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 (~ 43/3) 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 1.381023 J/K 151K a 64 PC 64 4.83106 Pa 2 k BTC 1.381023 J/K 151K 29 3 b 5 . 4 10 m 8PC 8 4.83106 Pa 27 k BTC a 64 PC 2 2 3.8 1049 J 2 /Pa b1/ 3 3.861010 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 lnV Nb const 2 3/ 2 V Nb 2 m Nb 5 Sideal Nk B ln1 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 ln1 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 lnV 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 lnV 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 2.2 102 m3 20 V0 Patm 5 266.2 2.2 102 1 S R ln R ln 5 . 24 10 26 25.5 J/K 3 3 2 273 110 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 TS VP 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): PV Nb RT PV RT PNb H U PV CV RT 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 dFT ,N SdT PdV dNT ,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 TSvdW T

Download
# Lecture 15. The van der Waals Gas (Ch. 5)

homework writing help

for a fair price!

check it here!