Clausius Clapeyron Equation

6 Key excerpts on "Clausius Clapeyron Equation"

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  Fundamentals of Thermodynamics

  • Claus Borgnakke, Richard E. Sonntag(Authors)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  We will also consider such related matters as generalized charts and equations of state. 12.1 THE CLAPEYRON EQUATION In calculating thermodynamic properties such as enthalpy or entropy in terms of other prop- erties that can be measured, the calculations fall into two broad categories: differences in properties between two different phases and changes within a single homogeneous phase. In this section, we focus on the first of these categories, that of different phases. Let us assume that the two phases are liquid and vapor, but we will see that the results apply to other differences as well. Consider a Carnot-cycle heat engine operating across a small temperature difference between reservoirs at T and T − ΔT. The corresponding saturation pressures are P and P − ΔP. The Carnot cycle operates with four steady-state devices. In the high-temperature heat-transfer process, the working fluid changes from saturated liquid at 1 to saturated vapor at 2, as shown in the two diagrams of Fig. 12.1. From Fig. 12.1a, for reversible heat transfers, q H = Ts fg ; q L = (T − ΔT )s fg W-27 FIGURE 12.1 A Carnot cycle operating across a small temperature difference. T T P P T – Δ T T T – Δ T P – Δ P P – Δ P P s v 1 2 3 (a) (b) 4 1 2 3 4 so that w NET = q H − q L = ΔTs fg (12.1) From Fig. 12.1b, each process is steady-state and reversible, such that the work in each process is given by Eq. 7.15, w = − ∫ v dP Overall, for the four processes in the cycle, w NET = 0 − ∫ 3 2 v dP + 0 − ∫ 1 4 v dP ≈ − ( v 2 + v 3 2 ) (P − ΔP − P) − ( v 1 + v 4 2 ) (P − P + ΔP) ≈ ΔP [( v 2 + v 3 2 ) − ( v 1 + v 4 2 )] (12.2) (The smaller the ΔP, the better the approximation.) Now, comparing Eqs.

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  Physical and Chemical Equilibrium for Chemical Engineers

  • Noel de Nevers(Author)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  (2) .

  (5.2)

  So for this change

  (5.3)

  Using the derivatives in Eq. 4.33 we can write

  (5.4)

  Factoring gives

  (5.5)

  This is the Clapeyron equation, which is rigorously correct for any phase change of a single pure chemical species. All of the curves in Figures 1.7 , 1.8 , 1.9 , and 5.2 agree with the Clapeyron equation. If your experimental data do not agree with it, then you have made an experimental error!

  Example 5.1 Compute the value of dP/dT for the steam– water equilibrium at 212°F using the Clapeyron equation, and compare it with the value in the steam table [5].

  From the steam table, we have , , and . Thus,

  (5.A)

  Using the nearest adjacent steam table entries for vapor pressure, we have

  (5.B)

  As in Example 4.1, this agreement does not demonstrate the correctness of the Clapeyron equation; the authors of the steam tables used the Clapeyron equation in making up that table. (If the values did not agree, it would show an error in the steam table!) In the Clapeyron equation we normally use P for pressure, because the Clapeyron equation applies to any equilibrium between any two phases of the same pure substance. If one of the phases is a gas or vapor, then the (dP/dT ) relation is a vapor-pressure curve, and we normally use p for vapor pressure, so the above derivatives would be written as dp/dT . This distinction is not very important, but the use of P for pressure in general and p for vapor pressure is common.

  5.4 The Clausius–Clapeyron Equation

  The Clapeyron equation is rigorous and exact . It applies not only to gas–liquid equilibrium, but to any two-phase equilibrium of a pure species (e.g., liquid–solid, gas–solid) or change between two different crystal forms (e.g., graphite to diamonds, see Example 4.2). By adding some simplifications, we find the Clausius–Clapeyron (C-C) equation, which is only approximate but is a surprisingly good

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  Phases of Matter and their Transitions

  Concepts and Principles for Chemists, Physicists, Engineers, and Materials Scientists

  • Gijsbertus de With(Author)
  • 2023(Publication Date)
  • Wiley-VCH

    (Publisher)

  When the phase transition of a substance is between the gas phase and a condensed phase (C) and occurs at T ≪ T cri of that substance, the specific volume of the gas phase V G is much larger than that of the condensed phase V C . Therefore, one may approximate ΔV = V GC = V G (1 − V C /V G ) by V G . For sufficiently low pressure the perfect gas law can be invoked so that V G = RT/P with R, as usual, the gas constant. Inserting in the Clapeyron equation leads to the Clapeyron–Clausius equation: dP dT = Δ vap H TΔV ≅ PΔ vap H RT 2 or d ln P d(1∕T) = − Δ vap H R (12.7) We deal with the Clapeyron–Clausius equation further in Section 13.1. 12.3 The Mosselman Solution for the Clapeyron Equation The Clapeyron equation is not an exact differential equation, and in order to arrive at an explicit relation between the pressure P and temperature T, one has to integrate the equation. Because the enthalpy and volume changes involved are both functions of T and P, this cannot be done directly. It is generally believed that the Clapeyron equation is unsolvable without preliminary substitution of a simple expression for ΔH(T) and ΔV (T). However, Mosselman et al. [4] found the integration factor ΔV /T that transforms the Clapeyron equation into an exact differential equation. Using this integrating factor, one can write T −1 ΔV (T, P) dP + Δ vap H(T, P) dT −1 = 0 satisfying Euler’s condition for integrability Δ vap H(T, P)∕P = [ΔV (T, P)∕T]∕T −1 The solution is T −1 ∫ P P 0 ΔV (T, P) dP + ∫ T −1 T 0 −1 ΔH(T, P 0 ) dT −1 = 0 T 0 −1 ∫ P P 0 ΔV (T 0 , P) dP + ∫ T −1 T 0 −1 ΔH(T, P) dT −1 = 0 (12.8) 12.3 The Mosselman Solution for the Clapeyron Equation 377 Because Δ vap H/T = ΔC P , integration of Eq.

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  Thermodynamics

  • H J Kreuzer, Isaac Tamblyn;;;(Authors)
  • 2010(Publication Date)
  • WSPC

    (Publisher)

  Phase Transitions 185 derivative, the magnetic susceptibility, is discontinuous. There is no latent heat in continuous transitions and two phase co-existence is not possible. 10.2 Clausius-Clapeyron equation As we shall see, in addition to the location of phase transitions, the slope of a transition (with respect to another thermodynamic variable) can have important implications for equilibrium structures. We will now calculate this slope. We again start with the phase co-existence condition (10.2) and take the derivative on both sides with respect to T , remembering that along the co-existence line P = P ( T ) ∂μ 1 ∂T + ∂μ 1 ∂P dP dT = ∂μ 2 ∂T + ∂μ 2 ∂P dP dT (10.12) Because ∂μ/∂T | P = -s and ∂μ/∂P | T = v we get dP dT = s 1 -s 2 v 1 -v 2 (10.13) in terms of the molar entropies and volumes of the two phases, or, in terms of the latent heat (10.10) dP dT = q T ( v 2 -v 1 ) (10.14) This is the Clausius-Clapeyron equation: it determines the change of pressure as a function of temperature along the co-existence curve. Invert-ing this equation dT dP = T ( v 2 -v 1 ) q (10.15) We see that with increasing pressure a liquid (phase “1” with molar vol-ume v 1 ) increases because the molar volume of the gas phase is always much larger that that of the liquid, v 2 >> v 1 . This is again in agreement with Le Chatelier’s principle because as the pressure increases more gas parti-cles transfer to the liquid thus reducing the pressure, or in Le Chatelier’s language, the system counteracts the externally applied constraint (higher pressure). This enhancement of the boiling point with pressure is the phe-nomenon underlying the pressure cooking invented by Count Rumford. Applying the inverted Clausius-Clapeyron equation to freezing we see that the freezing temperature will be increased or decreased depending on whether the liquid volume v 1 is smaller or larger than the solid volume v 2 . For most materials v 1 > v 2 , with the notable exception of water where

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  Thermodynamics, Kinetics, and Microphysics of Clouds

  • Vitaly I. Khvorostyanov, Judith A. Curry(Authors)
  • 2014(Publication Date)
  • Cambridge University Press

    (Publisher)

  1 ∑ = = x d a k k k c (3.6.11) In particular, for an aqueous solution that contains only two components (c = 2), water and solute, this relation is ln ln . = - x d a x d a s s w w (3.6.12) This equation allows determination of solute activities and concentrations with measurements of the water activities or relative humidities over solutions of various concentrations. It has been used in studies of aerosol liquid-solid phase transitions, as will be described in Chapter 11. 3.7. The Clausius–Clapeyron Equation 73 3.7. The Clausius–Clapeyron Equation An important application of the general equilibrium Equation (3.6.7) is the temperature-pressure relations for the three systems: liquid-ice, vapor-liquid, and vapor-ice. 3.7.1. Equilibrium between Liquid and Ice Bulk Phases Consider the equilibrium of the water and ice bulk phases, where phase 1 is water (subscript “w”), and phase 2 is ice (“i”). The radii r 1 = r 2 = ∝ for the bulk substances, the phase transition is melting, and the difference of the molar enthalpies is the molar melting latent heat, ˆ - = = h h L M L w i m w m , with L m being the specific melting heat [cal g −1 ], the molar volumes are those of liquid, v k0,1 = v w = M w /r w , and of ice, v k0,2 = v i = M w /r i . No foreign substance is present, and the misfit strain is absent, e = 0. No salt is present and both activities a k,1 = a k,2 = 1. Under these conditions, (Eqn. 3.6.7) is ˆ ( ) 0 . 2 - + - = L T dT v v T dp m w i (3.7.1) The number of components in this system is c = 1, the number of phases j = 2, and, according to the phase rule (Eqn. 3.3.6), the variance N w = 2 + 1 − 2 = 1. We have one independent variable and can study dependence of the melting pressure p m on the temperature T.

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  An Introduction to High-Pressure Science and Technology

  • Jose Manuel Recio, Jose Manuel Menendez, Alberto Otero de la Roza, Jose Manuel Recio, Jose Manuel Menendez, Alberto Otero de la Roza(Authors)
  • 2016(Publication Date)
  • CRC Press

    (Publisher)

  The boundaries between regions associated to different phases represent particular states of a system in which two phases may coexist. For a given temperature, the pressure at which one phase transforms reversibly into an-other is called the transition pressure, p tr ( T ) . Along the two-phase equilibrium curves, the value of G ( p, T ) per mole is the same for both phases. By using this result, one can calculate properties of the curve such as its slope at a given point, which is given by the Clapeyron equation: dp tr dT = Δ S tr Δ V tr = Δ H tr T Δ V tr , (1.48) where Δ S tr and Δ V tr are the differences in entropy and volume between both phases at that point. During most phase transitions, energy is absorbed or released by the system and the volume increases or decreases without change in temperature and pressure. This “latent heat” (or the volume change) is used to drive the transition from one phase to the other. The latent heat is given by Δ H tr . From a thermodynamic point of view, phase transitions can be classified according to the derivatives of G that are discontinuous across the transforma-tion [30]. First order phase transitions involve latent heat and are characterized by an infinite heat capacity at the transition temperature and pressure as well as by discontinuities in the properties that are derivatives of the Gibbs func-tion: V and -S . Most phase transitions belong in this category; boiling water is an example of a first order phase transition. In contrast, high order or continuous phase transitions do not involve la-tent heat. In second-order phase transitions, there is a discontinuity in the heat capacity, but it does not diverge. The isothermal compressibility and the thermal expansion coefficient, which contain also second derivatives of G with respect to T and/or p , are discontinuous too. The first derivatives of G , volume and minus entropy, show changes in their slopes, but they are con-tinuous across the transition.

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