## Posts Tagged ‘Carnot’s principle’

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:

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

– – – –

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).

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, with energy 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.

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

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.”

– – – –

P Mander August 2014

The Clapeyron equation

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:

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

– – – –

Applications of the Clapeyron equation

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

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

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

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

– – – –

The Clausius-Clapeyron equation

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

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

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

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

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

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

or

and its integrated form

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

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

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

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

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

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

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

– – – –