Posts Tagged ‘Rudolf Clausius’

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Reversible change is a key concept in classical thermodynamics. It is important to understand what is meant by the term as it is closely allied to other important concepts such as equilibrium and entropy. But reversible change is not an easy idea to grasp – it helps to be able to visualize it.

Reversibility and mechanical systems

The simple mechanical system pictured above provides a useful starting point. The aim of the experiment is to see how much weight can be lifted by the fixed weight M1. Experience tells us that if a small weight M2 is attached – as shown on the left – then M1 will fall fast while M2 is pulled upwards at the same speed.

Experience also tells us that as the weight of M2 is increased, the lifting speed will decrease until a limit is reached when the weight difference between M2 and M1 becomes vanishingly small and the pulley moves infinitely slowly, as shown on the right.

We now ask the question – Under what circumstances does M1 do the maximum lifting work? Clearly the answer is visualized on the right, where the lifted weight M2 is as close as we can imagine to the weight of M1. In this situation the pulley moves infinitely slowly (like a nanometer in a zillion years!) and is indistinguishable from being at rest.

This state of being as close to equilibrium as we can possibly imagine is the condition of reversible change, where the infinitely slow lifting motion could be reversed by an infinitely small nudge in the opposite direction.

From this simple mechanical experiment we can draw an important conclusion: the work done under reversible conditions is the maximum work that the system can do.

Any other conditions i.e. when the pulley moves with finite, observable speed, are irreversible and the work done is less than the maximum work.

The irreversibility is explained by the fact that observable change inevitably involves some dissipation of energy, making it impossible to reverse the change and exactly restore the initial state of the system and surroundings.

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Reversibility and thermodynamic systems

The work-producing system so far considered has been purely mechanical – a pulley and weights. Thermodynamic systems produce work through different means such as temperature and pressure differences, but however the work is produced, the work done under reversible conditions is always the maximum work that a system can do.

In thermodynamic systems, heat q and work w are connected by the first law relationship

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What this equation tells us is that for a given change in internal energy (ΔU), both the heat absorbed and the work done in a reversible change are the maximum possible. The corresponding irreversible process absorbs less heat and does less work.

It helps to think of this in simple numbers. U is a state function and therefore ΔU is a fixed amount regardless of the way the change is carried out. Say ΔU = 2 units and the reversible work w = 4 units. The heat q absorbed in this reversible change is therefore 6 units. These must be the maximum values of w and q, because ΔU is fixed at 2; for any other change than reversible change, w is less than 4 and so q is less than 6.

For an infinitesimal change, the inequality in relation to q can be written

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and so for a change at temperature T

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The term on the left defines the change in the state function entropy

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Since reversible conditions equate to equilibrium and irreversible conditions equate to observable change, it follows that

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These criteria are fundamental. They are true for all thermodynamic processes, subject only to the restriction that the system is a closed one i.e. there is no mass transfer between system and surroundings. It is from these expressions that the conclusion can be drawn – as famously stated by Clausius – that entropy increases towards a maximum in isolated systems.

Rudolf Clausius (1822-1888)

Rudolf Clausius (1822-1888)

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Die Entropie der Welt strebt einem Maximum zu

Consider an adiabatic change in a closed system: dq = 0 so the above criteria for equilibrium and observable change become dS = 0 and dS > 0 respectively. If the volume is also kept constant during the change, it follows from the first law that dU = 0. In other words the volume and internal energy of the system are constant and so the system is isolated, with no energy or mass transfer between system and surroundings.

Under these circumstances the direction of observable change is such that entropy increases towards a maximum; when there is equilibrium, the entropy is constant. The criteria for these conditions may be expressed as follows

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Note:
The assertion that entropy increases towards a maximum is true only under the restricted conditions of constant U and V. Such statements as “the entropy of the universe tends to a maximum” therefore depend on assumptions, such as a non-expanding universe, that are not known to be fulfilled.

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P Mander March 2015

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The Clapeyron equation

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This equation is exact and applies to equilibrium processes between any two phases, i.e melting, boiling and sublimation. It gives the slope of the coexistence curve at (saturated) vapor pressure P and temperature T.

The original derivation by Clapeyron is a geometrical method in which an expression for the PV work done in a Carnot cycle is equated with Carnot’s principle δW = δQ (δT)/T. This classic proof is given in Appendix I.

Derivation

If two phases A and B of the same pure substance are in equilibrium with each other, and GA and GB are the respective molar Gibbs free energies, then the condition of equilibrium is GA = GB (if this were not the case, substance in the phase with higher molar free energy would pass into the phase with lower molar free energy, which is a non-equilibrium condition).

At equilibrium dG = 0, which is also the condition for the thermodynamically reversible change dGA = dGB. Since dG = VdP – SdT, we can equate the differential expressions for the two phases:

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Since we are dealing with a thermodynamically reversible process at constant pressure and temperature, ΔS =(Qp)/T which we can write ΔS = ΔH/T where ΔH is the molar enthalpy of phase transition. Hence

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Applications of the Clapeyron equation

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1. Calculate the enthalpy of phase transition ΔH
If both dP/dT and ΔV at temperature T are known, ΔH can be calculated. Scroll down to Worked example 1

2. Estimate the effect of pressure change on the boiling point temperature
If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the boiling point temperature. Scroll down to Worked example 2

3. Estimate the effect of pressure change on the melting point temperature
If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the melting point temperature. Scroll down to Worked example 3

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Worked Example 1

The vapor pressure of water changes by 27.17 mm Hg from 372.15K to 373.15K, and the corresponding volume change (water>vapor) is 3.0114 x 10-2 m3mol-1. Estimate the enthalpy of vaporization ΔHvap of water at 373.15K.

Strategy

Use the Clapeyron equation to solve for ΔHvap

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The SI units are:
P (vapor) = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
T = K
V = m3mol-1
H = Jmol-1

Calculation

1. At 373.15 K, the vapor pressure of water changes by 27.17 mm Hg per kelvin. This may be taken as the value of dP/dT at 373.15K.

2. Convert dP/dT from mm Hg per kelvin to Nm-2K-1

dP/dT = 27.17 x (105/760) = 3575 Nm-2K-1

3. Compute ΔH = (dP/dT) . TΔV

ΔH = 3575 x 373.15 x 3.0114 x 10-2 = 40172 Jmol-1

The enthalpy of vaporization of water ΔHvap at 373.15 is estimated to be 40.17 kJmol-1.

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Worked Example 2

The enthalpy of vaporization ΔHvap of water at 373.15K is 40,657 Jmol-1 and the corresponding volume change (water>vapor) is 3.0114 x 10-2 m3mol-1. Estimate the boiling point temperature at 770 mm Hg.

Strategy

Use the inverted Clapeyron equation to solve for dT/dP

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The SI units are:
T = K
P = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
ΔV = m3mol-1
ΔH = Jmol-1

Note that P is the vapor pressure. Since T is the boiling point temperature, the vapor pressure and the externally applied pressure will be the same.

Calculation

1. Compute dT/dP

dT/dP = (373.15 x 3.0114 x 10-2)/40657 = 2.76 x 10-4 K(Nm-2)-1

2. Compute the pressure increase in Nm-2

Water boils at 373.15K at a pressure of 760 mm Hg. The pressure increase under consideration is therefore 770-760 = 10 mm Hg.

10 mm Hg = 10 x (105/760) Nm-2 = 1316 Nm-2

3. Put the two facts together

dT/dP tells us that the boiling point is raised by 2.76 x 10-4 K for 1 Nm-2 increase in vapor pressure.

So a 1316 Nm-2 increase in vapor pressure will raise the boiling point temperature by 1316 x 2.76 x 10-4 = 0.36K

The boiling point of water at 770 mm Hg is estimated at 373.15 + 0.36 = 373.51K

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Worked Example 3

At 273.15K the molar enthalpy of fusion of water ΔHfusion is 6 x 103 Jmol-1 and the corresponding volume change (ice>water) is –1.6 x 10-6 m3mol-1. Estimate the melting point of ice at a pressure of 150 atmospheres.

Strategy

Use the inverted Clapeyron equation to solve for dT/dP

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The SI units are:
T = K
P = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
ΔV = m3mol-1
ΔH = Jmol-1

Note that P is the vapor pressure. Since T is the melting point temperature, the vapor pressure and the externally applied pressure will be the same.

Calculation

1. Compute dT/dP

dT/dP = (273.15 x –1.6 x 10-6)/(6 x 103) = –7.28 x 10-8 K(Nm-2)-1

2. Compute the pressure increase in Nm-2

Ice melts at 273.15K at a pressure of 1 atmosphere. The pressure increase under consideration is therefore 150-1 = 149 atmospheres.

149 atmospheres = 1.49×107 Nm-2

3. Put the two facts together

dT/dP tells us that the melting point is lowered by –7.28 x 10-8 K for 1 Nm-2 increase in vapor pressure.

So a 1.49×107 Nm-2 increase in vapor pressure will lower the melting point by

(1.49 x 107) x (–7.28 x 10-8) = –1.08K

The melting point of ice at a pressure of 150 atmospheres is estimated at 273.15 – 1.08 = 272.07K

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The Clausius-Clapeyron equation

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The Clausius-Clapeyron equation is not exact, since approximations are used in its derivation. Its use is confined to processes involving vapor phase equilibrium, but nevertheless the equation is very useful.

Derivation

Beginning with the Clapeyron equation

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use is made of the fact that for sublimation and vaporization processes, there is generally a large difference in molar volume between the two phases. In this case, ΔV can be approximated as the molar volume of the vapor Vvap.

It is then assumed that the behavior of the vapor approximates that of an ideal gas, so that the molar volume Vvap can in turn be replaced with RT/P where P is the vapor pressure, giving

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This is the Clausius-Clapeyron equation, which approximates the slope of the liquid-vapor coexistence curve at vapor pressure P and boiling point temperature T. Using the mathematical identity

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the Clausius-Clapeyron equation can also be written in the form

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If we further assume that ΔHvap is independent of temperature, integration of the above equation can be performed

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This is the integrated form of the Clausius-Clapeyron equation. If the vapor pressure P1 is known at boiling point temperature T1, this equation can be used to estimate the boiling point temperature T2 at another pressure P2. Alternatively, if the boiling point temperatures are known at vapor pressures P1 and P2, the enthalpy of vaporization ΔHvap can be estimated.

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Applications of the Clausius-Clapeyron equation

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or

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and its integrated form

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1. Estimate the enthalpy of vaporization ΔHvap
If both dP/dT and the vapor pressure P are known at boiling point temperature T, the enthalpy of vaporization ΔHvap can be estimated. The value of R needs to be known. Scroll down to Worked example 4

2. Estimate the effect of pressure change on the boiling point temperature
If the vapor pressure P1 is known at boiling point temperature T1, the boiling point temperature T2 at another pressure P2 can be estimated. The values of R and ΔHvap need to be known. Scroll down to Worked example 5

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Worked Example 4

dP/dT is 27.17 mm Hg per kelvin for water at 373.15K. Estimate the enthalpy of vaporization ΔHvap of water at this temperature. The vapor pressure of water at 373.15K is 105 Nm-2, and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the Clausius-Clapeyron equation to solve for ΔHvap

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The SI units are:
T = K
P (vapor) = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
R = JK-1mol-1
ΔH = Jmol-1

Calculation

1. Convert dP/dT from mm Hg per kelvin to Nm-2K-1

dP/dT = 27.17 x (105/760) = 3575 Nm-2K-1

2. Compute ΔHvap = 1/P x dP/dT x RT2

ΔHvap = 10-5 x 3575 x 8.3145 x (373.15)2 = 41388 Jmol-1

The enthalpy of vaporization of water ΔHvap at 373.15K is estimated to be 41.39 kJmol-1 (which is about 2% higher than the experimental value of 40.66 kJmol-1).

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Worked Example 5

Water boils at 373.15K at a pressure of 1 atmosphere. At what temperature will water boil in a pressure cooker operating at a pressure of 2 atmospheres? The enthalpy of vaporization ΔHvap of water at 373.15K is 40657 Jmol-1 and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the integrated Clausius-Clapeyron equation to solve for T2

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The SI units are:
T = K
R = JK-1mol-1
ΔH = Jmol-1

The units of P1 and P2 are immaterial, so long as they are the same.

Note that P1 and P2 are vapor pressures. Since T1 and T2 refer to boiling point temperatures, the vapor pressures P1 and P2 will be the same as the externally applied pressures.

Calculation

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Water in a pressure cooker operating at a pressure of two atmospheres boils at 394K, or 121°C. This explains why vegetables cook so fast in these devices.

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Appendix 1

The Clapeyron equation – Clapeyron’s proof

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Émile Clapeyron was the first to use the PV diagram as an analytical tool

Émile Clapeyron’s 1834 paper Mémoire sur la Puissance Motrice de la Chaleur (Memoire on the Motive Power of Heat) contains the first analytical representation of the Carnot cycle, depicted above as the quadrilateral ABCD. By assigning infinitely small values to the variations of volume V and vapor pressure P during the four successive operations of the cycle, Clapeyron renders the quadrilateral a parallelogram.

He then demonstrates that the parallelogram ABCD, representing the work done in a complete cycle, is equal in area to BCEF since both stand on the same base BC and lie between the same parallels T and T–δT.

The base-altitude product BF.PQ therefore equates to the work done in a complete cycle.

Now BF=δP is the increase in vapor pressure per δT rise of temperature at constant volume, while PQ=δV equates to δQ/Lv where δQ is the heat absorbed and Lv is the latent heat of expansion per unit increase in volume

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Since by Carnot’s principle the work δW done in a complete cycle is δQ (δT)/T

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This is the famous Clapeyron equation.

In modern notation Lv is ΔH/ΔV and since vapor pressure is independent of volume the subscript can be omitted and the equation written

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In textbooks dealing with thermodynamics, the story is often recounted that Carnot’s groundbreaking memoir Réflexions sur la Puissance Motrice du Feu (Reflections on the Motive Power of Fire) went unnoticed by the scientific world when it was published, and lay forgotten for 24 years until Lord Kelvin seized on its significance and brought Carnot’s seminal work – in which lay the seeds of the second law of thermodynamics – into the spotlight of publicity.

This popular account, although broadly accurate, is incomplete in one important respect. It neglects the vital role played by Clapeyron.

Benoît Émile Clapeyron (1799-1864)

Benoît Émile Clapeyron (1799-1864)

Sadi Carnot (1796-1832) and Émile Clapeyron (1799-1864) were contemporaries. Both were engineers by training, and both had studied at the École Polytechnique in Paris. When Carnot published his memoir (at his own expense) in 1824, Clapeyron had already left Paris and was teaching at an engineering school in St. Petersburg, Russia. He returned to Paris just two years before Carnot’s untimely death during a cholera epidemic in 1832. By that time, Sadi Carnot’s little book had long since disappeared from the booksellers’ shelves. But Clapeyron had evidently obtained a copy, and had clearly recognised the importance of his compatriot’s work. Clapeyron’s own Mémoire sur la Puissance Motrice de la Chaleur (Memoir on the Motive Power of Heat), published in Journal de l’École Polytechnique in 1834, contains a full restatement of Carnot’s theoretical principles – albeit in a more analytical form.

While Carnot’s privately published booklet quickly faded into obscurity, Clapeyron’s memoir published in an academic journal did not. It was read in scientific circles and remained available in scientific libraries, and so kept Carnot’s powerful ideas alive. In 1837, Clapeyron’s memoir was translated into English and formed one of the papers presented in Volume I of Taylor’s Scientific Memoirs. This was how Lord Kelvin became aware of Carnot’s work – we know this because the following footnote appears in the historic paper On an Absolute Thermometric Scale which Lord Kelvin published in Philosophical Magazine in October 1848:

“Carnot’s Theory of the Motive Power of Heat – Published in 1824 in a work entitled Réflexions sur la Puissance Motrice du Feu, by M. S. Carnot. Having never met with the original work, it is only through a paper by M. Clapeyron, on the same subject, published in the Journal de l’École Polytechnique, Vol. XIV, 1834, and translated in the first volume of Taylor’s Scientific Memoirs, that the Author has become acquainted with Carnot’s Theory.”

William Thomson, later Lord Kelvin (1824-1907)

William Thomson, later Lord Kelvin (1824-1907)

At the end of 1848, Lord Kelvin succeeded in obtaining a copy of Carnot’s original work, and the following year published a lengthy paper entitled An account of Carnot’s Theory of the Motive Power of Heat in the Transactions of the Edinburgh Royal Society, XVI, 1849.

Curiously, just as Lord Kelvin came to discover Carnot’s masterwork through the agency of Émile Clapeyron’s memoir, so did another key figure in the development of thermodynamics – Rudolf Clausius.

In his first and most famous paper On the Motive Power of Heat, and on the Laws which can be deduced from it for the Theory of Heat published in Annalen der Physik  in 1850, Clausius cites Carnot’s memoir on the opening page and adds this footnote:

Réflexions sur la Puissance Motrice du Feu, par S. Carnot. Paris, 1824. I have not been able to obtain a copy of this work, and am acquainted with it only through the work of Clapeyron and Thomson [Lord Kelvin]”

Rudolf Clausius (1822-1888)

Rudolf Clausius (1822-1888)

Clapeyron’s memoir had been translated into German in 1843 and published in the same journal in which Clausius’ 1850 paper appeared – Annalen der Physik – so it likely to be this translation to which Clausius refers. It is doubtful whether Carnot had much inkling of how crucially important his ideas would be to the development of thermodynamics. But there is no doubt that Clapeyron immediately saw value in them. In the introduction to his memoir, Clapeyron describes Carnot’s ideas as ‘both fertile and beyond question‘ and ‘worthy of the attention of theoreticians‘.

He was right. The two foremost theoreticians of the day certainly found Carnot’s fertile ideas worthy of attention. And we have Émile Clapeyron to thank for enabling Lord Kelvin and Rudolf Clausius to discover them.