aN
P
Nk
T
V
Nk
T
V
V
aN
T
V
P B
P
B
PP

























PT T
V
P
S
















P
B
H C
Nb
Tk
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

21. Joule-Thomson Process for the vdW Gas (cont.)
cooling
heating 0
2
 b
Tk
a
INVB
The upper inversion temperature:
(at low densities)
C
B
INV T
bk
a
T
4
272

Substance
TINV
(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: 0
2
 b
Tk
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: 0
2
 b
Tk
a
B
    PNbTCPNbTRCPVUHPNbRTPVRTNbVP PV 
Thus, the vdW gas can be liquefied by
compression only if its T < 27/4TC.

22. 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
NbV
RT
VV
aNdV
V
aN
NbV
RT
PdVF
V
V
V
V 1
2
21
2
2
2
ln
112
1
2
1
  






21
2 11
VV
aNU TvdW
compare with TSUF 
 TvdWST