Lecture 15. The van der Waals Gas (Ch. 5)
Nobel 1910
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!
The van der Waals Model
short-distance
repulsion
The main reason for the transformation of
gas into liquid at decreasing T and (or)
increasing P - interaction between the
molecules.
4
U(r)
3
Energy
2
1
Two ingredients of the model:
0
-1
-2
-3
1.5
2.0
2.5
3.0
3.5
4.0
r
distance
r=
   
U r       
r r
12
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.
long-distance
6 attraction
Peff
-
N 2a
 P 2
V
the constant a is a measure of the longrange attraction
the strong short-range repulsion: the molecules are rigid: P   as soon as the
molecules “touch” each other.
Veff  V  Nb
-
the constant b (~ 43/3) is a measure of the short-range
repulsion, the “excluded volume” per particle

N 2a 
 P  2 V  Nb   Nk BT
V 

Nk BT
N 2a
P
 2
V  Nb V
The vdW equation
of state
The van der Waals Parameters
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).
a’
b’
(J. m3/mol2)
(x10-5 m3/mol)
Air
.1358
3.64
Carbon Dioxide (CO2)
.3643
4.27
Nitrogen (N2)
.1361
3.85
Hydrogen (H2)
.0247
2.65
Water (H2O)
.5507
3.04
Ammonia (NH3)
.4233
Helium (He)
Freon (CCl2F2)
Substance
Pc
(MPa)
3.77 2
NA
7.39
Tc
(K)
133 K
a
K
304.2
a'
N Ab

b'
3.39
126.2 K
3.73
 J  m6 
a' 
6
mol 2 


J

m
1.30
33.2 K

a

2
2 
22.09
647.3
K
molecules 
 molecule  
 NA

11.28
406 K
mole


.00341
2.34
0.23
5.2 K
1.078
9.98
4.12
385 K

N 2a 
When can  P  2 V  Nb   Nk BT be reduced to PV  Nk BT ?
V 

- low densities
Nb  V
Na 

PV  N  k BT 

V


k BT 
Na
V
- high temperatures
(kinetic energy >> interaction energy)
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?

N 2a 
 P  2 V  Nb   Nk BT
V 

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
NA2a / V2P = 0.2%
Nk BT
aN 2
P

V  Nb V 2
The vdW Isotherms
Nk BT  2 aN 2
abN3

V   Nb 
V
0
V 
P 
P
P

3
The vdW isotherms in terms of the gas density n:
0
 n 
P
8
T

 3 
PC
 n  T
 nC 
3   1 C
 nC 
N·b
T / TC
P / PC
n / nC
2
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
The critical point is the unique point where both (dP/dV)T = 0
and (d2P/dV 2)T = 0 (see Pr. 5.48)
Nk BT  2 aN 2
abN3

V   Nb 
V
0
V 
P
P
P


3
Three solutions of the equation should merge into one at T = TC:
V VC 3  V 3  3VCV 2  3VC 2V VC3  0
Critical parameters:
 ˆ 3  ˆ 1  8k BT
 P  ˆ 2 V   
3
3
V 

VC  3Nb
PC 
1 a
27 b 2
k BTC 
8 a
27 b
- in terms of P,T,V normalized by the critical parameters:
- the materials parameters vanish if we introduce the proper scales.
RTC 8
KC 
  2.67
PCVC 3
- the critical coefficient
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
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.
1 a
PC 
27 b 2
8 a
k BTC 
27 b

k BTC
b
8PC
27 k BTC 
27 1.381023 J/K 151K
a

64 PC
64
4.83106 Pa
2
k BTC 1.381023 J/K 151K
 29
3
b


5
.
4

10
m
8PC
8  4.83106 Pa
27 k BTC 
a
64 PC
2

2
 3.8 1049 J 2 /Pa
b1/ 3  3.861010 m  3.86 A
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:
U vdW  U ideal 
 U 
  dV

V T
T const 
 S 
 U 
 S 
 P  
 P 

  dU  TdS  PdV   T 
  P  
 
   T
 P

V

V

V

T

T

T

T
T 
V 

V

(see Pr. 5.12)
From
2
2
Nk
T
aN
Nk B
1
N 2 a   U 
 P 
N
a
B
the vdW P 
 2
 
  P 

 
 
2 
V  Nb T 
V   V T
V  Nb V
V2
 T V
equation:
U vdW  U ideal
f
N 2a
 U 
  
 dV  N k BT 

V
2
V
T
T const 
U vdW  U ideal  N
Na
V
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
1 1 
isothermal expansion of the vdW gas:
2
U vdW T  N
a  
V V 
f 
 i
Isothermal Process for the vdW
gas at low n and/or high T
U vdW T
1
1 
 N 2a  
V V 
f 
 i
1 1 
 V f  Nb 
 Nk BT N 2 a 
  N 2 a   
W    PdV    
 2  dV   Nk BT ln
V V 
V  Nb V 
f 
 Vi  Nb 
Vi
Vi 
 i
 V2  Nb 



Q

Nk
T
ln
U   Q   W
1 2
B H
 V1  Nb 
Vf
Vf
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.
WvdW T  9m ol2  0.138 kJ  l2 
Upot
1
1 
J
 0.17l  3m ol 0.0385l / m ol

 310K ln
  3m ol 8.3

m ol  3.4l 0.17l 
K  m ol
 3.4l  3m ol 0.0385l / m ol 
 6.94kJ  31.64kJ  24.7kJ
 
Wideal T   Nk BT ln V2   3mol 8.3 J  310K  ln 0.17l   23.12kJ
mol K
 3.4l 
 V1 
r
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.
V = const
U
Uideal
Uideal
U
UvdW
N 2a

V
T = const
T
CV vdW  CV ideal
UvdW
N 2a
U ideal
V
f
N 2a
U vdW  N k BT 
2
V
N 2a
 CV idealT 
V
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.
2
2
N
a
5
N
a
For 1 mole of the “vdW” Nitrogen (diatomic gas)
U vdW T0 ,V0   U ideal 
 RT0 
V0
2
V0
5
Tf is the temperature
For 1 mole of the “ideal” Nitrogen (after expansion): U ideal  RT f
2
after expansion
The internal energy of the gas is conserved in this
process (Q = 0,  W = 0), and, thus, U = 0:
5
N 2a 5
RT0 
 RTf
2
V0
2
2 N 2a
T f  T0 
5 RV0
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.
S, F, and G for the monatomic van der Waals gas
S
C
f
dT
Nk B
 S 
 S 
 P 
dS  

dV
 dT  
 dV  V dT  
 dV  Nk B
T
2
T
V  Nb
 T V
 V T
 T V
(see Pr. 5.12)
SvdW
S vdW 
f
Nk B ln T  Nk B lnV  Nb   const
2
3/ 2
  V  Nb   2  m
 Nb 
  5 
 Sideal  Nk B ln1    N k B ln 
 2 k BT    
V
N


 h
  2 
 
F
FvdW
3
N 2a
 U  TS  Nk BT 
 N k BT
2
V
- the same “volume” in the
momentum space, smaller
accessible volume in the
coordinate space.
3/ 2

  V  Nb   2  m

  5
ln
k
T




 

B
2
N
h
2





 

3/ 2
 V  Nb   2  m
aN 2
 
  N k BT ln 
 2 k BT    Nk BT 
V
 h
 
 N
 Nb  N a
 Fideal  N k BT ln1 

V  V

2
FvdW
G
GvdW
PvdW
Nk BT aN 2
 F 
 
 2
 

V
V

Nb
V

T
3/ 2
 V  Nb   2  m
2aN 2 Nk BTV
 
 F  PV   N k BT ln 

 2 k BT    Nk BT 
N
h
V
V  Nb

 

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
e  1
A
D
QH  cV TA  TD 
A-D
B
 QC
 QH
e  1
QC  cV TB  TC 
B-C
TB  TC 
TA  TD 
C
V1
S vdW 
V2
V
The relationship between TA, TB, TC, TD – from
the adiabatic processes B-C and D-A
f
Nk B ln T  Nk B lnV  Nb   const
2
Tf
V f  Nb
f
Nk B ln  Nk B ln
0
2
Ti
Vi  Nb
2/ f
SvdW 
f
2
Tf
V f  Nb
f
Nk B ln  Nk B ln
2
Ti
Vi  Nb
T V  Nb   const
 V  Nb 
 V  Nb 
  TD  1

TA  1
V  Nb 
T  T 
QC
 V2  Nb 
e  1
 1 B C  1  2
TA  TD 
QH
TA  TD
2/ f
adiabatic process
for the vdW gas
 V  Nb 

 1   1
 V2  Nb 
2/ f
 V  Nb 

 1   1
 V2  Nb 
R / cV
Problem
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.
S vdW 
f
R ln T  R lnV  Nb   f ( N , m)
2
Tf
Tf
Vf
 Vf  5


5




S  R ln
 R ln

R
ln

R
ln


2
T0
T0
 V0  Nb  2
 V0  VC / 3 
RTf
9 TCVC
T f  T0 
 266.2 K
Vf 
 2.2 102 m3
20 V0
Patm


5
266.2
2.2 102
1

S  R ln
 R ln


5
.
24

10
 26  25.5 J/K
3
3 
2
273
 110  0.04 10 
P
T = const (< TC)
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).
V1
V2
V
F
V
F1
(liquid)
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:
V  V1 Nliquid V2  V
N  N gas  Nliquid


N
V2  V1
N
V2  V1
F
F
V  V2
V V
F  1 Nliquid  2 N gas  F1
 F2 1
N
N
V1  V2
V1  V2
N gas
F2
(gas)
- we recognize this as the common tangent line.
V < V1
V1 < V < V 2
V2 < V
As usual, the minimum free energy principle controls
the way molecules are assembled together.
P
Pvap
7
T < TC
Phase Separation in the vdW Model (cont.)
3
Since the tangent line F(V) maintains the same slope
between V1 and V2, the pressure remains constant
between V1 and V2:
 F 
6
4
2
1
5
V
F
V
Fliq

  P

V

T , N
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.
Fgas
P
P
critical
point
Pvap(T1)
triple
point
Vliq
Vgas
V
T1
T
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.
Phase Separation in the vdW Model (cont.)


T / TC
P / PC
n / nC
n / nC
P / PC
  3n 
 9
 n
1
 3nC 
 F 
C 

     k BT ln
 ln
 1 
 k BTC  

 N T ,V
 n
 3nC / n  1 4
  nQ 
 nC 
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.
P
The Maxwell Construction
T < TC
7
3
Pvap
6
[finding the position of line 2-6 without analyzing F(V)]
2
4
1
5
On the one hand, using the dashed line on the F-V
plot:
V
 F 
 F 
Fgas  Fliquid   
dV



 V2  V6 
V T , N
 V T , N
V6 
2
V
F
V
Fliq
  Pvap V2  V6 
On the other hand, the area under the vdW isoterm
2-6 on the P-V plot:
Fgas
 F 
PdV


V
V  V T , N dV   Fgas  Fliquid 
6
6
V2
G
4
5
3
Thus,
7
V2
 P V dV P V
vdW
2,6
vap
2
 V6 
V6
the areas 2-3-4-2 and 4-5-6-4 must be equal !
1
the lowest branch
represents the stable
phase, the other branches
are unstable
V2
P
Problem
P
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?
V
Vliq
Vgas
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.
V1
V2
V
P
Phase Diagram in T-V Plane
T < TC
Pvap
V
F
V
Fliq
Fgas
T
Single Phase
TC
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:
condensation:
unstable
interface:
V1(T)
P
2F
 0,
0
V
V 2
V2(T)
V
N1  4

3
  R / V1 
N 3

 4 R2
Total balance:
VC
 F
 F   4 R 2  F
F
N1  4

3
  R / V1 
N 3

RC
R
heating
Joule-Thomson Process for the vdW Gas
cooling
The JT process corresponds to an isenthalpic expansion:
 H 
 H 
H  
 T  
 P  0

T

P

P

T
 H 

  CP
 T  P
 H 
 V 

T

 V

T

P

T



T  
P

 
CP
CP
 P  H
 H 
CP T  
 P
 P T
H  TS  VP
 S 
 V 
   

 P T
 T  P
 H 
 S 

  T   V
 P T
 P T
(see Pr. 5.12)
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...).
We’ll consider the vdW gas at low densities:

N 2a 
 P  2 V  Nb   Nk BT
V 

N 2 a  V 
 V 
P
  2 
  Nk B

T
V

T

P

P
N 2a
P 
V2
V  Nb
N 2a

...P
PV 
 PNb  Nk BT
V
T
Nk B
 V 

 
2
 T  P P  N a
V2
 T 

 

P

H
2
Na
 Nb
k BT
CP
Joule-Thomson Process for the vdW Gas (cont.)
Cooling:
2a
b  0
k BT
Heating:
2a
b  0
k BT
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):
PV  Nb  RT
PV  RT  PNb
H  U  PV  CV  RT  PNb  CPT  PNb
The upper inversion temperature:
(at low densities)
heating
cooling
2a
b  0
k BTINV
TINV 
2a
27

TC
k Bb
4
(TC – the critical temperature
of the vdW gas, see below)
Substance
TINV
(P=1 bar)
CO2
(2050)
CH4
(1290)
O2
893
N2
621
H2
205
4He
51
3He
(23)
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
dFT ,N   SdT  PdV  dNT ,N  PdV
 V2  Nb 
 RT
N 2a 
1
2  1





F    PdV    
 2 dV N a     RT ln
V  Nb V 
 V1 V2 
 V1  Nb 
V1
V1 
V2
compare with
V2
F  U  TS

U vdW T  N 2 a  1 
 V1
1

V2 
TSvdW T
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Lecture 15. The van der Waals Gas (Ch. 5)