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Lecture 15. The van der Waals (Ch. 5)

The simplest model of a -gas transition - the van der Waals model of “real” – grasps some essential features of this phase transformation. (Note that there is no such transformation in the model). This will be our attempt to take intermolecular interactions into account. Nobel 1910 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 acceptance speech, van der Waals saw the qualitative agreement of his 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

4 gas into liquid at decreasing T and (or)

U(r) 3 increasing P - interaction between the 2 . 1

0 Two ingredients of the model: -1 the weak long-range attraction: the -2 long-range attractive between the -3 1.52.02.53.03.54.0molecules tend to keep them closer dirt together; these forces have the same effect r = σ as an additional compression of the gas. long-distance 2 aN 12 6 attraction eff PP += 2 ⎛ ⎞ ⎛σσ⎞ V ()rU ∝ ⎜ ⎟ − ⎜ ⎟ - the constant a is a measure of the long- ⎝ ⎠ ⎝ rr ⎠ range attraction

the strong short-range repulsion: the molecules are rigid: P →∞as soon as the molecules “touch” each other. - the constant b (~ 4πσ3/3) is a measure of the short-range eff −= NbVV repulsion, the “excluded ” per particle ⎛ 2aN ⎞ TNk 2aN ⎜ P + ⎟()=− TNkNbV P = B − The vdW equation ⎜ 2 ⎟ B 2 ⎝ V ⎠ ()− NbV V of state The van der Waals Parameters b – roughly the volume of a , (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).

Substance a’ b’ Pc Tc (J. m3/mol2) (x10-5 m3/mol) (MPa) (K) Air .1358 3.64 3.77 133 K 2 A ⇔ ' A ⇔ bbNaaN ' Carbon Dioxide (CO2) .3643 4.27 7.39 304.2 K ⎡ J m 6 ⎤ Nitrogen (N2) .1361 3.85 3.39 126.2 K ⋅ a' ⎢ 2 ⎥ ⎡ J ⋅ m 6 ⎤ mol Hydrogen (H2) .0247 2.65 1.30 33.2 K ⎣ ⎦ a ⎢ 2 ⎥ = 2 (H O) .5507 3.04 22.09 647.3 K 2 ⎣ molecule ⎦ ⎛ molecules ⎞ ⎜ N A ⎟ Ammonia (NH ) .4233 3.73 11.28 406 K 3 ⎝ ⎠ Helium (He) .00341 2.34 0.23 5.2 K

Freon (CCl2F2) 1.078 9.98 4.12 385 K ⎛ 2aN ⎞ When can ⎜ P + ⎟()=− TNkNbV be reduced to = TNkPV ? ⎜ 2 ⎟ B B ⎝ V ⎠ << VNb - low densities Na ⎛ ⎞ Na - high temperatures ⎜ BTkNPV −≈ ⎟ Tk >> ⎝ V ⎠ B V ( >> interaction energy) Problem

2 6 -2 -5 3 -1 The vdW constants for N2: NA a = 0.136 Pa·m ·mol , NAb = 3.85·10 m ·mol . How accurate is the assumption that Nitrogen can be considered as an ideal gas at normal P and T?

⎛ 2aN ⎞ ⎜ P + ⎟()=− TNkNbV ⎜ 2 ⎟ B ⎝ V ⎠

-2 3 -1 1 mole of N2 at T = 300K occupies V1mol ≈ RT/P ≈ 2.5 ·10 m ·mol

-5 3 -1 NAb = 3.9·10 m ·mol NAb/V1mol ~1.6%

2 2 6 -2 -2 3 -1 2 2 2 NA a / V = 0.135 Pa·m ·mol /(2.5 ·10 m ·mol ) = 216 Pa NA a / V P = 0.2% TNk aN 2 The vdW Isotherms P = B − ()− NbV V 2 ⎛ TNk ⎞ aN 2 abN 3 3 ⎜ NbV +− B ⎟V 2 V =−+ 0 ⎝ P ⎠ P P

The vdW isotherms in terms of the gas density n: 0 2 P 8 T ⎛ n ⎞ = − 3⎜ ⎟ N·b PC ⎛ n ⎞ TC ⎝ nC ⎠ 3⎜ ⎟ −1 2 n . ⎝ C ⎠ 1

=

C

T / T / T A real isotherm cannot have a negative slope dP/dn (or T C

@ a positive slope dP/dV), since this corresponds to a

s a hegative . The black isotherm at P / P g C l a T=T (the top panel) corresponds to the critical e C d i 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. n / nC The critical isotherm represents a boundary between those isotherms along which no such 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 2 2 and (d P/dV )T = 0 (see Pr. 5.48) ⎛ TNk ⎞ aN 2 abN 3 3 ⎜ NbV +− B ⎟ V 2 V =−+ 0 ⎝ P ⎠ P P

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

3 3 2 2 3 ( C ) C 33 C VVVVVVVV C =−+−=− 0 1 a 8 a Critical parameters: = 3 PNbV = Tk = C C 27 b2 CB 27 b ⎛ 3 ⎞⎛ 1 ⎞ 8 Tk ⎜ Pˆ + ⎟⎜Vˆ − ⎟ = B - in terms of P,T,V normalized by the critical parameters: ⎝ Vˆ 2 ⎠⎝ 3⎠ 3 - the materials parameters vanish if we introduce the proper scales.

RT 8 substance H2 He N2 CO2 H20 K C ==≡ 67.2 C RT /P V 3.0 3.1 3.4 3.5 4.5 VP CC 3 C C C TC (K) 33.2 5.2 126 304 647 - the critical coefficient PC (MPa) 1.3 0.23 3.4 7.4 22.1 Problems For Argon, the critical point occurs at a PC = 4.83 MPa and TC = 151 K. Determine values for the vdW constants a and b for Ar and estimate the diameter of an Ar atom.

2 1 a 8 a Tk CB 27 ( Tk CB ) PC = 2 Tk CB = b = a = 27 b 27 b 8PC 64 PC

2 −23 2 27 ()Tk CB 27 ( 1038.1 ×⋅ 151J/K K ) − 249 a = = 6 ⋅= 108.3 /PaJ 64 PC 64 ⋅1083.4 Pa

−23 Tk CB 1038.1 ×⋅ 151J/K K − 329 3/1 −10 b == 6 ⋅= 104.5 m b =⋅= 86.31086.3 Am 8PC ⋅× 104.838 Pa Per mole: a=0.138 Pa m6 mol-2; b=3.25x10-5 m3 mol-1 Energy of Let’s start with an ideal gas (the ⎛∂U ⎞ dilute limit, interactions can be UU += ⎜ ⎟ dV the vdW Gas neglected) and compress the gas vdW ideal ∫ =constT ⎝ ∂V ⎠T (low n, high T) to the final volume @ T,N = const: ⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎡⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎤ ⎛ ∂P ⎞ ⎜ ⎟ []=−== TPdVTdSdU ⎜ ⎟ P =− ⎢⎜ ⎟ = ⎜ ⎟ ⎥ = T ⎜ ⎟ − P ⎝ ∂V ⎠ T ⎝ ∂V ⎠ T ⎣⎝ ∂V ⎠ T ⎝ ∂T ⎠V ⎦ ⎝ ∂T ⎠V (see Pr. 5.12) From 2 2 2 BTNk aN ⎛ ∂ P ⎞ Nk 1 ⎛ aN ⎞ ∂U aN the vdW P = − = B ⎜ P += ⎟ ⎛ ⎞ 2 ⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎟ = 2 equation: ()− NbV V ⎝ ∂ T ⎠ V − TNbV ⎝ V ⎠ ⎝ ∂V ⎠ T V

⎛ ∂U ⎞ f 2 aN Na UU += dV = TkN − vdW ideal ∫ ⎜ ⎟ B vdW ideal −= NUU = constT ⎝ ∂V ⎠ T 2 V 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 isothermal expansion of the vdW gas: ⎛ 11 ⎞ ()2aNU ⎜ −=Δ ⎟ vdW T ⎜ ⎟ ⎝ VV fi ⎠ Isothermal Process for the vdW ⎛ 11 ⎞ ()=Δ 2 aNU ⎜ − ⎟ gas at low n and/or high T vdW T ⎜ ⎟ ⎝ VV fi ⎠ V V f f ⎛ TNk 2aN ⎞ ⎛ − NbV ⎞ ⎛ 11 ⎞ δW PdV −=−= ⎜ B − ⎟ −= TNkdV ln⎜ f ⎟ 2aN ⎜ −+ ⎟ ∫∫ ⎜ − NbV V 2 ⎟ B ⎜ − NbV ⎟ ⎜ VV ⎟ Vi Vi ⎝ ⎠ ⎝ i ⎠ ⎝ fi ⎠ ⎛ − NbV ⎞ ⎜ 2 ⎟ δ +=Δ δ WQU δ −21 = TNkQ HB ln⎜ ⎟ ⎝ 1 − NbV ⎠

2 2 For N2, the vdW coefficients are N a = 0.138 kJ·liter/mol 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.

2 ⋅lkJ ⎛ 1 1 ⎞ J ⎛ ⋅− /0385.0317.0 mollmoll ⎞ ()WvdW mol ⋅=Δ 138.09 ⎜ − ⎟ mol⋅− 3.83 ⋅ K ln310 ⎜ ⎟ T 2 ⎝ 4.3 17.0 llmol ⎠ ⋅molK ⎝ ⋅− /0385.034.3 mollmoll ⎠ =+−= 7.2464.3194.6 kJkJkJ ⎛V ⎞ J ⎛ 17.0 l ⎞ ()−=Δ TNkW ⎜ 2 ⎟ mol⋅−= K⋅⋅ ln3103.83ln = 12.23 kJ ideal T B ⎜ ⎟ ⎜ ⎟ ⎝V1 ⎠ ⋅Kmol ⎝ 4.3 l ⎠ U pot 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 ΔW < ΔW r vdW ideal 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 T = const Uideal U U ideal U f 2 aN U = TkN − vdW 2 B V UvdW UvdW 2 aN = TC − 2 aN idealV T V V 2 aN U ideal − V V vdW = CC V ideal

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 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 For 1 mole of the “vdW” Nitrogen (diatomic gas) aN 5 aN vdW (), 00 UVTU ideal RT0 −=−= V0 2 V0 5 For 1 mole of the “ideal” Nitrogen (after expansion): = RTU Tf is the temperature ideal 2 f after expansion The of the gas is conserved in this 5 2aN 5 2 2aN process (δQ = 0, δ W = 0), and, thus, ΔU = 0: RT0 =− RTf f TT 0 −= 2 V0 2 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 ⎞ ⎛ ∂S ⎞ CV ⎛ ∂P ⎞ f dT Nk B S dS = ⎜ ⎟ dT + ⎜ ⎟ dV dT += ⎜ ⎟ dV = Nk B + dV ⎝ ∂T ⎠V ⎝ ∂V ⎠T T ⎝ ∂T ⎠V 2 T − NbV f (see Pr. 5.12) S = ln + ln()+− constNbVNkTNk vdW 2 B B - the same “volume” in the ⎧ 2/3 ⎫ ⎛ Nb⎞ ⎪ ⎡()− NbV ⎛ 2π m ⎞ ⎤ 5⎪ momentum space, smaller vdW ideal NkSS B ⎜1ln −+= ⎟ = kN B ⎨ln⎢ ⎜ BTk ⎟ ⎥ + ⎬ 2 accessible volume in the ⎝ V ⎠ ⎩⎪ ⎣⎢ N ⎝ h ⎠ ⎦⎥ 2⎭⎪ coordinate space.

2/3 3 2aN ⎪⎧ ⎡()− NbV ⎛ 2π m ⎞ ⎤ 5⎪⎫ F vdW BTNkTSUF −−=−= BTkN ⎨ln⎢ ⎜ 2 BTk ⎟ ⎥ + ⎬ 2 V ⎩⎪ ⎣⎢ N ⎝ h ⎠ ⎦⎥ 2⎭⎪ 2/3 ⎡()− NbV ⎛ 2π m ⎞ ⎤ aN 2 −= BTkN ln⎢ ⎜ 2 B ⎟ ⎥ BTNkTk −− ⎣⎢ N ⎝ h ⎠ ⎦⎥ V Nb 2aN 2 ⎛ ⎞ ⎛ ∂F ⎞ BTNk aN vdW ideal −= BTkNFF ⎜1ln − ⎟ − P −= ⎜ ⎟ = − V V vdW 2 ⎝ ⎠ ⎝ ∂V ⎠T − NbV V ⎡ 2/3 ⎤ 2 ()− NbV ⎛ 2π m ⎞ 2aN BTVNk G vdW −=+= BTkNPVFG ln⎢ ⎜ 2 B ⎟ ⎥ BTNkTk +−− ⎣⎢ N ⎝ h ⎠ ⎦⎥ V − NbV 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. δ Q P A e 1−= C δ QH

B A-D Δ = ( −TTcQ ) D DAVH ( −TT ) e 1−= CB ()−TT DA B-C Δ = ( −TTcQ CBVC ) C V V The relationship between TA, TB,TC,TD –from V1 2 the adiabatic processes B-C and D-A

f f Tf f − NbV SvdW = B ln + B ln()+− constNbVNkTNk SvdW =Δ NkB ln + NkB ln 2 2 Ti i − NbV

f f Tf f − NbV 2 NkB ln + NkB ln = 0 ()=− constNbVT adiabatic process 2 Ti i − NbV for the vdW gas /2 f /2 f ⎛ − NbV ⎞ ⎛ − NbV ⎞ T ⎜ 1 ⎟ −T ⎜ 1 ⎟ A ⎜ ⎟ D ⎜ ⎟ /2 f / cR V ΔQC ()−TT CB ⎝ 2 − NbV ⎠ ⎝ 2 − NbV ⎠ ⎛ 1 − NbV ⎞ ⎛ 1 − NbV ⎞ e −= 11 −= 1−= 1−= ⎜ ⎟ 1−= ⎜ ⎟ ΔQH ()−TT DA −TT DA ⎝ 2 − NbV ⎠ ⎝ 2 − NbV ⎠ 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 (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 .

f S = lnln ()+−+ mNfNbVRTR ),( vdW 2

5 T f ⎛ V f ⎞ 5 T f ⎛ V f ⎞ =Δ RS + R lnln ⎜ ⎟ = R + R lnln ⎜ ⎟ 2 T0 ⎝ 0 − NbV ⎠ 2 T0 ⎝ 0 −VV C 3/ ⎠

9 VT CC RT f − 32 f TT 0 −= = 2.266 K V f ⋅== 102.2 m 20 V0 Patm 5 2.266 ⎛ ⋅102.2 −2 ⎞ =Δ RS ln + R ln⎜ ⎟ −1 ≈+⋅−= 5.25261024.5 J/K ⎜ −3 −3 ⎟ 2 273 ⎝ ⋅−⋅ 1004.0101 ⎠ Phase Separation in the vdW Model T = const (< TC) P 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 V V 1 2 V 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 F tangent line lies below the free energy curve, molecules can V minimize their free energy by refusing to be in a single homogeneous phase in the region between V and V , and by F1 1 2 (liquid) preferring to be in two coexisting phases, gas and liquid: N N F2 gas V −V1 liquid V2 −V (gas) N = N gas + Nliquid = = N V2 −V1 N V2 −V1

F1 F2 −VV 2 1 −VV F Nliquid gas =+= FN 1 + F2 N N −VV 21 −VV 21 - we recognize this as the common tangent line.

As usual, the minimum free energy principle controls V < V1 V1 < V < V2 V2 < V the way molecules are assembled together. P 7 T < TC Phase Separation in the vdW Model (cont.) 3 Since the tangent line F(V) maintains the same slope 6 2 between V1 and V2, the pressure remains constant Pvap 4 between V1 and V2: ⎛ ∂F ⎞ 1 ⎜ ⎟ −= P 5 ⎝ ∂V ⎠ ,NT V In other words, the line connecting points on the PV F plot is horizontal and the two coexisting phases are in a V mechanical equilibrium. For each temperature below

Fliq TC, the phase transformation occurs at a well-defined pressure Pvap, the so-called vapor pressure.

Fgas Two stable branches 1-2-3 and 5-6-7 correspond to different phases. Along branch 1-2-3 V P P critical is large, P is small, the density point is also small – gas. Along branch 5-6-7 V is small, P is large, the density is large – Pvap(T1) liquid. Between the branches – the gas-liquid phase triple transformation, which starts point even before we reach 3 moving along branch 1-2-3. Vliq Vgas V T1 T n C PP / ⎞ ⎟ ⎟ ⎠ C n n ⎛ ⎜ ⎜ ⎝ CB Tk 4 9 − ⎤ ⎥ ⎥ ⎦ μ 1/3 − C nnn nn . The outer two values of 1 / C μ order transition. When we move st + ⎞ ⎟ ⎠ 1 rds the critical point, transition −− C n 3 ⎛ ⎜ ⎝ with the same ln

μ n ⎞ ⎟ ⎟ ⎠ C Q n n 3 ⎛ ⎜ ⎜ ⎝ and at the critical point, abruptness disappears. es which are in equilibrium with each other. C ln ⎡ ⎢ ⎢ ⎣ TT / Tk B = curve is a signature of the 1 VT , ( V ) ⎞ ⎟ ⎠ G F N ∂ ∂ ⎛ ⎜ ⎝ = C Phase Separation in the vdW Model (cont.)

μ , there are three values of

nn

C

2

.

1 / C

=

T

/

T

@ < T

s s

a

g

l

a T e d i The kink on the correspond to two stable phas along the gas-liquid coexistence curve towa becomes less and abrupt, For C PP / P 7 T < TC The Maxwell Construction

3 [finding the position of line 2-6 without analyzing F(V)] 6 P 2 vap 4 On the one hand, using the dashed line on the F-V 1 plot: 5 V 2 ⎛ ∂F ⎞ ⎛ ∂F ⎞ FF =− ⎜ ⎟ dV =⎜ ⎟ ()−VV V gas liquid ∫ ∂V ∂V 62 F V6 ⎝ ⎠ ,NT ⎝ ⎠ ,NT V vap ()−−= VVP 62 Fliq On the other hand, the area under the vdW isoterm 2-6 on the P-V plot: F V V gas 2 2 ⎛ ∂F ⎞ PdV −= ()−−= FFdV ∫∫⎜ ⎟ gas liquid ⎝ ∂V ⎠ ,NT G 3 V6 V6 4 V2 7 Thus, ()−= (VVPdVVP ) 5 2,6 ∫ vdW vap 62 V6 the areas 2-3-4-2 and 4-5-6-4 must be equal ! 1

the lowest branch P represents the stable phase, the other branches are unstable 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 T < TC Phase Diagram in T-V Plane

At T >TC, the N molecules can exist in a single phase in Pvap 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 V F for a homogeneous phase. V There are two regions within the two-phase “dome”:

Fliq 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, Fgas though the system would decrease the free energy by phase separation, it should overcome the potential T barrier first. Indeed, when small droplets with radius R are Single Phase initially formed, an associated with the surface energy term TC tends to increase F. The F loss (gain) per droplet:

N1 ⎛ 4 3 ⎞ m condensation: − ΔF ⎜ π R /V1 ⎟ e N 3 t ⎝ ⎠ le unstable a b s 2 a t t a interface: γ 4π R s b a t le e

m Total balance: δ F RC R VC V V (T) V2(T) N ⎛ 4 ⎞ 1 δ F = γ 4π R2 − ΔF 1 ⎜ π R3 /V ⎟ ∂P ∂2 F N 3 1 = 0, = 0 ⎝ ⎠ ∂V ∂V 2 heating Joule-Thomson Process for the vdW Gas cooling The JT process corresponds to an isenthalpic expansion:

⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎛ ∂H ⎞ H =Δ ⎜ ⎟ T +Δ ⎜ ⎟ P =Δ 0 ⎜ ⎟ = P P TCC −=Δ ⎜ ⎟ ΔP ⎝ ∂T ⎠P ⎝ ∂P ⎠T ⎝ ∂T ⎠P ⎝ ∂P ⎠T

⎛ ∂H ⎞ ⎛ ∂S ⎞ ⎛ ∂H ⎞ ⎛ ∂V ⎞ Δ+Δ=Δ PVSTH ⎜ ⎟ = T⎜ ⎟ +V − ⎜ ⎟ T ⎜ ⎟ − V ∂P ∂P ⎛ ΔT ⎞ ⎝ ∂P ⎠ ⎝ ∂T ⎠ ⎝ ⎠T ⎝ ⎠T ⎜ ⎟ = T = P ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜ ⎟ −= ⎜ ⎟ ⎝ ΔP ⎠ H C P C P (see Pr. 5.12) ⎝ ∂P ⎠T ⎝ ∂T ⎠P 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...). 2 aN We’ll consider the vdW gas at low densities: P >> >> NbV V 2 ⎛ 2 aN ⎞ 2 aN ∂ ⎜ P + ⎟ =− PVTNkNbV =−+ TNkPNb ... ⎜ 2 ⎟()B B ()P ⎝ V ⎠ V ∂T

Na 2 2 − Nb ⎛ ∂V ⎞ aN ⎛ ∂V ⎞ ⎛ ∂V ⎞ Nk B P⎜ ⎟ − ⎜ ⎟ = Nk ⎜ ⎟ = ⎛ ΔT ⎞ BTk 2 B 2 ⎜ ⎟ ≈ ⎝ ∂T ⎠ P V ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P aN P − ⎝ ΔP ⎠ H C P V 2 Joule-Thomson Process for the vdW Gas (cont.) 2a 2a Cooling: b >− 0 Heating: b <− 0 BTk BTk 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):

()=− = − = + = ()V + + = P + PNbTCPNbTRCPVUHPNbRTPVRTNbVP

The upper : (at low densities) T Substance INV 2a (P=1 bar) heating b =− 0 CO2 (2050) Tk INVB

CH4 (1290)

cooling a 272 O2 893 TINV == TC B bk 4 N2 621

H2 205 (TC – the critical temperature 4He 51 of the vdW gas, see below) 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 .

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

()dF ,NT −= (− + μdNPdVSdT ),NT = −PdV

V2 V2 ⎛ RT 2aN ⎞ ⎛ 11 ⎞ ⎛ − NbV ⎞ PdVF −=−=Δ ⎜ − ⎟ 2aNdV ⎜ −= ⎟ − RT ln⎜ 2 ⎟ ∫∫ ⎜ − NbV V 2 ⎟ ⎜ VV ⎟ ⎜ − NbV ⎟ V1 V1 ⎝ ⎠ ⎝ 21 ⎠ ⎝ 1 ⎠

compare with = −TSUF ⎛ 11 ⎞ ()2aNU ⎜ −=Δ ⎟ ( ΔST ) vdW T ⎜ ⎟ vdW T ⎝ VV 21 ⎠