the van der waals gas
TRANSCRIPT
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.
TNkNbVVaNP 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 4.0distance
Ene
rgy
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 (~ 43/3) is a measure of the short-range repulsion, the “excluded volume” per particle
612
rrrU
r =
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 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
molemolecules
molm J
moleculem J
A
AA
N
aa
bbNaaN
'
''2
TNkNbVVaNP B
2
2
When can TNkPV Bbe 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?
TNkNbVVaNP 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 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)
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
21VaNP
TNbVNk
TP 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
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
fiTvdW VV
aNU 112
fii
fB
V
V
BV
V VVaN
NbVNbV
TNkdVVaN
NbVTNkPdVW
f
i
f
i
11ln 22
2
NbVNbVTNkQ HB
1
221 lnWQU
kJkJkJmollmollmollmollK
molKJmol
llmollkJmolW TvdW
7.2464.3194.6/0385.034.3/0385.0317.0ln3103.83
17.01
4.31138.09 2
2
kJllK
KmolJmol
VVTNkW BTideal 12.23
4.317.0ln3103.83ln
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.
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 (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
00
2
00 25,
VaNRT
VaNUVTU idealvdW
fRTVaNRT
25
25
0
2
0
Tf is the temperature after expansion
0
2
0 52RV
aNTT f
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 G
S, F, and G for the monatomic van der Waals gas
F
VaNTNkTk
hm
NNbVTkN
Tkhm
NNbVTkN
VaNTNkTSUF
BBB
BBBvdW
22/3
2
2/3
2
2
2ln
252ln
23
2
2
VaN
NbVTNk
VFP B
TvdW
S dVNbV
NkTdTNkfdV
TPdT
TCdV
VSdT
TSdS B
BV
V
TV
2
252ln1ln
2/3
2 Tkhm
NNbVkN
VNbNkSS BBBidealvdW
- 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
B-C CBVC TTcQ
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
H
C
QQe
1
DA
CB
TTTTe
1
constNbVNkTNkfS BBvdW lnln2 NbV
NbVNk
TT
NkfSi
fB
i
fBvdW
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
3/
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
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.
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: P
VF
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
VF
Fliq
Fgas
Cnn / CPP /
C
CBC
C
Q
CB
VT nnTk
nnnn
nnTk
NF
49
1/3113ln3ln
,
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
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-V plot:
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 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:
metastable
met
asta
ble unstable
V1(T) V2(T)
0,0 2
2
VF
VP
condensation:
1
31 /34 VR
NNF
interface: 24 R
R F
1
312 /344 VR
NNFRF
Total balance: RC
P
V
Pvap
T < TC
VF
Fliq
Fgas
Joule-Thomson Process for the vdW Gas
0
PPHT
THH
TP
The JT process corresponds to an isenthalpic expansion:
PPHTCC
TH
TPP
P
VPST
PHPVSTH
TT
(see Pr. 5.12)P
P
P
T
H C
VTVT
CPH
PT
We’ll consider the vdW gas at low densities: NbVVaNP 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
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
Substance TINV
(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 ,,
NbV
NbVRTVV
aNdVVaN
NbVRTPdVF
V
V
V
V 1
2
21
22
2
ln112
1
2
1
21
2 11VV
aNU TvdW
compare with TSUF TvdWST