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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 be- comes: 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 atmo- spheric 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 con- stant, 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.
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