Latent Heat and Critical Temperature

PHYSICS ESSAYS 26, 4 (2013) Latent heat and critical temperature: A unique perspective Kent W. Mayhewa) 68 Pineglen Cres., Ottawa, Ontario K2G 0G8, Canada (Received 12 January 2011; accepted 14 October 2013; published online 30 December 2013) Abstract: We are going to introduce the concept of “kinematic numbers” and their application to the probability function. We shall then see how this may help explain both the latent heat of vaporization and the critical temperature. Moreover, our intuitive approach allows us to understand both phenomena based upon kinetic theory, and the Boltzmann factor. We shall also discuss the limitations of Avogadro’s hypothesis, and show a unique interpretation for the Clausius–Clapeyron C 2013 Physics Essays Publication. [http://dx.doi.org/10.4006/0836-1398-26.4.604] equation. V Résumé: Nous allons introduire la notion de “nombres cinématiques” et leur application à la fonction de probabilité. Nous verrons alors comment cela peut aider à expliquer la chaleur latente de vaporisation, et la température critique. En outre, notre approche intuitive nous permet de comprendre les deux phénomènes fondée sur la théorie cinétique, et le facteur de Boltzmann. Nous allons également discuter des limites de l’hypothèse d’Avogadro, et montrer une interprétation unique de l’équation de Clausius–Clapeyron. Key words: Latent Heat of Vaporization; Boiling Temperature; Critical Temperature; Kinematic Numbers; Probability; Avogadro’s Hypothesis. to Laplace Theory, that one-half of the latent heat must be equal to the molar surface energy.” They go on to argue that the surface energy should be proportional to: V3/2. Agrawal and Menon3 perform a quantitative analysis by adding a surface energy term, which differs from previous values, to the work required for vaporization. The problem with such an analysis is that surface energy is ambiguous, unless it is the energy associated with a tensile surface, which is a function of surface area.4 In which case, we must accept that a flat tensile layer’s surface area does not necessarily change during vaporization. It is obvious that the reasoning and value of the surface energy lacks consensus. Interestingly, Garai1 has slightly modified the surface energy argument into one wherein the vaporizing molecule must overcome the surface resistance. Consider a molecule in condensed matter, which is in thermal equilibrium. The discrete probability of that molecule being in a given state of energy (E) is5 I. INTRODUCTION The latent heat (L) of vaporization is: “the energy that has to be supplied to the system in order to complete the liquid–vapor phase transformation.”1 It is traditionally envisioned in terms of an isothermal, isobaric, process wherein the absorbed energy increases the system’s internal energy (U) and performs work (W),1 and is often referred to as the enthalpy of vaporization (DH). Letting the subscripts “l” and “g,” respectively, indicate the liquid and gas/vapor state, and using the “arrow” to indicate direction, we shall write the latent heat of vaporization as: Lðl!gÞ : In which case, the traditional interpretation would be written in the following form: Lðl!gÞ ¼ dU þ W: (1) The traditional interpretation for work is one of isobaric volume change, i.e., W ¼ PgdVg, hence Eq. (1) becomes Lðl!gÞ ¼ dU þ PgdVg: P0 ðEÞ ¼ BebE ; (2) where B is a normalization constant, ebE is the Boltzmann factor, b ¼ 1=ðkTÞ, k is Boltzmann’s constant, and T is the absolute temperature. We also used an apostrophe and subscript “(E)”to indicate that we are dealing with a probability. Frisch and Salsburg (Ref. 6, p. 263) concerning the Boltzmann factor state: “All assumptions required to effect derivation, are basically three in number: the truncation of interactions of higher order than binary collisions, the condition of molecular chaos…, and slow secular variation…” They then go on to discuss all three concerning gases, but fail to discuss higher order than binary collision in condensed states. Green7 does discuss the prospect of triple Although Eq. (1), and/or Eq. (2), are used on a daily basis, there remains a lack of consensus as to the exact reasoning behind the latent heat of vaporization, with many researchers believing that the surface energy play a significant role.1 Newman and Searle (Ref. 2, p. 209) state that “the internal latent heat of a liquid is, presumably, a measure of the work done against the internal pressure; and that done by the molecules in reaching the surface—i.e., half-way from the interior to outside—is measured by the potential energy acquired as surface energy. From this it is argued, according a) [email protected] 0836-1398/2013/26(4)/604/8/$25.00 (3) 604 C 2013 Physics Essays Publication V Phys. Essays 26, 4 (2013) 605 collision for gases, however, he fails to break it down into its simplest construct. A main issue concerning this paper is the fact that the Boltzmann factor is limited to energy exchange during binary collisions in the liquid state. In the gaseous/vaporous state, the kinetic energy (Ek) of  which is based upon a monatomic molecule is5,8 3kT=2, equipartition theory, i.e., the mean kinetic energy being  exchanged along each of the three orthogonal axis is kT=2. In a liquid state, the vibrational energy of a molecules is:5,8  which is based upon the same fundamental principles, 3kT, except in the liquid state,9 each molecule is bonded to its neighboring molecules, therefore, there exists both kinetic, and potential, energy. Thus, the total mean energy, accessi ble along each orthogonal axis becomes kT. The vaporization rate [Jðl!gÞ ] is traditionally written in the following form: Jðl!gÞ ¼ Jn exp½blðl!gÞ ; (4) wherein Jn is the normalization constant, or if you prefer: Jn ¼ NC, which is in terms of the normalization factor (N) and the concentration (C). Mayhew10 has argued that if one compares the energy of gas with its ability to perform work, then one attains a maximum possible efficiency factor of: 2/3, i.e., 66.67%. This can be understood by performing a reversal of the way we calculate the energy associated with a gas, in kinetic theory,8 i.e., by starting off with the kinetic energy of a gas and then reversing the classical analysis until we arrive at the ability of a gas to do work. The critical temperature is the temperature above which molecules can no longer exist in a liquid state, irrelevant of the pressure. Currently, there is no classical explanation for this. Frisch and Salsburg (Ref. 6, p. 13) states: “All thermodynamic functions have singularities at the critical point… the exact nature of the singularities have been the subject of much careful and difficult experiment but are still imperfectly known.” Contemporary theories generally are statistically based upon particle interactions11 with concentrated efforts being placed upon Bose gases, wherein numerous techniques and conclusions are postulated.12 The theory of Bose–Einstein Condensation (BEC) is based upon Einstein’s prediction concerning boson gas undergoing phase transition at low temperature, wherein the thermal de Broglie wavelength is greater than the mean particle spacing.13 Anderson et al.14 have found BEC can exist in dilute vapors. Interestingly, BEC may help explain why low temperature gases do not necessarily obey kinetic theory. This paper is part of an ongoing attempt by this author, to demonstrate that a different classical thermodynamic perspective exists, of which he has written a yet to be published book. As such, a goal herein is to demonstrate an explanation for both the latent heat of vaporization and critical temperature. We shall also briefly discuss the limitations of Avogadro’s hypothesis, which states: “Equal volumes of different gases at the same temperature and pressure contain an equal number of molecules.”15 II. ISSUE OF PROBABILITY Consider that in order for a liquid molecule to vaporize, it must extract the latent heat from one of its neighboring molecules. If we write the energy required per molecule for vaporization, as lðl!gÞ ¼ Lðl!gÞ =ð6:02  1023 Þ, then based upon Eq. (3), the probability of the a given molecule extracting the required energy from one of its neighboring molecules is P0 ðvÞ ¼ Be½blðl!gÞ ; (5) where B is a normalization constant, and the subscript “(v)” indicates vaporization. What applies to a solitary molecule equally applies to a mole of molecules. Therefore, there is no transgression if we multiply both the numerator and denominator in exponential of Eq. (5) by Avogadro’s number. Thus, transforming Eq. (5) into the probability on a per mole basis, i.e., Eq. (5) becomes P0ðvÞ ¼ BeðDH=RTÞ ; (6) where DH is the molar enthalpy of vaporization, R is the molar ideal gas constant. An issue confounding the application of either: Eq. (6) or Eq. (5) is that the latent heat of vaporization is significantly greater than the mean thermal energy exchanged at the liquid’s boiling temperature, i.e., Lðl!gÞ  RTb , where Tb is the liquid’s boiling temperature. For example: At 1 atm pressure the boiling point of water is 373 K. The molar latent heat of vaporization [Lðl!gÞ ] for water is 40.65 KJ/mol at 298 K. Therefore, Lðl!gÞ =RTb ¼ (40,650 J)/[8.31 (J/mol K)] [373 (K)] ¼ 13.1. In order for Lðl!gÞ  RTb , the water’s temperature would need to be of the order of 4900 K. The above issue is traditionally disregarded by realizing that the normalization constant (B) allow us to normalize the probability. Specifically B is calculated and/or plotted, allowing Eq. (6) to be used in experimentation. Although widely accepted, this may have grave consequences, i.e., even if eðDH=RTÞ in Eq. (6) is incorrect, the normalization process still allows one to create a plot that often correlates/ approximates with one’s empirical data. It is a scary unheralded notion!! Moreover, it is hard to fathom that the energy required for vaporization is so much greater than the thermal energy associated with any one molecule. Accordingly, it is hard to explain why when blðl!gÞ ! 13:1 that water boils. Why not some other number, i.e., 7.6, 12.5, 1000, etc. Although not definitive, this should at least make you ponder. III. KINEMATIC NUMBERS AND THE LATENT HEAT OF VAPORIZATION AT Tb The molar latent heat of vaporization, and boiling temperature, for noble elements is given in the top of Table I. We start by determining the ratio of the energy required for a mole of noble molecules to vaporize, versus the mean amount of energy contained in a mol of molecules, at the liquid’s boiling point, 606 TABLE I. Phys. Essays 26, 4 (2013) Values for boiling temperature and experimental molar latent heat. Atomic number Element Latent heat of vaporization (kJ/mol) Tb (degrees, K) RTb (kJ/mol) Ratio: (latent heat)/RTb Tc (degrees, K) RTc (kJ/mol) Ratio: (latent heat)/RTc 2 10 18 36 54 86 He Ne Ar Kr Xe Rn 0.08 1.73 6.44 9.03 12.64 16.40 4.2 27.1 87.3 119.4 167 211 0.03 0.23 0.73 0.99 1.39 1.75 2.3 7.7 8.9 9.1 9.1 9.4 5.19 44.4 150.9 209.4 289.8 377 0.04 0.37 1.25 1.74 2.41 3.13 1.9 4.7 5.1 5.2 5.2 5.2 3 11 19 39 55 Li Na K Rb Cs 145.9 97.0 79.9 72.2 67.7 1620 1156 1047 961 951 13.46 9.61 8.70 7.99 7.90 10.8 10.1 9.2 9.0 8.6 26.78 21.38 18.47 17.39 16.10 5.4b 4.4b 4.3b 4.2b 4.2b 2 12 20 38 56 Mg Ca Sr Ba 127.4 153.6 144.0 140.3 1380 1757 1656 1913 11.47 14.60 13.76 15.90 11.1 10.5 10.5 8.8 NA NA NA NA 3 21 39 57 Sc Y La 332.7 365.0 402.1 3109 3618 3737 25.84 30.07 31.05 12.9 12.1 12.9 NA NA NA 4 40 Zr 573.0 4650 38.64 14.8 NA 5 41 73 Nb Ta 682.0 743.0 5200 5698 43.21 47.35 15.8 15.7 NA NA 6 74 W 824.0 5933 49.30 16.7 NA 7 75 Re 715.0 5900 49.03 14.6 NA 9 77 Ir 604.0 4800 39.89 15.1 NA 11 29 47 79 Cu Ag Au 300.4 258.0 324.0 2840 2485 3129 23.60 20.65 26.00 12.7 12.5 12.5 NA NA NA 12 80 Hg 59.2 630 5.24 11.3 14.54 4.1 13 13 49 81 Al In Tl 293.0 231.8 164.0 2740 2273 1730 22.77 18.89 14.38 12.9 12.3 11.4 NA NA NA 14 14 32 50 82 Si Ge Sn Pb 359.0 334.0 295.8 177.7 3538 3106 2543 2013 29.40 25.81 21.13 16.73 12.2 12.9 14.0 10.6 NA NA NA NA 15 83 Bi 104.8 1833 15.23 6.9 NA L 58 59 60 62 63 64 65 66 67 68 70 71 Ce Pr Nd Sm Eu Gd Tb Dy Ho Er Yb Lu 398.0 331.0 289.0 165.0 176.0 301.3 293.0 280.0 265.0 280.0 159.0 414.0 3716 3793 3347 2067 1802 3546 3503 2840 2973 3141 1469 3675 30.88 31.52 27.81 17.18 14.97 29.47 29.11 23.60 24.71 26.10 12.21 30.54 12.9 10.5 10.4 9.6 11.8 10.2 10.1 11.9 10.7 10.7 13.0 13.6 NA NA NA NA NA NA NA NA NA NA NA NA Group Noble gases 18a Noble (approximateideal gas) Non-noble gases 1 a From Ref. 1. Exceptions are the anoble gases and bcritical temperatures, whose values are from Wolfram Research.18 b 3223 2573 2223 2093 1938 1750 Phys. Essays 26, 4 (2013) Lðl!gÞ =RTb : 607 (7) Ex. calculation for a mole of Kr: Ratio ¼ 9029 (J/mol)/ [{8.31(J/mol K)}{119.4 (K)}] ¼ 9.1. Performing the same calculations, we find that the ratio value of 9 correlates fairly well, for Ar, Kr, and Xe, as is shown in Table I. Now consider “9” as being “the kinematic number for the latent heat of vaporization”. Interestingly, 9 does not correlate particularly well for the two smallest noble gases, those being He and Ne. We shall give a brief plausible explanation for this, after we discuss the critical temperature. Let us reexamine the logic of Eq. (3), hence Eq. (6). Namely, does it make sense to limit our consideration to, the discrete energy that a molecule can attain from a single neighbor? Certainly, a given liquid molecule has six neighbors, with whom it is bonded. Since liquid molecules vibrate 1013 times a second, then the odds of all six neighbors colliding/interacting with a vaporizing molecule, over a short duration, should have a reasonable likelihood of occurrence. Consider that all six neighboring molecules collide/interact with the vaporizing molecule at some instant, and each pro If one accepts this, then the disvides a mean energy of kT. crete energy required from any one neighboring molecule would be lðl!gÞ =6, hence we could rewrite Eq. (5), as P0ðvÞ ¼ Be½blðl!gÞ=6 : (8) Is Eq. (8) suffice? Well the ratio for noble elements was 9, not 6. Therefore, the preferred probability for a noble liquid molecule vaporizing might be P0ðvÞ ¼ Be½blðl!gÞ=9 : (9) In order to explain Eq. (9) we need to determine a plausible path of which there are many, so this author is going to discuss just a few. Consider the vaporization of molecule #6 in Fig. 1, at Tb . At some instant, it could access a mean energy of kTb from each of molecules #2,5,10, 7, as well as the two molecules perpendicular to the page that are not shown. Consider that the vaporizing molecule has a mean energy of 3kTb , then an at first glance an eloquent solution is 6kTb þ 3kTb ¼ 9kTb . What remains problematic is that molecule #6 must possess a mean kinetic energy of 3kTb =2, after vaporization when in the gaseous state. Hence, the logic of simply adding the mean energy of molecule #6 to the accessible energy from its six neighbors remains awkward, unless a path is expressed. Perhaps, the vaporizing molecule cannot contribute 3kTb . Specifically, the potential energy exists due to the bond between the vaporizing molecule and its neighbors! Therefore, if we associate this potential energy with its neighbors then all the vaporizing molecule can possibly contribute is its kinetic energy, i.e., 3kTb =2. Does this mean that the energy to break all the liquid’s bonds is 6kTb þ 3kTb =2 ¼ 15kTb =2? And then the unbound molecule has numerous collisions with other molecules, thus reattaining a mean kinetic energy of 3kTb =2. Thus, 15kTb =2 þ 3kTb =2 ¼ 9kTb . Again, this is mathematically eloquent but we have omitted the concept of work. FIG. 1. Liquid molecules near an interface. Specifically, latent heat of vaporization generally involves an isobaric volume increase: W ¼ PgdVg. As Mayhew10 points out, this represents a displacement of the atmosphere and only 2/3 of an expanding gas’s energy can be extracted for work. Therefore, the energy requirement for the vaporizing molecule to perform work is 3kTb =2. Consider that 6kTb breaks the bonds, leaving the vaporizing molecule with 3kTb =2 to escape the liquid, and then another 3kTb =2 is extracted from the vaporizing molecules surroundings as it expands the system, i.e., performs work. At first the above may seem odd but it is not. Think of it this way, the vaporizing molecule starts off with 3kTb =2. As it pushes upwards expanding the gaseous system, it loses energy due to work. Yet every time it collides with molecules that constitute its surroundings, it regains energy. Therefore, we have 6kTb þ 3kTb =2 þ 3kTb =2 ¼ 9kTb ¼ lðl!gÞ , which again is our desired result. Although this author is comfortable with this path, he acknowledges that other plausible paths may exist that warrants consideration. Interestingly, Eq. (8) was contemplated for all six neighbors colliding with the vaporizing molecule. However, it should now be obvious that all the energy associated with latent heat is not limited to the bonds, thus we should rethink our probability. This author prefers P0ðvÞ ¼ BeðTb=TÞ : (10) Understandably, Eq. (10) can be obtained from Eq. (9), by considering that for noble elements: lðl!gÞ ¼ 9kTb , but it is not that simple. Equation (10) implies, as T ! Tb , then P0ðvÞ ! 100%, when B ¼ 1, i.e., all liquid molecules would try to vaporize at once. What prevents this is the fact that B correlates to the likelihood of all six neighbors passing on their discrete energies at the same instant plus the likelihood of the unbound vaporizing molecule actually escaping the liquid. Interesting consequences of Eq. (10) are that we now can better explain both the boiling process, and why molecules which vaporizes at Tb are not likely to be located upon the surface, i.e., surface molecules only have five neighbors. This brings us to the critical temperature. 608 Phys. Essays 26, 4 (2013) FIG. 2. Sketch for the latent heat of water versus temperature. Shown are the boiling temperature, Tb and critical temperature, Tc. V. LOW TEMPERATURE GASES AND AVOGADRO’S HYPOTHESIS IV. KINEMATIC NUMBERS AND THE CRITICAL TEMPERATURE We realize that as the liquid’s temperature increases then the latent heat of vaporization decreases, as sketched in Fig. 2. Furthermore, it is traditionally accepted that the latent heat of vaporization approaches zero as the liquid’s temperature approaches the critical temperature. Keeping it simple, we can assert that the energy required to break the bonds is fairly constant, hence the reason that the latent heat decreases is mainly due to the fact that the actual energy associated with the liquid molecules increases with temperature. So rather than strictly adhering to traditional dogma, let us open our minds and investigate for the critical temperature (Tc), the following ratio: Lðl!gÞ =RTc : (11) Ex. calculation for a mole of Kr: [9029 (J/mol)]/[{8.31(J/mol K)}(209 K)] ¼ 5.2. We used the latent heat at the boiling point, which may upset some readers. However, interestingly in Table I, we can see that the ratio [Eq. (11)] for noble substances approximates 5, with the exception being: He. Consider, molecule #2 that resides on top of the tensile layer in Fig. 1. It only has five neighboring molecules from which it can extract thermal energy. If the energy required from each neighboring molecule is: lðl!gÞ =5  kTc, then we realize that even if molecule #2 possessed absolutely no thermal energy, then the mean accessible thermal energy of its five neighbors still would be: 5kTc , which is sufficient for molecule #2 to vaporize. Thus explaining why no tensile layer can possibly exist when: T > Tc , irrelevant of the pressure. Furthermore, there must exist numerous paths wherein the vaporizing molecule, plus its neighbors, possesses enough energy, for vaporization as: T ! Tc . Each path would possess its own probability. Therefore, the total probability [P0ðvÞtotal ] of vaporization at Tc would be P0ðvÞtotal ¼ P0ðcÞ1 þ P0ðcÞ2 þ P0ðcÞ3 … P0ðcÞN00 ; FIG. 3. This figure shows that the five gases only obeys Avogadro’s hypothesis when the gas temperatures are sufficiently above absolute zero. (12) where the subscripts “1, 2, 3, …, N0 ” each signify a unique path. The explanation for why He and Ne both give poor correlations to the kinematic number for vaporization is likely based upon their low boiling temperatures (Tb). Specifically, Fig. 3 shows the “Number of Moles of Gas versus Temperature” for five gases, namely: H2, He, Ar, N2, and O2. The data for the plot were made by taking the density of gases at various temperatures,16 and dividing by their atomic weight, i.e., Table II. If all gases obeyed Avogadro’s hypothesis in all temperature regimes, then the plot would show a simple linearly decreasing function of temperature. Obviously, the closer to absolute zero a gas is, then the less that gas obeys Avogadro’s hypothesis. At such low temperatures, gas molecules possess significantly less kinetic energy allowing other forces to prevail. The problem with claiming EM based attraction/repulsion would be that both noble elements (He, Ar) behave similarly to non-noble elements. Hence the explanation maybe as simple as gravity becomes relevant. Or perhaps the problem is best resolved by quantum-based arguments, i.e., BEC. Whatever the reason, we must conclude that the validity of Avogadro’s hypothesis decreases when temperatures approach absolute zero. Since our kinematic numbers are based upon a unique consideration of Boltzmann factor and kinetic theory, of which Avogadro’s hypothesis is a result, we understand that there will be issues with correlation for both: He and Ne. It is interesting to note that for the correlation with critical temperature, only He critical temperature was too low to correlate with its kinematic number. VI. WALLS AND AVOGADRO’S HYPOTHESIS If two unbound molecules start off with the same velocity and then collide, then conservation of momentum states ~ v2 ¼ ðM1 =M2 Þ~ v1 ; (13) where M, ~ v, respectively, signify mass and velocity, and the subscripts differentiate molecule 1 from 2. Based upon v1 ¼ M1 =M2 . But Eq. (13), one would expect for gases: ~ v2 =~ different gases in an isothermal system possess the same kinetic energy: Ek ¼ Mv2 =2, irrespective of their size, hence adhere to ðv2 Þ2 =ðv1 Þ2 ¼ M1 =M2 . Phys. Essays 26, 4 (2013) TABLE II. 609 Density of gases at various temperature and atomic weights. Density at 1 bar pressure (kg/m3) Number of mol/m3 At 1 bar pressure (mol/m3) T ( K) 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 H2 He 1.32 0.62 0.41 0.30 0.24 0.20 0.17 0.15 0.13 0.12 0.11 0.10 0.09 0.09 0.08 0.08 2.50 1.20 0.80 0.60 0.48 0.40 0.34 0.30 0.27 0.24 0.22 0.20 0.19 0.17 0.16 0.15 Ar 4.92 4.06 3.46 3.02 2.68 2.41 2.19 2.01 1.85 1.72 1.60 1.50 N2 4.38 3.44 2.84 2.42 2.12 1.88 1.69 1.52 1.41 1.30 1.20 1.12 1.05 O2 3.94 3.25 2.77 2.42 2.15 1.93 1.75 1.61 1.48 1.38 1.28 1.20 AW H2 2.016 He 4.002 Ar 39.95 N2 28.014 O2 32 656.25 305.41 201.09 150.55 120.78 100.20 85.86 75.10 66.77 60.07 54.61 50.10 46.58 42.93 40.06 37.56 624.19 299.85 200.05 150.10 120.11 100.12 85.83 75.11 66.77 60.09 54.65 50.10 46.25 42.93 40.08 37.58 123.03 101.58 86.64 75.60 67.09 60.30 54.80 50.19 46.31 42.98 40.13 37.60 156.31 122.69 101.38 86.53 75.53 67.04 60.26 54.40 50.15 46.30 42.98 40.09 37.59 123.16 101.63 86.69 75.63 67.09 60.31 54.78 50.19 46.31 43.00 40.13 37.59 Notes: AW signifies atomic weight. mol/m3 calculated by: Density*1000/AW. Data for density vs temperature are from Ref. 16. Furthermore, the kinetic energy of gases is simply tem perature dependant (3NkT=2) thus obeying the ideal gas law: PV ¼ NkT, which can be rewritten as V ¼ NkT=P: (14) Equation (14) defines the gas’s volume in terms of its temperature, at a given pressure (P). Hence, Avogadro’s hypothesis holds over temperatures wherein Eq. (14) remains valid, i.e., adheres to: ðv2 Þ2 =ðv1 Þ2 ¼ M1 =M2 , which contravenes Eq. (13). The above argument exists because kinetic theory of gases is restricted to dilute gases. Specifically, dilute implies that the gas molecules will tend to collide with the system walls rather than each other. Compared with gas molecules, walls possess infinite mass and surface area. Moreover, walls  act like an engine pumping a mean kinetic energy of: kT=2 along each axis, onto any gas molecules colliding with it. In brief, so long as the gas molecules collide more often with the walls than with other gas molecules, then those gas molecules will adhere to kinetic theory hence both Avogadro’s hypothesis and the ideal gas law will hold true. Note: By walls, we mean surfaces of condensed matter. Remember thermodynamics is primarily based upon empirical data obtained from experimental systems with walls. It is accepted that high-density gases do not adhere to the ideal gas law.1 This author reasons that such gas molecules scatter amongst themselves, more than with the walls hence, they would adhere to Eq. (13) rather than Avogadro’s hypothesis, or the ideal gas law. This author apologizes for the limited discussion concerning this, leaving the full argument for his upcoming book, wherein pages are at less of a premium. VII. LATENT HEAT OF VAPORIZATION FOR NON NOBLE GASES Let us now consider the latent heat of vaporization of substances that are not noble. Garai1 published a table with the boiling points, and experimental latent heats of vaporization, for 45 elements. Under the heading “Non-noble gases” in Table I, we used his data in order to determine the ratio [Eq. (7)], i.e., [Lðl!gÞ =RTb ]. When doing so, we found that the resultant ratio tends to be between 9 and 16, with the majority of elements having ratios below 12, whilst a few elements do fall out of these ranges. Accepting the prospect that most liquid molecules adhere to the same path, then we need to explain why: lðl!gÞ =kTb > 9, or at least tends to be so. When dealing with liquids vaporizing into nonideal gases, we must concern ourselves with the bonding energy in both the liquid and gaseous states. This is traditionally accomplished by saying that the term “dU” in Eq. (2) covers the potential energy change in going from the liquid to gaseous state. That is fine, but let us now alter our visualization of this process. Envision, 9 as the kinematic number for vaporization with six neighbors, encompassing both the binding energy in the liquid state and kinetic energy in the gaseous state. Why is that additional energy required overcome any bonding potential in the gaseous state, which causes the ratio to be higher than: 9? Perhaps we should rephrase the question: Why not the gas molecules simply occupy a lesser volume, so that all that is required is: 9kTb ? The reason resides in this author’s assertion that walls force the gas molecules to obey Avogadro’s hypothesis. Specifically, isothermal gas molecules in a system with walls must occupy the same volume [22.4 l, at 273 K, and 1 atm 610 Phys. Essays 26, 4 (2013) (Ref. 15)], which explains why extra energy is needed to overcome the bonding potentials. And this energy is likely extracted from the system after the vaporizing molecule escapes from the liquid. For clarification purposes, we could calculate the energy to form a cloud of charged particles (Ref. 17, p. 193), they could use Energy ¼ 3Q2 =20pe0R: (15) If we simply plug in Q for a mol of electronlike molecules ¼ (1.6  1019 C)(6.02  1023), and consider a sphere whose mol volume is: 22.4 l, hence R ¼ 0.175 m, we end up with an energy that is way too large, to explain the energy difference. The probable reality is that for most gases, the net charge that needs to be contemplated is less than that of an electron. It should also be stated that noble gases only approximate ideal, thus we can surmise that the reason their ratios [Eq. (7)] are slightly greater than 9 is due to their approximation as ideal, i.e., they too will have small binding energies in the gaseous state that must be overcome. Pv ¼ C0 Pb exp½Pb Vb =RT: (18) Equation (18) may look familiar to the reader as it is similar to the Clausius–Clapeyron equation, Pv ¼ P0 exp½Lðl!gÞ =RT; (19) where Pv ¼ vapor pressure, P0 ¼ constant, R ¼ gas constant ¼ 8.31 J/mol K. Lðl!gÞ ¼ DHvap ¼ latent heat for a mole of molecules: This author’s view is that Eq. (17) may become a useful equation for experimental data, wherein: C0 is calculated based upon data. Moreover, the Clausius–Clapeyron equation [Eq. (19)] suffers the detriment: Lðl!gÞ > RT. The point made herein is that the probability as defined by Eq. (10) may also readily explain what we witness in experiment, which is also discussed at length in this author’s unpublished book. VIII. CRITICAL TEMPERATURE FOR NONIDEAL GASES Table I also shows that for other elements the ratio as defined by Eq. (11) is reasonably close to 5. Due to a lack of data for the critical temperature, more elements were not analyzed at this time as can be indicated by the NA (data not available). The critical temperature for water is: 647 K. If we calculate the ratio [Eq. (11)] for water, we obtain: 7.6. We can surmise that for various substances, there may be other considerations involved, i.e., perhaps bonding potentials in the gaseous state. IX. NORMALIZATION CONSTANT, PROBABILITY, AND VAPORIZATION RATE Can Eq. (10) equally explain empirical results? Consider that we want to calculate the vaporization rate through a unit surface area. Based upon Eq. (10), we could write Jðl!gÞ ¼ N½CeðTb=TÞ ; (16) where “N” is the normalization factor and “[C]” is the concentration. If Eq. (16) signifies the rate of vaporization, then we can surmise that in equilibrium it also equates to the rate of vapors condensing. Certainly, the rate of vapor molecules condensing must be proportional to the pressure exerted by those vapors. Accordingly, based upon Eq. (16), for a pure liquid ([C] ¼ 1), we can write Pv ¼ C0 PbeðTb =TÞ ; (17) where Pv is the vapor pressure, C0 is a constant that allows us to express the equation in terms of the pressure at the liquid’s boiling point (Pb , Tb ). Based upon the ideal gas law for a mol of molecules at the boiling point: Tb ¼ Pb Vb =R, Eq. (17) becomes X. CONCLUSIONS The theory given, herein, allows us to explain both the latent heat of vaporization and the critical temperature. It is based upon the simple consideration of what happens if vaporization requires all of the neighboring liquid molecules colliding with the vaporizing molecule, at some instant of time. The explanation does not require surface energy based arguments, which become cumbersome. We deduced that for the vaporization of noble liquid molecules occurs when we have a total energy of: 9kTb, hence we called “9” the kinematic number for vaporization. And a plausible unique path was given which explains the number 9. Certainly, this explains both, the latent heat of vaporization and why molecules from below the tensile layer are the ones, which tend to vaporize in simple terms. Finally, our new probability allowed us to calculate an equation, which was similar to the Clausius–Clapeyron equation. Extrapolating the same path dependent logic allowed us to explain why no tensile layer can exist at the critical temperature (Tc). This is based upon a surface molecule, only having five neighbors, and each of those neighbors having a mean accessible energy of kTc , which led to noble substances having the kinematic number: “5” for the critical temperature. We also discussed how experimental walls (i.e., surfaces of condensed matter) might influence our experimental findings, by forcing gases to obey Avogadro’s hypothesis. We discussed why Avogadro’s hypothesis is not valid at low temperatures and provided some unexpected insights. 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