Applications

PHY474

Applications

Ulrich Z¨urcher∗ Physics Department, Cleveland State University, Cleveland, OH 44115 (Dated: November 14, 2008)

PACS numbers: 05.70

I. PHASE TRANSFORMATION OF PURE not coexist (in equilibrium): ice directy sublimates into SUBSTANCES vapor. Sublimation of dry ice [carbondioxide]. the triple point lies above atmospheric pressure. Most substances: The curve of pressure vs volume for a quantity of applying more pressure raises the melting temperature. matter at constant temperature is determined by the Ice is different: applying more pressure lowers its melting Gibbs free energy of the substance. The curve is called temperature: this is related to the fact that density of ice isotherm. We consider the isotherms of a real gas in is lower than density of liquid. which atoms associate together in a liquid or solid phase. Liquid-gas phase boundary always has a poisitive A phase is a prortion of the system that is uniform in slope: if you increase temperature you must increase composition. pressure to prevent liquid from vaporizing. As pressure Two phases may coexist with a definite boundary be- increases, density of gas is increasing so gas and liquid tween them. An isotherm may show a region in the PV become indistinguishable. Eventually there is a point when there is no longer a discontinuous transition: crit- diagram in which liquid and gas coexist in equilibrium ◦ with each other. There are isotherms at low tempera- ical point. For water it is 374 C and 221 bars. Close to tures for which solids and liquids coexist and isotherms the critical point we call the system “fluid.” for which solids and gases coexist. Everything we say The thermodynamic condition for the existence of of for liquid-gas equilibrium equilibrium also holds for the two phases are the conditions for the equilbrium of two solid-gas and solid-liquid equilibrium. systems that are in thermal, diffusive, and mechanical We use vapor for a gas that coexist with its liquid or contact. The conditions Tl = TG and Pl = Pg are auto- solid form. Liquid and vapor coexist on a section of an matic. Diffusive equilibrium gives µl = µg. The chemical isotherm only if the temperature of the isotherm lies be- potentials are evaluated at the common temperature and low a critical temperature. Above the critical tempera- pressure, ture only a single phase - the fluid phase - exist no matter how great the pressure. No reason to call it either liuqid µl(T,P ) = µg(T,P ). (1) or gas, therefore fluid. For water, H2O, the critical tem- ◦ ◦ perature is 641 K, or 374 C and for CO2 Tc = 31 C. At a general point in the PT palne, the two phases do Liquid and gas will never coexist along entire isotherm not coexist. If µl < µg then the liquid phase is stable, from zero to infinite pressure. For fixed temperature and and if µg < µl the gas phase alone is stable. Metastable number of atoms, there is a finite volume above which phases may occur by superheating or supercooling. A all atoms present are in the gas phase. A small drop of metastable phase may have a transient existence, some- water placed in an evacuated jar at room temperature times brief, sometimes long at a temperature for which will evaporate entirely leaving the jar filled with H2O another phase of the same substance has a lower chemical gas at some pressure. A drop of water exposed to air not potential. entirely saturated with moisture may evaporate entirely. There is a concentration of water, however, above which the atoms from the vapor will bind themselves into a liquid drop. A. Clausius-Clapeyron equation The lines on a phase diagram represent conditions un- der which two different phases can coexist. At 1 atm, ice We consider Consider phase boundar between liquid and water coexist at 0◦C. The pressure at which a gas and gas. At phase boundary, liquid and gas are in diffu- can coexist with its solid or liquid phase is called vapor sive equilibrium so their chemical potentials must be the pressure. The vapor pressure of water at room temper- same: ature is approximately 0.03 bar. At T = 0.01◦C and P = 0.06 bar, all three phases can coexist: this is called µ P ,T ) = µ (P ,T ), at phase boundary, (2) the triplet point. At lower pressures, liquid water can- g 0 0 l 0 0 and also

∗ Electronic address: [email protected] µg(P0 + dP, T0 + dT ) = µl(P0 + dP, T0 + dT ). (3) 2

3 Now use a series expansion on both sides At atmospheric pressure vg/vl ' 10 . We thus set ∆v ' vg. We assume that the gas law can be applied to ∂µ ∂µ g g the gas phase: PVg = NgkT so that µg(P0,T0) + dP + dT + ... ∂P T ∂T P Vg kT ∂µl ∂µl ∆V ' vg = = . (12) = µl(P0,T0) + dP + dT + ... (4) Ng P ∂P T ∂T P This gives This gives: dP l ∂µg ∂µl ∂µg ∂µl = P, (13) − dP = − − dT dT kT 2 ∂P T ∂P T ∂T P ∂T P (5) or This gives the differential equation for the coexistence d ln P l curve of vapor pressure curve: = 0 . (14) dT kT 2 ∂µg ∂µl ∂T − ∂T Here we set l = l0 assuming that the latent heat doesn’t dP P P = (6) change significantly. Integration gives dT ∂µl ∂µg ∂P − ∂P T T l ln p = − 0 + const. (15) We can relate the derivatives to physical quantities. kT Since G = Nµ, we have If L0 is the latent heat per mole, we get ∂µ V ∂µ S = = v, = − = s. (7) L ∂P N ∂T N P (T ) = P exp − 0 . (16) T P 0 RT Thus the differential equation for the existence curve becomes: From the Final exam in F07: 4.) You are invited next summer to a climb the famous dP s − s = g l . (8) Matterhorn in the Swiss Alps with height of 4,478 m dT vg − vl above sea level. It’s generally known that the atmospheric pressure decreases on mountain tops which, in Here sg −sl is the increase in entropy of the system when turns, affects the boiling temperature of water. Estimate we transfer one molecule from the liquid to the gas phase, the boiling temperature of water on top of Matterhorn! and vg − vl is the increase in volume of the system when Make reasonable approximations along your calculation, we transfer one molecule from the liquid to the gas phase. and explain. You can assume, for example, that the It is important to understand that dp/dT is not simply temperature of the atmosphere does not change with taken from the equation of state of the gas [such as the the height above sea level. Our calculation is in two steps. ideal gas law PV = NkT ]. The derivative refers to the very special interdependent change of P and T in which a) Calculate the pressure at the summit. the gas and liquid will continue to coexist. The number of molecules in the gas and liquid will vary, subject to We have for the pressure dependence: Ng + Nl = N = const. The change in entropy is related to heat that must be mgz Mgz P (z) = P exp − = P exp − added to the system to transfer one molecule reversibly 0 kT 0 RT from the liquid to the gas while keeping temperature constant, where M = 28.8 × 10−3 kg/mol is the molar mass. Thus dQ = T (s − s ) = l. (9) ! g l 28.8 × 10−3 kg/mol · 9.8 m/s2 · 4, 478 m P = 1.0 atm exp − This heat is called the latent heat of vaporization and can 8.3 J/(K · mol) · 273 K be measured with calorimetry. The change of volume: That is P = 0.57 atm. ∆v = vg − vl. (10) b.) Now, estimate the boiling temperature of water at We then have the summit. dP l = . (11) dT T ∆v We have Clausius-Clapeyron equation: This is the Clausius Clapeyron equation or vapor pres- dP L = , sure equation. dT T (Vg − Vl) 3

3 where we can set T ' 373 K. For N2 we have a ' 2.5 eV · A . We have for the latent heat of vaporization: L = 22.6 × This gives 105 J/kg, or per mole NkT aN 2 J P = − 2 . (20) L = 4.07 × 104 . V − Nb V mol This is van der Waals. For the molar volumes, we have for the liquid Vl = We introduce 18 ml = 1.8 × 10−5 m3. Use the ideal gas law, for the 1 a 8 a gas V = 3Nb, P = , kT = , (21) c c 27 b2 c 27 b 373 K 1.0 atm V ' 22.4 L · · = 50.0 L = 5.0 × 10−2 m3. g 293 K 0.57 atm and dimensionless variables P V T Note that Vg >> Vl so that p = , v = , t = . (22) Pc Vc Tc dP 4.1 × 104 J Pa ' = 2.2 × 103 , The van der Waals equation follows dT 373 K · 5.0 × 10−2 m3 K 3 1 8 We thus have for the change in the boiling temperature, p + v − = t, (23) v2 3 3 dP −1 Pa−1 ∆T = ∆P = 2.2 × 103 ·−4.3 × 104 Pa 'or −20 K. dT K (8/3)t 3 ◦ p = − . (24) That is, the water on the summit boils at 80 C [or v − 1/3 v2 176◦F] and not at the standard 100◦C [or 212◦F]. That is, different substances have different values of Pc, Note: This is not a real problem for making a soup, of Vc and Tc. in terms of reduced variables p, v and t all course. However, do not attempt to cook a nice bowl of real gases behave the same: corresponding states of the spaghetti at the summit! substances. For high temperatures t alomost the same as ideal gas. As the temperature is lowered we get a loop: We start at low volume and high pressure, pressure falls as volume B. Van der Waals Equation of State increases, dp/dv = 0 at point 5, pressure increases as volume increases between 5 and 3, dp/dv = 0 at 3, and then Liquid -gas transformation can be understood from the pressure decreases as volume increases. For mechanical van der Waals equation: stability we need that the pressure decreases as volume increases: aN 2 P + (V − Nb) = NkT. (17) dp V 2 < 0, mechanical stability. (25) dv In a real gas, atoms or molecules interact with each other: So we see that the lobe 5 to 3 is unphysical. Actually short range repulsion and long range attraction. Two more is wrong. modifications from ideal gas law. (1) Short range repul- In addition to mechanical stability we require that the sion tells us that a fluid cannot be compressed all the way system is thermal equilibrium: Gibbs free energy is min- to zero: minimum value is Nb. The parameter b repre- imized. At constant number of particles and at constant sents the minimum volume occupied by a single molecule. 3 temperature dG = V dP so that For small molecules like N2 and H2O we have b ' (4 A) . Now the attractive part of the interaction. The potential ∂G ∂P energy of one molecule due to interactions with all its = V . (26) ∂V ∂V neighbors is proportional to the density N/V . Since we N,T N,T have N molecules, the total potential energy is Integration gives the Gibbs free energy aN 2 total potential energy = − . (18) (NkT )(Nb) 2aN 2 V G = −NkT ln(V − Nb) + − + c(T ). V − Nb) V Negative sign because the interaction is attractive. The (27) pressure follows from P = −∂U/∂V or Plot of G vs P shows a loop: points 2 and 6 coincide: that is system will go from point 2 (gas) to 6 (liquid) with an d aN 2 aN 2 abrupt change in the volume: phase transformation. P 0 = − − = − . (19) dV V V 2 At intermediate volumes, the thermodynamically stable 4 state is a combination of gas and liquid: coexistence of We write the chemical potential in terms of the (par- liquid and gas phase. tial) pressure We note that there is no change in the Gibbs free en- o o ergy as we move from 2 to 6: µv = µv + kT ln(P/P ), (34)

Z Z ∂G Z where P o is a reference pressure. We take P 0 as the 0 = dG = dP = V dP. (28) nominal vapor pressure, that is for a liquid with ﬂat sur- loop loop ∂P loop N,T face, then RH = P/P o is called the relative humidity and o That is the area above and below the horizontal line must µv = µl. This Thus, for the critical radius be equal: Maxwell construction. Note that this deter- 2σvl mines the vapor pressure. rc = , (35) In PV diagram we have an area of coexistence between kT ln(RH) liquid and gas. Within that area there is a region where or for the relative humidity: liquid and gas is mechanically stable but thermodynamically unstable: supercooling and superheating. These are 2σv 2σV metastable phases. In PV -diagram: region of metasta- RH = exp l = exp l , (36) kT r RT r bility. Spinodal: line separating between mechanically c c stable and unstable phases. where Vl is the liquid volume per mole. This is known as the Kelvin equation. For water at 20◦ C:

II. NUCLEATION 2σV 2 · 0.073 J/m2 · 18 × 10−6 m3 l = = 1.08 nm. (37) RT 8.315 J/K · 293 K For a pure system G = Nµ thus for a composite, e.g., water and vapor So the Kelvin equation becomes

G = N µ + N µ = Nµ + N (µ − µ ), (29) 1.08 nm l l v v v l l v r = , RH = e1.08 nm/rc . (38) c ln(RH) where N = Nl + Nv is the total number of molecules. If vl is the volume per molecule in the liquid, then Nl is the Note that RH is must be greater than 100%. Humdidity ratio of the total volume of the liquid to vl. We assume in atmosphere is never much greater than 100%. Thus a spherical droplet: droplets smaller than 10 nm should never be stable ac- cording to this analysis. Note that a 10 nm droplet con- 3 4πr tains about 105 molecules so it could not form sponta- G = Nµv + (µl − µv). (30) 3vl neously. So there must be another process involving nucleation [crystallization]. The Gibbs free energy associated with the surface is the In fact, we have discussed here so far is homogeneous surface area times the area: nucleation. In application to atmospheric problem, e.g., 3 cloud formation, this is mostly irrelevant. There is an 4πr 2 G = Nµv + (µl − µv) + 4πr σ. (31) abundance of dust particles in the atmosphere on which 3v l vapor particles condensate: condensation nuclei. One

We have two cases. i) Consider µl > µv then the then speaks of heterogeneous condensation. Gibbs free energy is increasing for all radiis. ii) For µv > µl, then the surface term dominates for small radius dG/dr > 0, while the bulk term dominates for large III. SURFACE TENSION radius, dG/dr < 0. There is a critical radius rc for which dG/dr = 0. If r < rc then G(r) > G(0) which means that In the “bulk,” molecules are surrounded on all sides a droplet with radius r has a higher Gibbs free energy by same molecules. On surfaces, forces are diﬀerent and than no droplet at all, that is the droplet will evaporate. forces are not ‘balanced,’ i.e., they point “inwards.” Thus If r > rc then we can reduce the Gibbs free energy by particles [molecules] near surfaces have diﬀerent behavior making the droplet even bigger. That is, the droplet will than particles in the bulk. If we want to increase the grow until there’s no vapor left. We have surface area we have to exert a force F : work done is dW = F dx. if the surface has a width l [perpendicular 2 dG 4πrc to dx] the force F is proportional to l, 0 = = − (µv − µl) + 8πrcσ. (32) dr r vl c F = γl, (39) Thus solving for rc where γ is the surface tension. Thus, we get the work 2σvl rc = . (33) µv − µl dW = γldx = γdA. (40) 5

If the work is done reversibly, the work done equals the so that dA/dV = 2/r thus change in Gibbs free energy, dG = γdA. This can be 2γ used to define the surface tension as ∆P = liquid drop. (50) r ∂G γ = . (41) We want to derive this equation for a doplet by elemen- ∂A T,P tary means: The “inward” force due to the surface tension is Fin = −γ · 2πr. The “outward” force due to pres- This gives 2 sure difference is Fout = ∆P · 4πr . The net force is dG = −SdT + V dP + γdA. (42) zero: 2 Note that a film has two surfaces: F = Fin + Fout = −γ2πr + ∆P · 4πr = 0. (51) Note that for a spherical soap bubble we have an inner F = γ2l. (43) film and outer interface: Liquid γ (N/m) 4γ ◦ ∆P = soap bubble. (52) Benzene (20 C) 0.029 r Blood (37◦C) 0.058 Glycerin (20◦C) 0.063 Problem: Water is stuck between two parallel glass Mercury (20◦C) 0.47 plates. The distance between the plates is d, and the Water (20◦C) 0.073 diameter of the trapped water ‘disc’ is D >> d. What is Water (100◦C) 0.059 the force acting between the plates? The cross-sectional In chapter III, we consider two systems in thermal, edge of the disc of water is a semicircle of radius r = d/2. mechanical, and diffusive equilibrium. For mechanical This gives a pressure difference ∆P = 2γ/d. Thus the equilibrium we found force acting between the plates is: πD2 2γ dU1 = −P1dV1, dU2 = −P2dV2. (44) F = . (53) 4 d We considered bulk properties only and ignored “surface For D = 1 m and d = 1 mm, this force is: F ' 100 N. effects”:

dU = 0 = −P1dV1 − P2dV2 = (P2 − P1)dV1, (45) A. Young Dupr´eEquation since We consider a “sessile” drop,i.e., a liquid drop in equi- dV1 = −dV2. (46) librium on a (plane) solid surface. We have a solid S, liquid L and vapor V in equilibrium. We have three sur- We thus have P1 = P2. If we include the interface we faces: liquid-vapor ALV, liquid-solid ALS and solid-vapor include the change in internal energy due to the surface: ASV. The change in the Gibbs free energy is:

dUs = γdA. (47) dG = γLVdALV + γLSdALS + γSVdASV. (54)

We then have We have

dU = 0 = dU1 + dU2 + dUs dALS = −dASV. (55) = −P1dV1 − P2dV2 + γdA Thus = (P2 − P1)dV + γdA. (48) dG = γLVdALV + (γLS − γSV)dALS. (56) Thus Note that (γ − γ ) can be either positive or nega- LS SV ∂A tive. i) If solid-liquid interactions are negligible, the en- P1 − P2 = γ . (49) ∂V1 ergy associated with the liquid-solid interface is practi- cally the same as the sum of energies associated with This tells us how the pressure between region 1 [eg the in- the separate liquid-vapor and solid-vapor interfaces and side of a liquid] differs region 2 [the outside] when the liq- γLS ' γSV + γLV. The sessile drops then behaves as if uid changes its volume. This is called the Laplace-Young it is fully immersed in its vapor and adopts a spherical equation. This more familiar for spherical droplets: shape.ii) If γLS = γSV, then molecules of the liquid part of the interface behave just like molecules in the bulk in- dA = 8πrdr terface and the drop adopts the shape of a hemisphere. 2 dV = 4πr dr, iii) If γLS < (γSV − γLV) then the intermolecular forces 6 are so effective in reducing the energy of the system that This is the purpuse of adding surfactants [detergents and no equilibrium can be achieved and the liquid spreads soaps]. out on the solid surface, i.e, ALS = ALV → ∞. We consider a sessile drop with the shape of a spherical cap o height H and radius r: We have π B. Capillary Rise V = H(3r2 + H2) 6 2 2 ALV = π(r + H ) Capillary immersed in liquid. The intrface liquid-vapor 2 inside capillary is called meniscus. We use Laplace, the ALS = πr π pressure difference is maintained by the hydrostatic pres- dV = πHrdr + (r2 + H2)dH. (57) 2 sure:

At ﬁxed volume dV = 0, we thus have ∆P = ρgh. (63) r2 − H2 dA = 2πrdr (58) LV r2 + H2 We then get for the height h of the capillary: dALS = 2πrdr (59) 2γ h = . (64) We introduce the contact angle Θ: ρgr r2 − H2 Here r is the radius of the interface: r = R cos Θ, where cos Θ = 2 2 . (60) r + H Θ is the contact angle and R is the radius of the capillary. Thus We get

dG = [γ cos Θ + γ − γ ]2πrdr = 0. (61) 2γ cos Θ LV LS SV h = . (65) ρgR We solve for the contact angle:

γSV − γLS This relation can be used to determine the contact angle. cos θ = . (62) Note that the capillary height h increases as the radius γLV R decreases. Most liquids Θ < 90◦ and the meniscus is This is the Young-Dupre equation. If we want to wet a curved ‘upwards.’ Notable excepetion is mercury where surface, then cos Θ ' 1 and we need γSV − γLS ' γLV. meniscus is curved ‘downwards’: capillary depression.