Posts Tagged ‘Èmile Clapeyron’

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James Clerk Maxwell and the geometrical figure with which he proved his famous thermodynamic relations

Historical background

Every student of thermodynamics sooner or later encounters the Maxwell relations – an extremely useful set of statements of equality among partial derivatives, principally involving the state variables P, V, T and S. They are general thermodynamic relations valid for all systems.

The four relations originally stated by Maxwell are easily derived from the (exact) differential relations of the thermodynamic potentials:

dU = TdS – PdV   ⇒   (∂T/∂V)S = –(∂P/∂S)V
dH = TdS + VdP   ⇒   (∂T/∂P)S = (∂V/∂S)P
dG = –SdT + VdP   ⇒   –(∂S/∂P)T = (∂V/∂T)P
dA = –SdT – PdV   ⇒   (∂S/∂V)T = (∂P/∂T)V

This is how we obtain these Maxwell relations today, but it disguises the history of their discovery. The thermodynamic state functions H, G and A were yet to be created when Maxwell published the above relations in his 1871 textbook Theory of Heat. The startling fact is that Maxwell navigated his way to these relations using nothing more than a diagram of the Carnot cycle, allied to an ingenious exercise in plane geometry.

Another historical truth that modern derivations conceal is that entropy did not feature as the conjugate variable to temperature (θ) in Maxwell’s original relations; instead Maxwell used Rankine’s thermodynamic function (Φ) which is identical with – and predates – the state function entropy (S) introduced by Clausius in 1865.

Maxwell’s use of Φ instead of S was not a matter of personal preference. It could not have been otherwise, because Maxwell misunderstood the term entropy at the time when he wrote his book (1871), believing it to represent the available energy of a system. From a dimensional perspective – and one must remember that Maxwell was one of the founders of dimensional analysis – it was impossible for entropy as he understood it to be the conjugate variable to temperature. By contrast, it was clear to Maxwell that Rankine’s Φ had the requisite dimensions of ML2T-2θ-1.

Two years later, in an 1873 publication entitled A method of geometrical representation of the thermodynamic properties of substances by means of surfaces, the American physicist Josiah Willard Gibbs politely pointed out Maxwell’s error in regard to the units of measurement of entropy:

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Maxwell responded in a subsequent edition of Theory of Heat with a contrite apology for misleading his readers:

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– – – –

Carnot Cycle revisited

The centrepiece of the geometrical construction with which Maxwell proves his thermodynamic relations is a quadrilateral drawn 37 years earlier by Émile Clapeyron in his 1834 paper Mémoire sur la Puissance Motrice de la Chaleur (Memoir on the motive power of heat).

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When Émile Clapeyron drew this PV-plane representation of the Carnot cycle in 1834, heat was believed to be a conserved quantity. By the time Maxwell used the diagram in 1871, heat and work were understood to be interconvertible forms of energy, the latter being the conserved quantity.

This is the first analytical representation of the Carnot cycle, shown as a closed curve on a pressure-volume indicator diagram. The sides ab and cd represent isothermal lines, the sides ad and bc adiabatic lines. By assigning infinitely small values to the variations of volume and pressure during the successive operations of the cycle, Clapeyron renders this quadrilateral a parallelogram.

The area enclosed by the curve equates to the work done in a complete cycle, and Maxwell uses the following contrivance to set this area equal to unity.

Applying Carnot’s principle, Maxwell expresses the work W done as a function of the heat H supplied

W = H(T2 – T1)/T2

with T2 and T1 representing the absolute temperatures of the source and sink respectively.
Maxwell then defines

T2 – T1 = 1
H/T2 = 1

The conversion of heat into work is thus expressed as the product of a unit change in temperature T and a unit change in Rankine’s thermodynamic function Φ, equivalent to entropy S:

W = Δ1T . Δ1S = 1

Maxwell’s definitions also give the parallelogram the property that any line drawn from one isothermal line to the other, or from one adiabatic line to the other, is of unit length when reckoned in the respective dimensions of temperature or entropy. This is of central significance to what follows.

– – – –

Geometrical extensions

Maxwell’s geometric machinations consist in extending the isothermal (T1T2) and adiabatic lines (Φ1Φ2) of the original figure ABCD and adding vertical lines (pressure) and horizontal lines (volume) to create four further parallelograms with the aim of proving their areas also equal to unity, while at the same time enabling each of these areas to be expressed in terms of pressure and volume as a base-altitude product.

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As the image from Theory of Heat shown at the head of this article reveals, Maxwell did not fully trace out the perimeters of three (!) of the four added parallelograms. I have extended four lines to the arbitrarily labelled points E, F and H in order to complete the figure.

– parallelogram AKQD stands on the same base AD as ABCD and lies between the same parallels T1T2 so its area is also unity, expressible in terms of volume and pressure as the base-altitude product AK.Ak

– parallelogram ABEL stands on the same base AB as ABCD and lies between the same parallels Φ1Φ2 so its area is also unity, expressible in terms of volume and pressure as the base-altitude product AL.Al

– parallelogram AMFD stands on the same base AD as ABCD and lies between the same parallels T1T2 so its area is also unity, expressible in terms of pressure and volume as the base-altitude product AM.Am

– parallelogram ABHN stands on the same base AB as ABCD and lies between the same parallels Φ1Φ2 so its area is also unity, expressible in terms of pressure and volume as the base-altitude product AN.An

– line AD, which represents a unit rise in entropy at constant temperature, resolves into the vertical (pressure) and horizontal (volume) components Ak and Am

– line AB, which represents a unit rise in temperature at constant entropy, resolves into the vertical (pressure) and horizontal (volume) components Al and An

– in summary: ABCD = AK.Ak = AL.Al = AM.Am = AN.An = 1 [dimensions ML2T-2]

– – – –

Maxwell’s thermodynamic relations

Maxwell’s next step is to interpret the physical meaning of these four pairs of lines.

AK is the volume increase per unit rise in temperature at constant pressure: (∂V/∂T)P
Ak is the pressure decrease per unit rise in entropy at constant temperature: –(∂P/∂S)T

Recalling the property of partial derivatives that given the implicit function f(x,y,z) = 0

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Since AK = 1/Ak

(∂V/∂T)P = –(∂S/∂P)T

AL is the volume increase per unit rise in entropy at constant pressure: (∂V/∂S)P
Al is the pressure increase per unit rise in temperature at constant entropy: (∂P/∂T)S

Since AL = 1/Al

(∂V/∂S)P = (∂T/∂P)S

AM is the pressure increase per unit rise in temperature at constant volume: (∂P/∂T)V
Am is the volume increase per unit rise in entropy at constant temperature: (∂V/∂S)T

Since AM = 1/Am

(∂P/∂T)V = (∂S/∂V)T

AN is the pressure increase per unit rise in entropy at constant volume: (∂P/∂S)V
An is the volume decrease per unit rise in temperature at constant entropy: –(∂V/∂T)S

Since AN = 1/An

(∂P/∂S)V = –(∂T/∂V)S

– – – –

In his own words

I leave it to the man himself to conclude this post:

“We have thus obtained four relations among the physical properties of the substance. These four relations are not independent of each other, so as to rank as separate truths. Any one might be deduced from any other. The equality of the products AK, Ak &c., to the parallelogram ABCD and to each other is merely a geometrical truth, and does not depend on thermodynamic principles. What we learn from thermodynamics is that the parallelogram and the four products are each equal to unity, whatever be the nature of the substance or its condition as to pressure and temperature.”

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– – – –

P Mander August 2014

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Historical background

If you received formal tuition in physical chemistry at school, then it’s likely that among the first things you learned were the 17th/18th century gas laws of Mariotte and Gay-Lussac (Boyle and Charles in the English-speaking world) and the equation that expresses them: PV = kT.

It may be that the historical aspects of what is now known as the ideal (perfect) gas equation were not covered as part of your science education, in which case you may be surprised to learn that it took 174 years to advance from the pressure-volume law PV = k to the combined gas law PV = kT.

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The lengthy timescale indicates that putting together closely associated observations wasn’t regarded as a must-do in this particular era of scientific enquiry. The French physicist and mining engineer Émile Clapeyron eventually created the combined gas equation, not for its own sake, but because he needed an analytical expression for the pressure-volume work done in the cycle of reversible heat engine operations we know today as the Carnot cycle.

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The first appearance in print of the combined gas law, in Mémoire sur la Puissance Motrice de la Chaleur (Memoir on the Motive Power of Heat, 1834) by Émile Clapeyron

Students sometimes get in a muddle about combining the gas laws, so for the sake of completeness I will set out the procedure. Beginning with a quantity of gas at an arbitrary initial pressure P1 and volume V1, we suppose the pressure is changed to P2 while the temperature is maintained at T1. Applying the Mariotte relation (PV)T = k, we write

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The pressure being kept constant at P2 we now suppose the temperature changed to T2; the volume will then change from Vx to the final volume V2. Applying the Gay-Lussac relation (V/T)P = k, we write

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Substituting Vx in the original equation:

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whence

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– – – –

Differences of opinion

In the mid-19th century, the ideal gas equation – or rather the ideal gas itself – was the cause of no end of trouble among those involved in developing the new science of thermodynamics. The argument went along the lines that since no real gas was ever perfect, was it legitimate to base thermodynamic theory on the use of a perfect gas as the working substance in the Carnot cycle? Joule, Clausius, Rankine, Maxwell and van der Waals said yes it was, while Mach and Thomson said no it wasn’t.

With thermometry on his mind, Thomson actually got quite upset. Here’s a sample outpouring from the Encyclopaedia Britannica:

“… a mere quicksand has been given as a foundation of thermometry, by building from the beginning on an ideal substance called a perfect gas, with none of its properties realized rigorously by any real substance, and with some of them unknown, and utterly unassignable, even by guess.”

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Joule (inset) and Thomson may have had their differences, but it didn’t stop them from becoming the most productive partnership in the history of thermodynamics

It seems strange that the notion of an ideal gas, as a theoretical convenience at least, caused this violent division into believers and disbelievers, when everyone agreed that the behavior of all real gases approaches a limit as the pressure approaches zero. This is indeed how the universal gas constant R was computed – by extrapolation from pressure-volume measurements made on real gases. There is no discontinuity between the measured and limiting state, as the following diagram demonstrates:

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Experiments on real gases show that

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where v is the molar volume and i signifies ice-point. The universal gas constant is defined by the equation

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so for real gases

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The behavior of n moles of any gas as the pressure approaches zero may thus be represented by

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The notion of an ideal gas is founded on this limiting state, and is defined as a gas that obeys this equation at all pressures. The equation of state of an ideal gas is therefore

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– – – –

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William Thomson, later Lord Kelvin, in the 1850s

Testing Mayer’s assumption

The notion of an ideal gas was not the only thing troubling William Thomson at the start of the 1850s. He also had a problem with real gases. This was because he was simultaneously engaged in a quest for a scale of thermodynamic temperature that was independent of the properties of any particular substance.

What he needed was to find a property of a real gas that would enable him to
a) prove by thermodynamic argument that real gases do not obey the ideal gas law
b) calculate the absolute temperature from a temperature measured on a (real) gas scale

And he found such a property, or at least he thought he had found it, in the thermodynamic function (∂U/∂V)T.

In the final part of his landmark paper, On the Dynamical Theory of Heat, which was read before the Royal Society of Edinburgh on Monday 15 December 1851, Thomson presented an equation which served his purpose. In modern notation it reads:

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This is a powerful equation indeed, since it enables any equation of state of a PVT system to be tested by relating the mechanical properties of a gas to a thermodynamic function of state which can be experimentally determined.

If the equation of state is that of an ideal gas (PV = nRT), then

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This defining property of an ideal gas, that its internal energy is independent of volume in an isothermal process, was an assumption made in the early 1840s by Julius Robert Mayer of Heilbronn, Germany in developing what we now call Mayer’s relation (Cp – CV = PΔV). Thomson was keen to disprove this assumption, and with it the notion of the ideal gas, by demonstrating non-zero values for (∂U/∂V)T.

In 1845 James Joule had tried to verify Mayer’s assumption in the famous experiment involving the expansion of air into an evacuated cylinder, but the results Joule obtained – although appearing to support Mayer’s claim – were deemed unreliable due to experimental design weaknesses.

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The equipment with which Joule tried to verify Mayer’s assumption, (∂U/∂V)T = 0. The calorimeter at the rear looks like a solid plate construction but is in fact hollow. This can be ascertained by tapping it – which the author of this blogpost has had the rare opportunity to do.

Thomson had meanwhile been working on an alternative approach to testing Mayer’s assumption. By 1852 he had a design for an apparatus and had arranged with Joule to start work in Manchester in May of that year. This was to be the Joule-Thomson experiment, which for the first time demonstrated decisive differences from ideal behavior in the behavior of real gases.

Mayer’s assumption was eventually shown to be incorrect – to the extent of about 3 parts in a thousand. But this was an insignificant finding in the context of Joule and Thomson’s wider endeavors, which would propel experimental research into the modern era and herald the birth of big science.

Curiously, it was not the fact that (∂U/∂V)T = 0 for an ideal gas that enabled the differences in real gas behavior to be shown in the Joule-Thomson experiment. It was the other defining property of an ideal gas, that its enthalpy H is independent of pressure P in an isothermal process. By parallel reasoning

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If the equation of state is that of an ideal gas (PV = nRT), then

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Since the Joule-Thomson coefficient (μJT) is defined

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and the second term on the right is zero for an ideal gas, μJT must also be zero. Unlike a real gas therefore, an ideal gas cannot exhibit Joule-Thomson cooling or heating.

– – – –

Finding a way to define absolute temperature

But to return to Thomson and his quest for a scale of absolute temperature. The equation he arrived at in his 1851 paper,

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besides enabling any equation of state of a PVT system to be tested, also makes it possible to give an exact definition of absolute temperature independently of the behavior of any particular substance.

The argument runs as follows. Given the temperature readings, t, of any arbitrary thermometer (mercury thermometer, bolometer, whatever..) the task is to express the absolute temperature T as a function of t. By direct measurement, it may be found how the behavior of some appropriate substance, e.g. a gas, depends on t and either V or P. Introducing t and V as the independent variables in the above equation instead of T and V, we have

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where (∂U/∂V)t, (∂P/∂t)V and P represent functions of t and V, which can be experimentally determined. Separating the variables so that both terms in T are on the left, the equation can then be integrated:

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Integrating between the ice point and the steam point

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This completely determines T as a function of t.

But as we have already seen, there was a catch to this argumentation – namely that (∂U/∂V) could not be experimentally determined under isothermal conditions with sufficient accuracy.

– – – –

The Joule-Thomson coefficient provides the key

Thomson’s means of circumventing this problem was the steady state Joule-Thomson experiment, which measured upstream and downstream temperature and pressure, and enabled the Joule-Thomson coefficient, μJT = (∂T/∂P)H, to be computed.

It should be borne in mind however that when Joule and Thomson began their work in 1852, they were not aware that their cleverly-designed experiment was subject to isenthalpic conditions. It was the Scottish engineer and mathematician William Rankine who first proved in 1854 that the equation of the curve of free expansion in the Joule-Thomson experiment was d(U+PV) = 0.

William John Macquorn Rankine (1820-1872)

William John Macquorn Rankine (1820-1872)

As for the Joule-Thomson coefficient itself, it was the crowning achievement of a decade of collaboration, appearing in an appendix to Joule and Thomson’s final joint paper published in the Philosophical Transactions of the Royal Society in 1862. They wrote it in the form

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where the upper symbol in the derivative denotes “thermal effect”, and K denotes thermal capacity at constant pressure of a unit mass of fluid.

The equation is now usually written

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By the method applied previously, this equation can be expressed in terms of an empirical t-scale and the absolute T-scale:

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where C’P is the heat capacity of the gas as measured on the empirical t-scale, i.e. C’P = CP(dT/dt). Cancelling (dT/dt) and separating the variables so that both terms in T are on the left, the equation becomes:

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Integrating between the ice point and the steam point

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This completely determines T as a function of t, with all the terms under the integral capable of experimental determination to a sufficient level of accuracy.

– – – –

P Mander May 2014

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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

– – – –

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.

– – – –

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

– – – –

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.

– – – –

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

– – – –

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).

– – – –

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.

– – – –

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.