Posts Tagged ‘vapor pressure’

Dmitiri Konovalov (1856-1929) was a Russian chemist who made important contributions to the theory of solutions. He studied the vapor pressure of solutions of liquids in liquids and in 1884 published a book on the subject which gave a scientific foundation to the distillation of solutions and led to the development of industrial distillation processes.

On the subject of partially miscible liquids forming conjugate solutions, Konovalov in 1881 established the following fact: “If two liquid solutions are in equilibrium with each other, their vapor pressures, and the partial pressures of the components in the vapor, are equal.”

J. Willard Gibbs in America had already developed the concept of chemical potential to explain the behavior of coexistent phases in his monumental treatise On the Equilibrium of Heterogeneous Substances (1875-1878). Konovalov was unaware of this work, and independently found a proof on the basis of this astutely reasoned thought experiment:

«Consider Figure 77 shown above. Two liquid layers α and β in coexistent equilibrium are contained in a ring-shaped tube, and above them is vapor. If the pressure of either component in the vapor were greater over α than over β, diffusion of vapor would cause that part lying over β to have a higher partial pressure of the given component than is compatible with equilibrium. Condensation occurs and β is enriched in the specified component. By reason of the changed composition of β however, the equilibrium across the interface of the liquid layers is disturbed and the component deposited by the vapor will pass into the liquid α. The whole process now commences anew and the result is a never-ending circulation of matter round the tube i.e. a perpetual motion, which is impossible. Hence the partial pressures of both components are equal over α and β and therefore also their sum i.e. the total vapor pressure.»

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Equivalence of vapor pressure and chemical potential

Konovalov showed that the condition of equilibrium in coexistent phases was equality of vapor pressure p for each component. This is consistent with the concept of ‘generalized forces’, a set of intensive variables which drive a thermodynamic system spontaneously from one state to another in the direction of equilibrium. Vapor pressure is one such variable, and chemical potential is another. Hence Gibbs showed that chemical potential μ is a driver of compositional change between coexistent phases and that equilibrium is reached when the chemical potential of each component in each phase is equal. In shorthand the equilibrium position for partially miscible liquids containing components 1 and 2 in coexistent phases α, β and vapor can be stated as:

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P Mander, July 2020

The absolute humidity formula posted in 2012 on this blog has a range of -30°C to 35°C. To expand this range I have developed a new formula to compute absolute humidity from relative humidity and temperature based on a simple but little known polynomial expression (Richards, 1971) for the saturation vapor pressure of water, valid to ±0.1% over the temperature range -50°C to 140°C.

Formula for calculating absolute humidity

In the formula below, temperature (T) is expressed in degrees Celsius, relative humidity (rh) is expressed in %, and e is the Euler number 2.71828 [raised to the power of the contents of the square brackets]:

Absolute Humidity = 1013.25 × e^[13.3185t – 1.9760t^2 – 0.6445t^3 – 0.1299t^4] × rh × 18.01528
(grams/m^3)                                                   100 × 0.083145 × (273.15 + T)

where the parameter t = 1 – 373.15/(273.15 + T)

The above formula simplifies to

Absolute Humidity = 1013.25 × e^[13.3185t – 1.9760t^2 – 0.6445t^3 – 0.1299t^4] × rh × 2.1667
(grams/m^3)                                                                   273.15 + T

To cite this formula please quote: P Mander (2020), carnotcycle.wordpress.com/2020/08/01/compute-absolute-humidity-from-relative-humidity-and-temperature-50c-to-140c

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Study notes

Strategy for computing absolute humidity, defined as water vapor density in grams/m3, from temperature (T) and relative humidity (rh):

1. Water vapor is a gas whose behavior in air approximates that of an ideal gas due to its very low partial pressure.

2. We can apply the ideal gas equation PV = nRT. The gas constant R and the variables T and V are known in this case (T is measured, V = 1 m3), but we need to calculate P before we can solve for n.

3. To obtain a value for P, we can use the polynomial expression of Richards (ref) which generates saturation vapor pressure Psat (hectopascals) as a function of temperature T (Celsius) in terms of a parameter t

Psat = 1013.25 × e^[13.3185t – 1.9760t2 – 0.6445t3 – 0.1299t4]
where t = 1 – 373.15/(273.15 + T)

4. Psat is the vapor pressure when the relative humidity is 100%. To compute the pressure P for any value of relative humidity expressed in %, the expression for Psat is multiplied by the factor rh/100:

P = 1013.25 × e^[13.3185t – 1.9760t2 – 0.6445t3 – 0.1299t4] × rh/100

5. We now know P, V, R, T and can solve for n, which is the amount of water vapor in moles. This value is then multiplied by the molecular weight of water to give the answer in grams.

Absolute humidity (grams/m3) = Psat  ×  rh  ×  mol wt
                                                          100 × R × (273.15 + T)

Saturation vapor pressure Psat is expressed in hectopascals hPa
Relative humidity rh is expressed in %
Molecular weight of water mol wt = 18.01528 g mol-1
Gas constant R = 0.083145 m3 hPa K-1 mol-1
Temperature T is expressed in degrees Celsius

6. Summary:
The formula for absolute humidity is derived from the ideal gas equation. It gives a statement of n solely in terms of the variables temperature (T) and relative humidity (rh). Pressure is computed as a function of both these variables; the volume is specified (1 m3) and the gas constant R is known.

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Formula jpgs

decimal separator = .

decimal separator = ,

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P Mander, July 2020

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I dare say most of you will remember this classroom demonstration, in which water passes through a semi-permeable membrane and causes the liquid level to rise in the stem of the thistle funnel. The phenomenon is called osmosis, and at equilibrium the osmotic pressure is equal to the hydrostatic pressure.

Historical background

This experiment has its origins way back in the mid-18th century, when a French clergyman named Jean-Antoine Nollet tied a piece of pig’s bladder over the mouth of a jar containing alcohol and immersed the whole thing in a vat of water. What prompted him to do this is not known, but we do know the result of his experiment. The bladder swelled up and ultimately burst from the internal pressure.

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Jean-Antoine Nollet 1700-1770

Nollet published his findings in Recherches sur les causes de Bouillonement des Liquides (1748) in which he gave a correct interpretation of the phenomenon, which arises from the much more marked permeability of the bladder to water as compared with alcohol.

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Moritz Traube 1826-1894

The actual measurement of osmotic pressure had to wait for over a century, until the German chemist Moritz Traube showed in 1867 that artificial semipermeable membranes could be made using gelatin tannate or copper ferrocyanide. Traube’s compatriot Wilhelm Pfeffer, a botanist, succeded in depositing the latter in the walls of a porous jar, which when filled with a sugar solution, connected to a mercury manometer and then plunged into pure water, provided a means of measuring osmotic pressures.

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Wilhelm Pfeffer 1845-1920

Following Pfeffer’s osmotic pressure measurements using sucrose solutions, on which JH van ‘t Hoff based his famously flawed gaseous theory of solutions, there were two notable teams of experimentalists – one on each side of the Atlantic – which provided high quality osmotic pressure data to test the ideas of theoreticians. In the USA, Harmon Northrop Morse and Joseph Christie Whitney Frazer led a team at Johns Hopkins University, Baltimore, Maryland from 1901 to 1923. In Britain meanwhile, the aristocrat-turned-scientist Lord Berkeley and co-worker Ernald Hartley set up a private research laboratory near Oxford which operated (with gaps due to war service) from 1904 to 1928.

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Ernald Hartley (1875-1947) Besides being a research chemist, Hartley was an amateur clarinetist who played in the Oxford orchestra for many years. The photo dates from 1925.

While Morse and Frazer used the same principle as Pfeffer, albeit with a more advanced electrochemical method of depositing the membrane in the pores, Berkeley and Hartley reversed the arrangement of solvent and solution, applying measured pressure to the latter to attain equilibrium.

osm06

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Theoretical development in Europe

In Europe, the rigorous application of thermodynamics to the phenomenon of osmosis started in 1887 with Lord Rayleigh, who combined the use of the ideal gas law PV = nRT with the idea of a reversible isothermal cycle of operations in which the sum of work in the complete cycle is zero.

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Lord Rayleigh 1842-1919

Being essentially an attempt to provide hypothesis-free support to van ‘t Hoff’s troubled gaseous theory of solutions, the solute in Rayleigh’s cycle was a mole of ideal gas, which was first dissolved in the solution by applied pressure and then recovered from the solution by osmotic pressure to return the system to its original state.

Rayleigh’s approach, using a zero-sum cycle of operations, was thermodynamically sound and continued to form the basis of theoretical development in its next phase, which in Europe focused on vapor pressure following the influential papers of Alfred Porter in 1907 and Hugh Callendar in 1908.

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Alfred Porter 1863-1939

By 1928, the theoretical model in JAV Butler’s popular textbook The Fundamentals of Chemical Thermodynamics was close to the familiar classroom demonstration of osmosis shown at the head of this post, in which the hydrostatic pressure acting on the solution counteracts the tendency of the solvent to pass through the semi-permeable membrane. At equilibrium, the hydrostatic pressure P is equal to the osmotic pressure.

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JAV Butler 1899-1977

To obtain a thermodynamic relation for osmotic pressure in terms of vapor pressures, Butler uses Rayleigh’s idea of a reversible isothermal cycle of operations together with a semipermeable membrane in the form of a movable piston between the solution and the solvent:

osm10

The diagram shows a solution under hydrostatic pressure P which is equal to the osmotic pressure. Below the semi-permeable piston is pure solvent. Butler then applies the following argumentation:

1] Vaporize 1 mole of the pure solvent at its vapor pressure p0, and expand it reversibly so that the vapor pressure falls to p equal to the partial pressure of the solvent in the solution (Butler assumes that p is not affected by P applied to the solution). Condense the vapor into the solution. Since the work of vaporization and condensation cancel out, the only work done is the work of expansion. Assuming the vapor obeys the ideal gas law, the work (w) done is given by the textbook isothermal expansion formula

2] Now move the semi-permeable piston up against the pressure P until a quantity of solvent equivalent to 1 mole of vapor has passed through it. If the decrease in the volume of the solution is ΔV, the work done is PΔV.

The cycle is now complete and the system has returned to its original state. The total work done is zero and we may equate the two terms

(1)

where P is the osmotic pressure, ΔV is the partial molal volume of the solvent in the solution, p0 is the vapor pressure of the pure solvent and p is the vapor pressure of the solvent in the solution. This thermodynamically exact relation, which involves measured vapor pressures, is in good agreement with experimental determinations of osmotic pressure at all concentrations.

There is a great irony here, in that this equation is exactly the one that JH van ‘t Hoff found his way to in Studies in Chemical Dynamics (1884), before he abandoned his good work and went completely off-track with his gaseous theory of solutions.

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JH van ‘t Hoff 1852-1911

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Theoretical development in America

In the US, the theory of the semipermeable membrane and the ‘equilibrium of osmotic forces’ was the work of one supremely gifted man, Josiah Willard Gibbs, who more or less single-handedly laid the theoretical foundations of chemical thermodynamics in his milestone monograph On the Equilibrium of Heterogeneous Substances.

J Willard Gibbs 1839-1903

But before delving into the powerful idea he introduced, let us return to the subject of equilibrium in a system subject to osmotic pressure with a set-up that is slightly different to that used by Butler. In the diagram below, the piston supplies pressure Psoln to the solution which is just enough to stop solvent passing through the membrane and bring about equilibrium at constant temperature; the osmotic pressure is defined as the excess pressure Psoln – p01.

osm13

The question can now be asked: Does the condition of osmotic equilibrium coincide with equality of a thermodynamic variable on either side of the membrane? Clearly it cannot be pressure or volume, nor can it be temperature since constant temperature does not prevent osmotic disequilibrium.

The P, V, T variables do not provide an affirmative answer, but in his monumental masterwork, Gibbs supplied one of his own invention which did – the chemical potential, symbolized μ. It is an intensive variable which acts as a ‘generalized force’, driving a system from one state to another. In the present context the force drives chemical components, capable of passing through a membrane, from a state of higher potential to a state of lower potential.

So given a membrane dividing solution from solvent and permeable only to the latter, we can understand the osmotic force driving the solvent (designated by subscript 1) through the membrane into the solution in terms of movement to a region of lower potential since

Now the difference in potential can be calculated according to the textbook formula

(2)

where x1 is the mole fraction (<1) of the solvent in the solution. To achieve equilibrium, the chemical potential of the solvent in the solution must be increased by the amount –RTlnx1 (a positive quantity since lnx1 is negative). This can be done by increasing the pressure on any solution exhibiting ideal behavior since

is positive (V1 is the partial molar volume of the solvent in the solution).

The osmotic pressure is defined as the excess pressure Psoln – p01. As can be seen from the diagram below, this is the pressure required to raise the chemical potential of the solvent in the solution so that it becomes equal to the chemical potential of the pure solvent.

Since the slope is V1, it follows that

(3)

Combining (2) and (3) and designating the osmotic pressure by P gives the desired equilibrium relation

This is exactly equivalent to equation (1) derived by Butler, since by his terminology

The two methods of proof are thus shown to be equivalent – we can regard osmotic pressure as the excess pressure required to increase either the chemical potential or the vapor pressure of the solvent in the solution. But Gibbs saw an advantage in using potentials, which he voiced in an 1897 letter to Nature entitled Semi-Permeable Films and Osmotic Pressure:

“The advantage of using such potentials in the theory of semi-permeable diaphragms consists … in the convenient form of the condition of equilibrium, the potential for any substance to which a diaphragm is freely permeable having the same value on both sides of the diaphragm.”

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A closer look at the equations applied to osmotic systems

The foregoing analysis has reached its conclusions by assuming that solutions in osmotic systems exhibit ideal behavior. On this assumption equation 2

can be interpreted as an ‘osmotic’ equation showing that excess pressure equal to –RTlnx1 needs to be applied to the solution to bring the system into osmotic equilibrium by stopping the passage of solvent into the solution. But there are no membrane-dependent terms in this equation, which can be re-written as

This simply shows the effect of adding a solute to a solvent. It should be noted that the equation is only valid at 1 atmosphere pressure, and this provides the clue to how the equation can properly be extended to apply to osmotic systems. If excess pressure ΔP is applied to the system an extra term is required to account for the fact that µ(solvent) is a function of pressure

We now have an equation that can be applied to osmotic systems. Since (δµ1/δP)T is the partial molar volume of the solvent in the solution (V1) the equation becomes

If V1 is positive it is evident that osmotic equilibrium is obtained when the condition

is met. This equation of condition allows us to conclude that solvent passes spontaneously into the solution, and that this process can be arrested by applying to the solution an excess pressure ΔP which we call the osmotic pressure. Note that this pressure is external in origin and arises by virtue of the fact that a semi-permeable membrane exists between solvent and solution rather than through a property of the membrane itself or component interactions with it*.

But what if the partial molar volume of the solvent in the solution (V1) is negative? This can in fact happen due to ionic attraction because V1 is defined not as a solvent volume but the change in volume when solvent is added to the solution. Referring to the above equation of condition, it can be seen that since RTlnx1 is always negative, equilibrium when V1 is negative can only be obtained when ΔP is also negative. Reduced pressure on the solution is equivalent in its effect to excess pressure on the solvent. This allows us to conclude that solvent passes spontaneously out of the solution and that this process can be arrested by applying to the solvent an excess pressure ΔP.

This reversal of what we commonly understand as osmosis might seem incredible, but it does happen under these conditions. It is important however not to view this as the spontaneous unmixing of a mixture. The membrane, the solvent phase and the negative partial molar volume of the solvent in the solution are part of the description of the total system, which will spontaneously move only in the direction of thermodynamic equilibrium and never away from it.

* This is consistent with Wilhelm Ostwald’s incontrovertible proof that osmotic pressure is independent of the nature of the semi-permeable membrane.

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Suggested further reading

Eric M Kramer and David R Myers: Five popular misconceptions about osmosis, American Journal of Physics 80, 694 (2012).

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P Mander March 2019, revised and extended January 2022

ccequation2

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

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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

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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

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

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

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

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.

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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

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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

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

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.

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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

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

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jefferson st nyc subway L

NYC Subway flooding 2nd September 2021

The incidents of severe flooding in the United States, Germany, Belgium and several other places in the world during the summer of 2021 were traumatizing and ruinous to the many people affected and a shock to weather forecasters and local authorities. A wake-up call has been served and much meteorological research and analysis has been set in motion. (more…)