Posts Tagged ‘Clausius-Clapeyron equation’

The above image from the northern latitudes of Mars, taken by the Viking 2 Lander in May  1979, revealed the presence of a white substance on and around the red rocks. We get this white stuff on Earth too. It’s called frost.

I can remember when I first saw this image confirming the presence of water frost on the Martian surface, I took an instant step closer to believing in the possibility of Little Green Men. During the Martian summer, the temperature can rise above freezing point, and lo! there would be liquid water, the swimming pool of life.

After revelling in this idea for a while, the thought struck me that I had directly transferred the phase change behavior of water on Earth to another planet. On reflection however, this appeared perfectly valid. There are no planet-dependent terms in the Clausius-Clapeyron equation, so the phase diagram of water must look exactly the same up there as it does down here.

Sound enough reasoning. The phase diagram is a PT plane, and according to what I was taught, those  thermodynamic variables, along with state functions such as enthalpy of fusion, were considered to apply universally. But then duh! the realization dawned on me – I had failed to consider differences in atmospheric pressure.

At the time, I knew nothing at all about the atmosphere on Mars, apart from the fact that it had one. But at least I could calculate the minimum surface pressure that the atmosphere would need to exert in order for water to undergo a solid-liquid phase change at 0°C. The Magnus-Tetens approximation, which generates saturated vapor pressure (hPa) as a function of temperature (°C)

made things simple since when T=0 the answer is immediately 6.1094 hectopascals, or if you prefer, 6.1094 millibars.

Back then, I never did find out what the atmospheric pressure on Mars was, and my inquiry stalled. Years later I came upon the information by happenstance. The atmospheric surface pressure on Mars is just below 6.1 hectopascals.

Now a glance at the phase diagram for water shows that the triple point temperature is a gnat’s whisker from 0°C, so for practical purposes the triple point pressure can be taken as 6.1 hPa. This is significant because at no point in the PT plane below the horizontal drawn through the triple point does the liquid phase exist. Since the atmospheric surface pressure on Mars lies below this horizontal, the only possible phase change is between solid and vapor. No liquid water can exist on the surface of Mars.

Except. Well, there are craters on Mars, so if at the bottom of a sufficiently deep one the atmospheric pressure were to nudge above 6.1 hPa, and if the temperature crept above 0°C, and if there was surface frost to start with, then liquid water would form. That’s a lot of ifs, but a possibility all the same. There might not be Little Green Men down there, but there might be something else that answers the description of ‘life form’.

Photo credits: (top) Mars / NASA, (below) Little Green Men / Wikimedia
Triple point diagram credit: National Metrology Institute of Japan
(I have added the horizontal line through the triple point)

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

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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Barometric pressure and boiling point

Posted: August 4, 2012 in physical chemistry, thermodynamics
Tags: , , , , ,

Our pressure cooker was a Prestige Skyline from the 1950s. The regulator is sticking up at the back.

A familiar object in the kitchen of my youth was our pressure cooker. It cooked the vegetables in a fraction of the time, saving significant amounts of energy and therefore cost.

Our pressure cooker was a stainless steel device equipped with a regulator that maintained an internal pressure of 2 atmospheres. I was thinking back to it the other day, and started wondering about the temperature of the superheated water inside the container. This can be calculated using 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

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.

– – – –

Worked Example

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.