Entropy and the Second Law of Thermodynamics

7/30/2019 Entropy and the Second Law of Thermodynamics

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ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

1. Introduction .................................................................................... 1

2. Entropy and probability ................................................................. 1

3. Relationship between entropy and properties of an ideal gas... 5

4. The Second Law of Thermodynamics ......................................... 7

5. Examples of calculations of entropy changes ............................. 7


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ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

1. Introduction

Our motive for studying thermodynamics is to be able to predict whether areaction or change in a system will occur spontaneously. Most spontaneousreactions evolve heat (exothermic) but there are exceptions. KI dissolvesspontaneously in water but the water gets cooler and ice spontaneously melts at25C. Both these changes are endothermic. A second factor also contributes tothe spontaneity of the system; this factor is the tendency of the system to becomemore random (disordered). This factor is called entropy.

Systems tend to proceed spontaneously toward an equilibrium state. When a hotobject and a cold object are brought into contact, they proceed to a common

temperature. If a container of gas is opened to the atmosphere the gas becomesuniformly distributed through the atmosphere. Solutes or solvents always diffusethrough a membrane in such a way that the concentrations on both sides of themembrane become equal. Electrochemical cells always lose their ability to providecurrent.

Entropy, S, is a state function which is used as a measure of capacity for change. Itoriginally arose as part of a discussion of the efficiency of a theoretical heatengine. That discussion is beyond the scope of this course. In 1896, Boltzmann

explained entropy using the concept of probability and we will introduce entropyby this means, beginning with a very simple model of two molecules beingdistributed between imaginary compartments in two interconnected containers.

2. Entropy and probability

Reference: C.W. Castellan, Physical Chemistry,(Addison-Wesley, 1971) pp 195-200

Assume we have a system of two containers, interconnected so that molecules traymove from one to the other when a tap is opened. Assume that each container hastwo compartments, each of which may hold one molecule, and that initially bothof the molecules to be considered are in the one container.

i.e.


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2

Now the tap is opened and the two molecules are free to move wherever theychoose and to fill the two containers in all possible ways. The resultingdistinguishable arrangements for two identical molecules are:

There are six possible arrangements. The probability of any one arrangement is1/6.

If the number of compartments in the two containers is increased to four 1 you cansee from the diagram below that the total number of possible arrangements is 28.

The latter situation represents a state of greater disorder than the former in thatthere are more possible arrangements.

The entropy, S, of a system in a specified state can be defined in terms of the

number of possible arrangements of the particles composing the system using theequation (1), proposed by Boltzmann.

S = kln W (1)

where k = Boltzmann constant

W = probability or number of possible arrangements


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3

For example: For the first situation: For two molecules confined to the left handside container w = 1.

S1 = kln 1 = 0

For two molecules to be anywhere in the two containers

W=6

S2 = kln 6 = 1.79k

The entropy change associated with the expansion of the system from twocompartments to four compartments is

S = S2 - S1 = kln 6 for two molecules

=2

kln 6 for one molecule

For the expansion from two compartments to four:

S = S3 - S1 = kln 28 for two molecules.

Notice that Sis positive and increases as the number of possible arrangementsincreases. A change from a less probable state for a system to a more probablestate is associated with an increase in entropy.

Now we will generalize the discussion for the case of two molecules.

If the containers have N1 cells into which two molecules may be placed, there areN1 ways to place the first molecule and, for each choice of compartment for thefirst molecule there are (N1 - 1) choices for the second molecule, i.e. the totalnumber of arrangements is N1 (N1 - 1). However, since the molecules areindistinguishable, the number of distinguishably different arrangements is:

W = 2

1NN11

(2)


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4

S1 = kln [ N1 (N1 -1) ] (3)

If the number of compartments is increased by N2

S2 = kln [ N2 (N2 - 1)] (4)

S = S2 - S1

= kln

1NN

1NN

11

22 (5)

If N2 = 4, N1= 2 we have the situation studied above i.e. S = k ln 6.

Suppose N1 and N2 are very large:

then N ~ (N1 - 1)

N2 ~ (N2 - 1)

Then S = kln2

1

2

N

N

(6)

= 2kln1

2

N

Nfor two molecules (7)

Now let us assume that we are dealing with molecules of an ideal gas. The positionof a molecule at any time is the result of pure chance, and since molecules areindependent, the proximity of other molecules does not affect the chance of amolecule being where it is. We may imagine that we are dealing with imaginarycompartments of a given size, in which case the number of compartments wouldbe proportional to the volume available to the gas.

i.e. N1 V1

N2 V2

and1

2

1

2

V

V

N

N

Thus, for two molecules we may substitute in (7)


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5

S = 2kln1

2

V

V(8)

This treatment may be extended to the Avogadro number of molecules, providedthat the number of locations available for the molecules is very large compared to

the number of molecules. Equation (6) becomes:

S = kln A1

2 N

N

N

= kln A1

2 N

V

V

= NAkln1

2

V

V(9)

but you will remember from Section 2.4 that

k =AN

R

(9) becomes S = Rln 1

2

V

V

(10)

From the standpoint of a statistical definition of entropy, expansion of a gas froma volume V1 to a volume V2 increases entropy because there are more ways toarrange a given number of molecules in a larger volume than in a smaller volume.

3. Relationship between entropy and properties of an ideal gas

From the First Law of Thermodynamics:

dU = dq + dw (11)

For an ideal gas the internal energy is constant at constant temperature. Thus, foran isothermal reversible change

dU = 0


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6

and dqrev = -dwrev

but we have deduced in Section 2. that

dwrev= - nRT ln1

2

VV

dqrev = nRT ln1

2

V

V

For one mole,

1

2rev

V

VlnR

T

dq (12)

Substitute from (11) into (13) for an infinitesimal change

i.e. dS =T

dqrev (13)

Remember that wrev is the minimum amount of work done on a system.

Thus, for non-reversible process i.e. one taking place under other that equilibriumconditions

dS >T

dqirrev (14)

and in general

dS >T

dq(15)

Equation (15) is the defining equation for the entropy function. To assist us tounderstand entropy we have chosen a very special set of conditions to derive thisfunction. You should understand however that equation (15) is a general equation

which may be derived in other ways which are beyond the scope of this course.


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7

4. The Second Law of Thermodynamics

In this course we are particularly interested in equation (15) as a useful operationalstatement of the Second Law of Thermodynamics for a system not completelyisolated from its surroundings. A more general statement is possible for a system

which is totally independent, because in such a case there will be no exchange of

heat i.e. dqrev = 0. Then (15) becomes

dS 0 (16)

This equation maybe associated with the statements given on p.169 of G.M.Barrow, Physical Chemistry, (McGraw-Hill, 1973).

The two statements that constitute the second law are:

1. When a process is carried out reversibly the entropy change in the universe ofthe process is zero ..........

2. For processes that proceed irreversibly i.e. out of balance and thereforespontaneously, the entropy of the universe of the process increases."

i.e. Suniverse 0 (compare with the First Law Uuniverse = 0).

5. Examples of calculations of entropy changes

(1) Ideal gas expansionCalculate Sfor the isothermal reversible expansion of 2.00mol of an ideal gasfrom 10.0 to 12.0 L at 27C.

SolutionFor the isothermal reversible expansion of an ideal gas

S =T

qrev = nR ln1

2

V

V

where V2 is the final volume


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8

V1 is the initial volume

= 2 x 8.314ln0.10

0.12

= 3.03 JK-1 Ans.

(2) Phase changesCalculate the entropy change occurring when one mole of water is converted tosteam reversibly at its boiling point and 1atm pressure. (The heat of vaporizationof water = 40.67 kJ mol-1 ).

Solution

For a reversible process:

S =T

qrev

=373

40670

= 109JK-1 Ans.

(3) Chemical ReactionsSince entropy is a state function, Hess's Law also applies. Thus for a reaction suchas

2H2(g) + O2(g) 2H2O(l) at 25oC

Sreaction = 2So(H2O(l)) - 2S

o(H2(g)) - So(O2(g))

You will observe here that entropy content, S is given (compared to enthalpychange, H).This is possible because we have a reference point for entropy -absolute zero. The Third Law of Thermodynamics states a perfect crystal isperfectly ordered at absolute zero (i.e. s = 0).

As a consequence, S for elements is not zero at 25C. Tables in Unland also listS.

Sreaction


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9

= 2(69.6) - 2(130.6) - 205

= -326J.

Since the evolution of a gas is the biggest change in entropy of the phase changes,a useful approximation is if ngas> 0 then S > 0.

The emphasis in this section has been on the concept of entropy. The definitiondS= dqrev/T is fundamentally important but not very useful for calculation.

Entropy calculations for irreversible pathways are usually calculated by trying tomatch this irreversible pathway to a series of reversible pathways. The examples in

5. are the only ones we need to consider.


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10

LIQUID VAPOUR EQUILIBRIA FOR PURE SOLVENTS ANDSOLUTIONS

1. Introduction ................................................................................. 11

2. The vapour pressure of pure liquids .......................................... 11

3. Phase diagram for a pure substance ......................................... 13

4. The vapour pressure of solutions: Raoult's Law ...................... 14

5. Worked example ........................................................................... 16

6. Henry's Law ................................................................................. 177. Worked example ........................................................................... 18

8. Liquid-vapour equilibria for non-ideal solutions ....................... 19

9. Pressure-composition diagrams for ideal solutions ................. 20

10. Boiling point - composition diagrams for ideal solutions ........ 21

11. Vapour-pressure composition and boiling point-composition

diagrams for non-ideal solutions ....................................................... 23


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LIQUID VAPOUR EQUILIBRIA FOR PURE SOLVENTS ANDSOLUTIONS

1. Introduction

As you have seen previously, when pressure is sufficiently increased ortemperature sufficiently lowered, a real gas will condense to a liquid, still inequilibrium with a vapour or gaseous phase. Liquids represent a condensed phaseof matter for which a given amount has a fixed volume under given temperatureand pressure conditions. Liquids, like gases and unlike solids, take the shape oftheir container.

Some structure may be recognised in the liquid state, but basically the liquid isquite a disordered arrangement.

2. The vapour pressure of pure liquids

Like molecules in the gaseous phase, molecules in a pure liquid are in constantrandom motion with a distribution of velocities. Occasionally a molecule travellingat a relatively high velocity breaks through the surface of the liquid and enters thegaseous or vapour phase. This is the process ofevaporation. Increasing thetemperature of the liquid increases the kinetic energy of the molecules and favourevaporation. In the reverse process, when molecules of low kinetic energy in the

vapour phase collide with the liquid surface and are trapped, condensationoccurs.

If the liquid and vapour are enclosed and the space is not so large that the liquidevaporates completely, then eventually, at a fixed temperature, an equilibrium willbe established.

liquid vapour

For a particular liquid, the pressure exerted by the vapour is a function only of thetemperature and is called thevapour pressure of the liquid at that temperature.

As temperature is increased, so is the vapour pressure. Typical data for two liquidsis shown graphically below.

(Data is from Barrow, G.M., Physical Chemistry, (McGraw-Hill, 1973, page 542.)


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The quantity of heat absorbed in the conversion of one mole of a liquid into onemole of a gas is the molar heat (or enthalpy) of vaporization. The more stronglythe molecules of a liquid are attracted to one another, the more heat is required to

vaporize the liquid. Thus water, in which molecules are strongly attracted byhydrogen bonding in addition to normal van der Waal's forces, will have a muchhigher enthalpy of vaporization than will, say, diethyl ether where only van der

Waal's forces are significant. Actual data are:

Hvap (water) = 44.8kJ mol-1 at 0C

Hvap (diethyl ether) = 28.8kJ mol-1 at 0C

The temperature at which the vapour pressure of the liquid equals atmosphericpressure is the normal boiling point of the liquid. Thus the boiling point of pure

water is 100C and the boiling point of pure diethyl ether is 35C at 101,325Nm-2

in each case.

Of course at a given temperature the vapour pressure of a liquid with highcohesive forces between the molecules (non volatile liquid) will be lower than forone with weak cohesive forces (volatile liquid). Again, from the diagram above itcan be seen that the vapour pressure of water at 20C is 0.023atm while the

vapour pressure of diethyl ether at 20C is 0.581atm.

You should also remember that, since the molecules moving at the highestvelocities are the ones most likely to escape from a liquid the average kinetic


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energy of the remainder tends to fall, so that evaporation has a cooling effect.Evaporation of perspiration on the skin is an important mechanism for coolingthe body and the effect is even more noticeable if the evaporating liquid is more

volatile, one such as alcohol (in methylated spirits, for example).

When the escaping molecules are not trapped in contact with the liquid (as in avessel open to the atmosphere) equilibrium is not established, and evaporationgoes to completion.

Such processes are aided by mechanical removal of the vapour from the vicinity ofthe liquid, as when a wind blows over the surface of the liquid.

Apart from increasing the temperature, the rate of evaporation may be increasedby lowering the external pressure. While the normal boiling point of a liquid isdefined as the temperature at which the liquid boils under a pressure of oneatmosphere, the liquid may be made to boil at a lower temperature simply bylowering the external pressure e.g. by creating a vacuum above the liquid. Thus atan external pressure of 0.100 atmospheres, water will boil at 46C.

3. Phase diagram for a pure substance

The discussion above related only to liquid vapour equilibria. As temperature and

pressure are varied for a pure substance, three phases are of significance - solid,liquid and gaseous. A typical phase diagram relating all three is shown in thesketch below:

TC represents the liquid-vapour equilibrium discussed in the previous sections,while BT represents the solid-liquid equilibrium e.g. ice water and AT represents


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the solid-vapour equilibrium e.g. ice water vapour at temperature below 0C atone atmosphere pressure. T is known as the triple point, and represents the one set of

The freezing point of the solvent is the temperature at which solid and liquid arein equilibrium at one atmosphere pressure. For water this is 0C.

4. The vapour pressure of solutions: Raoult's Law

A solution may be defined as a homogeneous mixture of two or morecomponents that form a single phase. We will begin by considering so called idealsolutions consisting of two similar liquids mixed together.

An ideal solution is one in which it is assumed that all intermolecular forces are

equal. i.e. in a solution consisting of A and B it is assumed that A-A, A-B and B-Bintermolecular forces are equal. Such a situation is closely approached by mixturesof benzene (I) and toluene (II).

The thermodynamic consequences of ideality is that there are no enthalpy changesassociated with the mixing process. The free energy change G, is given generallyby

G = H -TS

If H = 0

Then G = -TS

Thus the free energy change associated with mixing arises only as a result ofentropy changes. It should be obvious that the mixing process will lead to anincrease in entropy for the system.

In the vapour phase above a solution there will be molecules of both species

involved i.e. each will contribute a partial pressure to the total pressure. Accordingto Dalton's Law, the total pressure P is given by


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P = P1 + P2

where P1 and P2 are the partial pressures of components 1 and 2 respectively. Foran ideal solution

P1 = X1. P 1o

P2 = X2. P 2o

where X1 and X2 are the mole fractions and P 1o and P 2

o are the vapour pressures

of the pure liquids respectively. These two equations express Raoult's Law i.e. thepressure of any component is equal to its mole fraction X multiplied by itspressure in the pure state.

Mole fraction is defined as the number of mole of the species underconsideration divided by the total number of mole of all species in the solution.

Thus, if there are n1 mole of component 1 and n2 mole of component 2, then

X1 =21

1

nn

n

X2 =21

2

nn

n

and X1 + X2 = 1

A plot for the benzene-toluene system which obeys Raoult's Law is shown below.The temperature is 20C.


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The above properties are reasonable on a qualitative basis. Since evaporation is asurface phenomenon, and since the area of surface occupied by each type ofmolecule will be proportional to its mole fraction in the solution, then the numberof molecules of each type in the vapour above the liquid should also be related tomole fraction.

5. Worked example

The vapour pressures of pure benzene and pure toluene at 60C are 5.13 x l04Nm-2 and 1.85 x 104 Nm-2 respectively. Calculate the partial pressures of benzeneand toluene, the total vapour pressure of the solution and the mole fraction oftoluene in the vapour above a solution with 0.60 mole fraction of toluene.

SolutionFrom Raoult's Law:

Pi = Xi P io

where Pi = partial pressure of component i

Xi = mole fraction of component i

P io = vapour pressure of pure i

For toluene,


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P(toluene) = 0.60 x 1.85 x 104

= 1.11 x 104 Nm-2 Ans.

Since X(toluene) + X(benzene) = 1.00

X(benzene) = 0.40

For benzene

P(benzene) = 0.40 x 5.13 x 104

= 2.05 x 104 Nm-2 Ans.

From Dalton's Law of Partial Pressures

P = P(toluene) + P(benzene)

= 3.16 x 104 Nm-2

Since mole fraction is proportional to partial pressure

X(toluene) =4

4

10x16.3

10x11.1

= 0.351Ans.

[Note that the vapour is not as rich in the less volatile component (toluene) as isthe solution.]

6. Henry's Law

When the solute in a dilute solution is volatile it is found that

P2 = k X2

where P2 = vapour pressure of the solute


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X2 = mole fraction of the solute

k = constant dependent on the nature of the solute andsolvent and the temperature.

This relationship is known as Henry's Law.

It is also convenient to write Henry's Law, as an alternative form:

P2 = km m2

where P2 = vapour pressure of the solute (2)

m2 = molality of component 2

km = constant dependent on the nature of the solute and solventand the temperature.

Henry's Law is of importance since it is particularly applicable to the dissolution ofsmall concentrations of gases in liquids. The effervescence observed when a bottleof drink is opened is an example of the decrease in the solubility of a gas as itspartial pressure is lowered. Another is the problem of dissolved nitrogen in the

blood of deep sea divers. At a point 40m below the surface of seawater the totalpressure is about 6atm. The solubility of nitrogen in plasma is then about3.7x10-3mol kg-1, nine times that at sea level. If the diver swims upward rapidlydissolved nitrogen gas will start boiling off and bubbles in the bloodstream willresult in dizziness and possibly death.

Henry's Law is really only applicable when there is no reaction between the gasand the solvent. Thus it is not applicable to oxygen in blood since the oxygenreacts with blood components such as haemoglobin.

7. Worked example

The Henry's Law constant k' for nitrogen in water at 298K is 1610atm mol-1 kg.What is the concentration of N2 in water in contact with air at 298K?

SolutionFrom Henry's Law:

2NP = k' 2Nm


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where2N

P = partial pressure of nitrogen

2Nm = molar concentration of N2

Since air contains 78% nitrogen by volume, its partial pressure in air is 0.78atm.

k

Pm 2

2

N

N

1610

78.0= 4.8 x 10-4 mol kg-1 of water

8. Liquid-vapour equilibria for non-ideal solutions

Raoult's Law provides a mathematical description for the behaviour of idealsolutions i.e. those solutions in which all intermolecular forces are equal. Ingeneral this situation does not hold.

If the force of attraction between different molecules in a solution is stronger thanbetween identical molecules, then the molecules of each type have more trouble inescaping from the solution than they would from an ideal solution. Thus the

vapour pressure due to each component and the total vapour pressure will be less

than that predicted from Raoult's Law.

If the force of attraction between unlike molecules is less than between identicalmolecules, then the vapour pressure due to each component and the total vapourpressure will be higher than predicted from Raoult's Law.

Note from the diagrams that in the first case there is a minimum in the totalvapour pressure-solution composition curve (negative deviation from Raoult's


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Law), while in the second case there is a maximum (positive deviation fromRaoult's Law).

9. Pressure-composition diagrams for ideal solutions

A vapour pressure-composition diagram is a device for showing, on one diagram,the composition of liquid and vapour phases which are in equilibrium over theentire range of solution composition from pure component 1 to pure component2.

Consider an ideal solution containing components 1 and 2. From Raoult's Law:

P1 = X1. P1

o P2 = X2. P2

o

where P1, P2 are vapour pressures due to components 1 and 2

X1 , X2 are mole fractions of components 1 and 2 in the liquid phase

P 1o , P 2

o are vapour pressures of pure 1 and 2.

Using Dalton's Law of Partial Pressures (pressure fraction is proportional to molefraction),

v

1X =21

1

PP

P

v2X =

21

2

PP

P

where v1X andv

2X are the mole fractions of components 1 and 2 in the vapour

phase. These equations may be rewritten:

v

1X =21

o

11

PP

PX

v2X =

21

o

22

PP

PX

The ratio of v1X tov

2X is


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V

2

V

1

X

X=

O

2

O

1

2

1

P

P

X

X

This equation may be used qualitatively to show that the vapour above a solutionof composition (X1 , X2) will be relatively richer in component 1

( V1X > X1) and poorer in component 2 (V

2X < X2) ifO

1P is greater thanO

2P .

Study Example 9.6 again and notice that you have actually followed the abovederivation there.

The equation may be used to calculate points for a vapour pressure-composition

diagram such as that shown below, once O1P and O2P are known.

In such a diagram, the X axis is used to represent both liquid and vapourcomposition. Thus the solution corresponding to point A on the liquid line has a

vapour pressure P and the composition of the vapour above the liquid may befound by extrapolating horizontally to the vapour line at point B and down to the

X axis. Note again thatV

2X is greater than X2 since component 2 is the most

volatile of the pair. Remember that the vapour line is always under the liquid line.

10. Boiling point - composition diagrams for ideal solutions

Since laboratory processes are more often performed at constant pressure (oneatmosphere) than at constant temperature, diagrams which show the relationshipbetween liquid composition vapour composition and temperature are of


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considerable practical significance. Such diagrams are difficult to deducemathematically, and are normally based on experimental data.

Again, using qualitative reasoning, we may see that the pure liquid withthe highest pressure will have the lowest boiling point, and vice versa. At constant

pressure, the liquid is the stable phase at low temperature, so the liquid line is nowbelow the vapour line.

Allow a liquid of composition X2 to evaporate at constant pressure. When thevapour above the liquid is condensed it has a composition X, which is richer in themore volatile component 2 than is the original liquid. If the liquid X werecollected and re-evaporated, it would be still richer in the more volatilecomponent (composition X1) .

This discussion introduces the subject of distillation - the separation of liquids ofdiffering volatility by heating the solution and collecting and condensing the

vapour which is richer in the more volatile component. By repeated application ofthe technique it is possible to separate completely mixtures whose behaviour isdescribed by the graph above.

In practice, the process of fractional distillation combines many hypotheticaldistillations into a single process. The type of apparatus used is shown in thediagram below.


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The high boiling fraction, rich in the least volatile component, is concentrated in

the pot, while the most volatile components concentrate at the collection point atthe top of the still. A gradient is established in between.

Fractional distillation is widely employed in industry and fractionating columns area particularly important part of any plant such as an oil refinery. Here thepetroleum is separated into fractions such as gasoline (boiling range 40 - 200C),kerosene (boiling range 175 - 325C), etc.

11. Vapour-pressure composition and boiling point-compositiondiagrams for non-ideal solutions

In general solutions are not ideal. The most commonly encountered situation hasa maximum in the vapour pressure-composition diagram and consequently aminimum in the boiling point-composition diagram. The diagrams are complexand are normally determined experimentally. A consequence of the existence ofminima or maxima in the curves is that at some point vapour and liquid inequilibrium have the same composition, and it is not possible to separatecompletely the two components of the mixture. Study the diagram below which istypical of those for mixtures with a minimum boiling point.


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Begin with a solution containing X1 mole fraction of component 1. If this liquid isheated it will begin to boil at t1 and if a sample of the vapour is collected and

condensed, the condensate will have composition X2. If a solution containing X3mole fraction of component 1 is heated it will begin to boil at t2 and the collecteddistillate will have composition X4. Note that in each case, repeated distillationprocesses will lead to a distillate which has the composition X corresponding tothe minimum point in the curve, while the solution in the pot becomes richer inthe pure component 2 in the first case (Xl to X2) and pure component 1 in thesecond case (X3 to X4).

If one started with a solution of composition X, the vapour in equilibrium will

have identical composition, so that, at a given pressure, separation of such asolution into its components is not possible. Such a solution is called anazeotrope. An azeotrope mixture distills unchanged at a given pressure (the

vapour has the same composition as the liquid).

If a solution with a composition lying to the left of the minimum in the abovediagram is distilled so that the distillate having the azeotropic composition iscollected, the mixture remaining in the pot will become richer and richer incomponent 2. Ultimately, the point may be reached where only component 2remains in the pot. This component may then be recovered in a pure state. From asolution with a composition lying to the right of the minimum in the diagram,pure component 1 may ultimately be isolated from the pot.

One of the best known examples of the above behaviour is the ethanol-watersystem. At normal atmospheric pressure, water boils at 100.0C and ethanol boilsat 78.5C. Ethanol and water form an azeotrope of boiling point 78.2C with acomposition of 95.6% ethanol and 4.4% water by weight. Fermentation results in

ethanol/water solutions containing, say 10% ethanol. Distillation is used toincrease the ethanol concentration, but instead of yielding 100% ethanol, 95%ethanol is obtained For many purposes this mixture is used directly


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NON-IDEAL BEHAVIOUR OF GASES

1. Introduction .................................................................................. 27

2. Condensation of gases and the critical point ........................... 27

3.1 Introduction 293.2 Correction for excluded volume 29

3.3. Correction for attractive forces between molecules 29


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NON-IDEAL BEHAVIOUR OF GASES

1. Introduction

Actual gases exhibit, to some extent, deviations from the behaviour predicted bythe ideal gas equation of state. These deviations are particularly important at

(i) high pressures(ii) low temperatures

To show deviations from ideality it is convenient to use a factor z called thecompressibility factor. With this factor

PV = znRT (1)

and z =nRT

PV(2)

For one mole of an ideal gas, z must have the value one. Examine Figure 2 whichshows plots of z versus pressure for hydrogen and for methane at two differenttemperatures. Notice that the gases show behaviour close to ideal at pressuresclose to 1atm and even at 100atm, the deviations for hydrogen at 0C andmethane at 100C are about 7% and 3% respectively. Notice also that the

deviations increase as pressure increases and as temperature decreases.

2. Condensation of gases and the critical point

If there were really no attractive forces between gas molecules, then gases wouldnever condense to liquids. They would always completely fill the containing vessel.

We are able to observe the onset of non-ideal behaviour by studying

pressure-volume isotherms of the type introduced in the discussion on Boyle'sLaw. Such isotherms are shown in Figure 3.

At high temperatures the isotherms show only slight deviations from ideal gasbehaviour. They also conform to ideal behaviour at low pressures and large

volumes. However, at low temperatures very marked deviations occur. Follow onesuch isotherm. From A to B, as pressure is increased volume decreasesapproximately in accordance with Boyle's Law. At point B the gas begins tocondense to a liquid and the pressure remains constant at the equilibrium vapour

pressure for the liquid at that temperature i.e. any attempt to increase the externalpressure simply results in the condensation of more liquid without change in


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pressure. By point C all gas has condensed to liquid and the pressure may againincrease. However, because it is very difficult to compress a liquid, very largeincreases in pressure result in only minute volume changes i.e. the isotherm isalmost vertical.

Figure 2: Plot of compressibility factor z for one mole of gas, where z =nRT

Pv

Note that the diagram shows the pressure-volume conditions over which thevarious phases are present.

Figure 3: Typical PV isotherms for a real gas in the region of the critical point (T l > T2 > T3)


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One isotherm is of particular significance. This is the so-called CRITICALISOTHERM which is the one which just shows a point of horizontal inflectionat the top of the zone which indicates equilibrium between the gas and liquidphases. The point of horizontal inflection is called the CRITICAL POINT.

The temperature corresponding to the critical isotherm is known as theCRITICAL TEMPERATURE. The particular significance of this temperatureis that it represents the highest temperature at which the gas may be condensed bythe application of pressure alone.

CRITICAL PRESSURE and CRITICAL VOLUME are the pressure andvolume per mole respectively at the critical point.

3.1 IntroductionIn 1873, Van der Waals attributed the failure of the ideal gas equation to duplicatethe behaviour of real gases to the neglect of

(i) Volume occupied by gas molecules(ii) Attractive forces between the molecules

3.2 Correction for excluded volumeWhen n mole of gas is placed in a container of volume V, the volume in which themolecules are free to move is equal to V only if the volume of the gas moleculesthemselves is negligible. The presence of molecules of finite size means that acertain volume - called the excluded volume is not available for movement by themolecules.

If the volume excluded by one mole of molecules is represented by b, thenP(V-nb)=nRT is a better equation than PV = nRT for the description of the

behaviour of real gases, since it gives the real free volume for movement bymolecules. Values of b must be measured empirically for each gas.

Note that nb becomes more significant for a given amount of gas as V decreasesi.e. as P increases. This is in accord with the experimental fact that deviations fromideality are most serious at high pressure.

3.3. Correction for attractive forces between moleculesThe presence of attractive forces between molecules is indicated by the fact thatall gases condense to liquids at low enough temperatures i.e. attractive forces


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between molecules are sufficiently large to overcome thermal motion due tokinetic energy. This motion normally keeps a gas filling its container rather thansettling in bulk under the influence of gravity. The nature of Van der Waals forceshas been studied in Inorganic Chemistry I.

The Van der Waals attractive forces act with the pressure to hold moleculestogether. Van der Waals saw these attractive forces as being particularly significantas a molecule was about to strike the walls of the container and contribute to thepressure. Attractive forces from within the body of the gas pull the molecule backfrom the wall, reducing pressure. These forces act throughout the body of the gas

and their effect is related to gas density expressed asV

n. Since there is a two way

interaction between gas molecules, Van der Waals made the correction term

proportional to2

V

n

with a constant of proportionality i.e. gas pressure, or the

total pulling together of molecules is increased by2

2

V

naby Van der Waal's forces.

The Van der Waal's equation of state for a real gas then becomes:

2

2

V

anP (v - nb) = nRT

These attractive forces will be most important when

(i) molecules are forced into close proximity i.e. at high pressure(ii) when molecules are moving along slowly and spend long periods adjacent to

one another i.e. at low temperature.

Example:Heald, C., Smith, A.C.K.,Applied Physical Chemistry, (MacMillan, 1974), p.13.

An autoclave of 1.00L capacity is filled with 644g of carbon tetrachloride and thenheated to 650K. It is desired to calculate the pressure developed. The Van der

Waals coefficients for carbon tetrachloride are

a = 2.04Nm4 mol-2 and b = 1.38 x 10-4 m3 mol-1 ; the molar mass of this substanceis 154g.


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Solution

number of mole of CCl4 =154

654= 4.18

Van der Waals equation is:

2

2

V

anP (V - nb) = nRT

Substitute the values given.

23

2

10x00.1

18.4x04.2P x (1.00x10-3 - 4.18x1.38x10-4)

= 8.314 x 4.18 x 650

i.e. (P + 3.56 x 107) 4.23 x 10-4 = 2.26 x 104

P = 17.9MNm-2 (177atm).

Calculate the pressure using the ideal gas equation and compare the two.


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THE GIBBS FREE ENERGY FUNCTION

1. Introduction .................................................................................. 33

2. The Gibbs free energy function ................................................... 33

3. Worked example ........................................................................... 35

4. Standard free energies ................................................................ 35

5. Relationship between free energy and pressure ....................... 37

6. Relationship between standard free energy change and theequilibrium constant ........................................................................... 39

7. Worked example .......................................................................... 40

8. Temperature dependence of free energy and equilibriumconstant ............................................................................................... 44

9. Le Chatelier's principle ............................................................... 45

10. Worked example .......................................................................... 46


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THE GIBBS FREE ENERGY FUNCTION

1. Introduction

We now have a criterion for spontaneous reactions viz Suniverse0 (Suniverse = 0

for reversible processes). However, entropy changes for the whole universe aredifficult to measure. Note that for reactions which proceed spontaneously butSsystem < 0 then there must be an entropy increase of equal or greater magnitudein the surroundings. When water is frozen in a fridge S < 0 but the motor givesoff heat, causing an increase in the entropy of the air molecules.

The tendency for chemical reactions to move towards equilibrium is influenced bytwo factors

(i) the tendency towards minimum energy(ii) the tendency towards maximum entropy.

A new state function, called Gibbs Free Energy (G) considers both the energy(enthalpy) and entropy components of a change (at constant pressure andtemperature.)

2. The Gibbs free energy functionFrom the Second Law, for any reaction at constant T

S2 - S1 q/T (1)

From the First Law

U2 - U1 = q + w (2)

S2 - S1T

wUU 12 (3)

if only pressure-volume work is considered

S2 - S1

T

VVPUU 1212 (constant T, P) (4)

(Since w = -PV at constant P)


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S2 - S1

T

PVUPVU 1122 (5)

but H = U + PV

S2 - S1 T

HH 12 (6)

for a spontaneous charge

T(S2 - S1) - (H2 - H1) 0 (7)

or (H2 - H1) - T(S2 - S1) 0 (8)

Define G by G = H - TS

at constant T, G = H -TS

(8) becomes

G < 0 spontaneous

G = 0 reversibleG > 0 no spontaneous reaction

Thus for a reaction to proceed spontaneously towards equilibrium in the directionwritten G must be negative. A reaction for which G is positive will not proceedspontaneously as written. In fact it should proceed spontaneously in the reverse ofthe direction being considered.

It must be noted here that thermodynamics says nothing about the speed withwhich a reaction will occur. Thus, while a spontaneous reaction may be predictedby the G value, the reaction may be so slow that there is no evidence of itsoccurrence.

Factors influencing rates of reactions will be studied in the Chemical Kineticscourse.

You can see that G is a state function because U, P, V, T and S are all statefunctions.


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3. Worked example

Barrow, G.M., Physical Chemistry, (McGraw-Hill, 1973) page 193. Calculate the freeenergy change for the process of converting 1mol of water at 100C and 1atm tosteam at the same temperature and pressure.

SolutionGibbs free energy G, is defined as

G = H-TS

For a change at constant pressure and temperature

G = H -TS

Since the process is carried out at constant pressure, the enthalpy change is equalto the heat absorbed

i.e. H = qp

If the change is carried out reversibly

S =T

q

T

q prev

G =H-TS

= qp- T T

qp

= 0

This means that since water and steam are in equilibrium at 100C, the free energychange for the conversion of one into the other must be zero (i.e. the process isreversible).

4. Standard free energies* Tables of Standard Free Energies are to be found in Unland.


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As a reference point, zero values of free energy are assigned to the stable form ofthe elements at one atmosphere pressure. These, and the free energies ofcompounds based on these references, are called standard free energies offormation. These free energies may be manipulated in the same way as standard

enthalpies of formation (usually quoted at 25C).

ExampleBarrow, G.M., Physical Chemistry, (McGraw-Hill, 1973), page 196. Calculate the freeenergy change for the reaction

H2C= CH2(g) + H2(g) CH3CH3(g)

Does the calculated free energy change indicate that products are favoured overreactants at standard condition?

SolutionFrom the tables we find that

G fo (H2C = CH2(g)] = + 68.12kJ mol

-1

Gfo [H2 (g)] = 0

Gfo [CH3CH3 (g)] = -32.89kJ mol

-1

For the reaction as written

Go298= G fo [Products] - G f

o [Reactants]

= -32.89 - 68.12

= -101.01kJ molAns.

Since Go298 is negative, products are favoured over reactants at standardconditions.


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5. Relationship between free energy and pressure

Gibbs standard free energy data has provided us with criteria for assessing thepossibility of a reaction occurring at 25C when all reagents are in their standardstates. For the data to be of real use it is necessary to be able to make calculationat other temperatures and pressures etc. We shall now consider the relationship

between free energy and pressure.

From the definition of free energy:

G = U+PV-TS (13)

Differentiate (1)

dG = dU + PdV + VdP - TdS - SdT (14)

From the First Law of Thermodynamics

dU = dq + dw (15)

From the Second Law of Thermodynamics

dS =T

dqrev (reversible process) (16)

From the definition of reversible PV work

dw = -PdV (17)

Combining (15), (16) and (17) gives

dU = TdS - PdV (18)

dU - TdS + PdV = 0 (19)

Substitute (19) into (14)

dG = VdP - SdT (20)


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at constant temperature

TdP

dG

= V (21)

To study the change in free energy for an ideal gas we may write:

2

1

G

G

dG = 2

1

P

P

V dP at constant T (22)

For an ideal gas:

V = PnRT

(23)


2

1

G

G

dG = nRT 2

1

P

PP

dP(24)

[g]1

2

GG

= nRT [nRT]1

2

PP

(25)

G2 - G1 = nRT l n1

2

P

P(26)

If we let state 1 be the standard state and state 2 be the state under considerationthen

G2 = G

G1 = G

P2 = P

P1 = 1 atmosphere

Then G = G + RT ln P (27)


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Notice that, contrary to our usual practice, pressure is specified in atmospheres inequation (27) rather than in Nm-2.

The full derivation of (26) and (27) has been included here for reference only. The

result, however, is needed to derive the very important relationship in 6.

6. Relationship between standard free energy change and theequilibrium constant

Consider a reaction involving four ideal gases, A, B, C and D all at temperature T:

a A + b B c C + d D

where a, b, c and d are the numbers of moles of each reagent involved.

Free energy of a mol of A = aGA = aGAo + aRT ln PA

Free energy of b mol of B = bGB = bGBo + bRT 1n PB

Free energy of c mol of C = cGC = cGCo + CRT ln PC

Free energy of d mol of D = dGD = dGDo + dRT ln PD

where GA, GB, ..., GAo , GB

o , ... are free energies of one mol of reagent A, B, etc.

The free energy change for the reaction is given by

G = Gproducts G reactants

= cGC + dGD - aGA - bGB

= cGCo + dGD

o - aGAo - bGB

o + RT ln bB

a

A

d

D

c

C

PP

PP

or G = Go + RT ln

b

B

a

A

d

D

c

C

PP

PPat constant T


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=Go + RT ln Q (where Q is the ratio of products to reactants)

In the case where the PA , PB , PC and PD values are partial pressures of thereagents at equilibrium then G = 0 so that

Go = -RT ln

b

B

a

A

d

D

c

C

PP

PP= -RT ln Q (28)

Since Go for a particular reaction at a fixed temperature is constant, then theterm on the right hand side must have a constant value that is independent of theindividual pressures. We write:

Kp =

b

B

a

A

d

D

c

C

PPPP (29)

and Go = -RT ln Kp (30)

(Q = Kp only at equilibrium)

where Kp is the equilibrium constant expressed in terms of partial pressures of thereagents involved in the reaction.

A similar result applies for reactions in solution i.e. Go = -RT ln K in general.

Remember that to use equation (30) pressures must be in atmospheres.

7. Worked example

1. Use tabulated standard free energies to decide whether or not it is worthseeking a catalyst to increase the rate of production of NO(g) from a mixtureof N2 (g) and O2 (g) at 298K and 1atm pressure.

The reaction is:

N2 (g) + O2 (g) 2 NO (g)


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From tables

Gfo [N2 (g) ] = 0.0kJ mol

-1

G fo

[O2 (g)] = 0.0kJ mol

-1

Gfo [NO (g)] = 86.69kJ mol-1

For reaction as written

G fo = 2G f

o [NO(g)] - G fo [N2 (g)] - G f

o [O2 (g)]

= 2 x 86.69 - 0 - 0

= 173.38kJ

Since Go is +173.4kJ the reaction as written will not proceed to a significantextent under standard conditions. Since a catalyst changes only the rate ofreaction and not the position of equilibrium, there is no point in seeking acatalyst.

2. Goates, J.R., Ott, J.B., Chemical Thermodynamics, (Harcourt Brace Jovanovich,1971), page 75.

For the reaction

2 CO (g) + O2 (g) = 2 CO2 (g)

write the equilibrium constant expression and calculate the value for Kp at25C.

SolutionAt equilibrium,

Go = -RT ln Kp (1)

Where

Kp =22

2

O2CO

2

CO

PP

P

(2)


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From (1)

ln Kp = -RT

G o(3)

From tables

Gfo [CO (g)] = -137kJ mol-1

Gfo [O2 (g)] = 0

Gfo [CO2 (g)] = -394kJ mol

-1

For the reaction

G fo = 2(-394) - 2(-137) - 0

= -514kJ

Substitute in (3)

ln Kp = -298x314.81000x514

= 207

Kp = 7.92 x 1089 atm-1

i.e. at equilibrium the reaction is heavily in favour of the products i.e. CO2.

3. Calculate Go and Kp for the following reaction at 298K.

N2 (g) + 3H2 (g) 2 NH3 (g).

Solution

From tables

G f

o

[N2 (g) ] = 0


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Gfo [H2 (g)] = 0

Ofo [NH3 (g)] = -16630J mol

-1

For the reaction as written

o298G = 2 (-16630) - 0 - 3 x 0

= -33260J Ans.

At equilibrium,

o

298G = -RT ln Kp

ln Kp = - 298x314.833260

= 13.42

Kp = 6.7 x 105 at 298K Ans.

(This is a very important reaction since it is the basis for the commercialproduction of ammonia.)

4. A mixture of N2(g), H2(g) and NH3(g) is introduced into a container inamounts which gave the following partial pressures at 298K.

P(N2 ) = 0.10atm

P(H2 ) = 0.10atm

P(NH3) = 10atm

Comment on what is likely to happen in such a mixture, using the data in theexample above. Calculate Kp for the reaction.

SolutionIn general G = Go + RT ln Q

G = Go +RT ln3

HN

2

NH

22

3

PxP

P


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= -33260 + 8.314 x 298 x l n3

2

10.0x10.0

10

= -33260 + 8.314 x 298 x l n(1.00 x 106 )

= 969J.

ln Kp = - G/RT = 33260/8.314 x 298 =13.4

Kp = e13.4 = 6.6 x 105

Note that K is not equal to Q.

From this result we see that such a mixture is not at equilibrium, and would

proceed towards equilibrium in a direction which is the reverse of that writtenabove i.e. ammonia would decompose to give nitrogen and hydrogen, despiteGobeing less than zero. It is G, not Go, which predicts the course of areaction. In this reaction G is positive and will decrease as ammoniadecomposes, until G=0 (at equilibrium) and Q = K.

8. Temperature dependence of free energy and equilibriumconstant

The quantitative relationship between K and T is easy to derive if we assume H

is independent of T (a reasonable assumption provided the temperature range isnot too great).

Go= Ho -TSo - RT lnK (1)

at two temperatures T1 and T2

ln K2 =RS

RTH

o

2

o

(2)

ln KI =R

S

RT

H o

1

o

(3)

Subtracting (3) from (2)

ln1

2

KK =

1

2

o

T1

T1

RH (4)


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(assuming So is also independent of T)

or ln K =

RT

H+ constant (5)

It is clear from (4) that the rate of change of equilibrium constant withtemperature depends on the standard enthalpy of the reaction. From (4) it may beshown that when a reaction is exothermic (H < 0), the equilibrium constantdecreases as temperature increases.

9. Le Chatelier's principle

A very valuable qualitative principle for discussing the effect of pressure andtemperature changes on the position of chemical equilibrium is Le Chatelier'sPrinciple which may be stated:

"If a change occurs in one of the factors, such as temperature or pressure, underwhich a system is in equilibrium, the system will tend to adjust itself so as to annul,as far as, possible, the effect of that change."

(Glasstone, S., "Textbook of Physical Chemistry", (Macmillan, 1962), page 831).

Changing the pressure is of great significance only in reactions involving one ormore gaseous species, and there only when the number of mole of gaseousreactants is different from the number of mole of gaseous products.

Consider the reaction:

N2 (g) + 3 H2 (g) 2 NH3 (g); H298 = -92.4kJ

There are four mole of gaseous reactants and two mole of gaseous products. If thereaction is at equilibrium and the pressure is increased, the system will move tocounteract the stress by reducing the number of molecules, i.e. N2 will react with 3H2 to form 2 NH3. Thus the forward reaction is favoured by increasing pressurei.e. Kp will increase.

We have seen in the previous section that an increase in temperature for anexothermic reaction results in a reduction of the equilibrium constant Thus


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ammonia would decompose to nitrogen and hydrogen if temperature wereincreased in the above case.

The third type of stress which is of interest is a change in the concentration of onespecies. Thus if more N2 (g) is added to the above system at equilibrium, the

system will respond by moving in the forward direction i.e. N2 (g) and H2 (g) willcombine to give NH 3 (g). We have met this effect in Example 3 in Section 8.

10. Worked example

In a study of the reaction

H2 (g) + I2 (g) 2 HI(g)

it was found that the average value of H over the temperature range 430 - 470Cwas - 12.5kJ.

(i) Predict the effect on the equilibrium position of an increase in temperature.(ii) Predict the effect on the equilibrium position of an increase in pressure.(iii) Predict the effect of the addition of HI to the mixture.

Solution(i) Since this is an exothermic reaction (i.e. heat is given out) and increase in

temperature will decrease the equilibrium constant i.e. HI(g) will decompose.(ii) Since there are equal number of moles of gaseous reactants and products,

pressure changes will not effect the position of equilibrium.(iii) Addition of HI imposes a stress on the reaction. This stress may be relieved if

HI decomposes to give H2 (g) and I2 (g).


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THERMODYNAMICS - THE FIRST LAW

1. Introduction .................................................................................. 49

2. Thermodynamic terms ................................................................. 49

3. State functions ............................................................................. 50

4. Pressure-volume work ................................................................. 50

5. Heat ............................................................................................... 615.1 Heat capacity 62

6. Summary ....................................................................................... 63

7. The First Law of Thermodynamics.............................................. 638. Application of the First Law under special conditions .............. 65

8.1 Processes occurring at constant volume 65

8.2 Processes occurring at constant pressure 66

8.3 Relationship between qP and qV 69

9. Heat capacities at constant pressure and constant volume ..... 71

10. Relationship between Cp and Cv for an ideal gas ....................... 72

11. Internal energy for an ideal gas ................................................... 74


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49

THERMODYNAMICS - THE FIRST LAW

1. Introduction

Thermodynamics is concerned with studying the energy changes associated withchemical and physical systems. The energy changes associated with chemicalreactions can be used to predict the equilibrium position of a chemical reaction. Thusthe principles of thermodynamics can be used to associate the thermal properties ofa system with the equilibrium state of a chemical system. Two points need to bemade initially about thermodynamics.

(i) it is concerned only with the macroscopic properties of a system - not with thebehaviour of individual atoms or molecules.

(ii) thermodynamics tells nothing about the rate of attaining equilibrium (this will bedealt with in the kinetics section).

2. Thermodynamic terms

Thermodynamics is concerned with systems, separated from the surroundings by aboundary. The system together with the surroundings constitute the universe. Thepositioning of the boundary is quite arbitrary (it need not be a physical boundary).

The system is open when mass passes across the boundary; closed when no masspasses across the boundary. The system is isolated when the boundary prevents anyinteraction with the surroundings.

Energy may be transferred between a system and its surroundings by:


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50

(i) performance of work(ii) exchange of heat

3. State functions

A state function is one which is used to describe the state of a system. It has themathematical property that the change in the function depends only on the initialand final state of the system and not on how the change was accomplished i.e. it is

path independent.

ExampleThe simplest chemical system to study is the one dealt with in sections 1 - 3 i.e. an

ideal gas. The state of the gas is specified by the state functions P, V, n, T.However, since they are related; by the Equation of State (PV = nRT), only three ofthese functions need to be specified to completely define the state of the gas.

Pressure, for example, is a state function since the change in pressure, P, goingfrom pressure P1 to P2 is simply P2 - P1 , regardless of how the pressure was changed.

The concept of state functions should become clearer when we consider two

non-state functions (or path dependent functions), work and heat.

4. Pressure-volume work

Work is defined as the product of a force and the distance through which the forceoperates i.e. w = fdx

where w = workf = force

dx = displacement

In the SI system of units, work, like energy, is measured in joules (J)

i.e. w = f x = N x m = J


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51

From our point of view we are particularly interested in pressure-volume work as isachieved when, say, a piston is displaced by an expanding gas. Note that the productof pressure and volume has the units of J.

P x V = Nm-2 x m3= Nm= J

Consider the work associated with the expansion or contraction of fluids. In thefollowing diagrams a gas is enclosed in a chamber by a frictionless, weightless piston.

The gas exerts a pressure P on the face of the piston and to keep the piston in placean external pressure Pe equal to P must be applied.

Note that the system is the gas in the piston and the boundary is the walls ofthepiston.

If Peis reduced by P, the gas inside the chamber will expand and the piston willmove until the pressure of the gas (P-P) again equals the external pressure (PC-AP).

During this process a displacement x occurs in the position of the piston.

Consider a minute (infinitesimal) change: Since, by definition,

Pressure = force/unit area


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52

i.e. Pe = f / A (1)

... f = PeA (2)

where f is the force exerted on the piston by Pe and A is the area of the face of thepiston.

Since w = f dx (3)

Substitute in (2)

w = Pe Adx (4)

At this stage we introduce a sign convention which states that work is positivewhen done on a system and negative when done by a system. An expanding gas iscapable of doing work, so that w in this case should be negative

w = - Pe dV (5)

where Adx = dV, the change in volume.

Now we are concerned with the calculation of work during an entire change and notjust during an infinitesimal change, so we shall spend a little time on mathematics.

If thepressure remains constant during an expansion from V1 to V2 , this

may be shown graphically as:


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53

The magnitude of the work is given by the product of pressure and change involume. This is simply given by the shaded area of the graph under the linecorresponding to Pe and between the limits of V1 and V2. In this case

w = - P(V2 - V1) (6)

Mathematically, the area under a curve is obtained by integration between the limitsof V1 and V2. In this case

dVPw eV

V

2

1 (7)

Since. Pe is constant, it may be taken outside the integral. Then (7) becomes

dVPw 21

V

Ve

1

2e V

VVP

12e VVP (8)

The above example could be expressed in an alternative way. Suppose the piston hada block, x, resting on it. When the block is removed the gas expands until gas - Pe(the external pressure) = P2


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54

w = - Pe(V2 - V1)

= - P2 (V2 - V1)

Suppose the change from state 1 to state 2 is now carried out a different way. Insteadof one block x, we have two blocks, each half the weight of x anti these are removedone at a time.


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55

Total work = work (1 1') + work (1 2) since both steps are constant pressuresteps

w = - P1'(V1'- V1 ) - P2 (V2 - V1')

Note that the shaded area is less than the first case, although the initial and finalstates are the same.

We could continue to divide the blocks and remove them one at a time (increasingthe work done) until we reach the limiting case - approximated below.


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56

Here x is a pile of sand and we remove the grains one at a time.

In this case, work is still given directly by the area under the P-V curve. We may seethat this is so by dividing the area into a number of rectangles, each of which is

similar to the rectangle in the previous diagram. It is obviously possible to assumethat the pressure changes by small jumps from P1 to P2 to P3, etc., and if all volumechanges are made equal at V then

w = -(P1V+P2V+P3V+..........)

VP

ii

(9)

The estimate of the area under the curve will improve as V is made smaller. Again,the true area is found by integration between the limits of V1 and V2. To make thispossible it is necessary to know the relationship between P and V. For an ideal gasthis is obtained from the equation of state.


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57

i.e. PV = nRT

V

nRTP

Since PdV 21

V

Vw

substitute for P from the equation of state for an ideal gas:

dVV

nRT 2

1

V

V

For a given amount of gas at constant temperature

V

dVnRT 2

1

V

Vw

1

2

VV

VlnnRT

1

2

V

VnRTln (10)

(The process of integration will be quite familiar to many of you and something of amystery to others. For the latter group, just think of integration as a mathematicaldevice to find the area under a curve between limits, as shown here. The integrals wehave used here viz.

1212

a

aaaa

axdx

2

1

and 1

2

1

2

a

aa

aln

a

axln

x

dx2

1


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58

whereln refers to natural logarithm or log to the base e, will keep on recurring,but will be the only ones necessary for this course.

The limiting case described above is the constant temperature, or isothermal, case

and represents the maximum workdone by the system. This case is a reversibleone, as the system is never more than an infinitesimal distance from equilibrium. Thesystem can be returned to equilibrium in a ideal reversible case with no observablechange in the surroundings. Obviously reversibility is an ideal case which can at bestonly be approximated. However, for an engine, say, it does represent the bestoperating conditions thermodynamically (in terms of energy flow). There is thedisadvantage that an ideal reversible system changes infinitely slowly (quite adrawback for an engine!) but remember thermodynamics says nothing about rates.

The above case demonstrates work is not a state function. The value variesdepending on the path, and is different in the three cases (although the initial andfinal states are the same). This can be illustrated in the following case.

Example0.4 moles of an ideal gas initially at 10 atmospheres and one litre expands to a final

volume of 10L and one atmosphere;

(i) at constant pressure(ii) in two steps; initially to a volume of 5L and 2atm(iii) isothermally and resersibly (at 25C).

Calculate the work done in each case.

(i) At constant pressure:

P1

= 10atm = 1,013,250 Pa

V1 = 1L = 10-3 m3 (remember SI units!)

P2 = 1atm = 101,325 Pa

V2 = 10L = 10-2 m3


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w = - P2 (V2 - V1) = -101,325 (0.01 - .001)

= - 912J (negative, since the system is doing work)

(ii) P1 = 1,013,250 1P = 202,650

V1 = 10-3 m3 V1 = 5 x 10

-3 m3

P2 = 101,325 Pa

V2 = 10-2 m 3

w = - 202,640(.005 - .001) - 101,325(.01 - .005)

= - 811 - 507 = -1318J

(iii) in the isothermal case equation 10 is used

w = - nRT ln V2/V1

= - 0.4 x 8.314 x 298 l n3

3

10x1

10x10

= - 2282J

A third type of process which is of interest is that which occurs at constant volume.In this case there is obviously no change in volume, so there is no work being doneon or by the system.

i.e. w = PdV 21

V

V

= 0 since dV = 0

In summary, we are concerned with processes occurring under three differentconditions:

At constant pressure


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w = PdV21

V

V

= -P(V2 - V1) (8)

At constant temperature

w = PdV21

V

V

= -nRTln2

1

V

V(ideal gas) (10)

At constant volume

0dVsince0

PdVw 21

V

V

Example 1Goatcs, J.R., Ott, J.B., Chemical Thermodynamics, (Harcourt, Brace, Jovanovich, 1971),p.13.

Calculate the work in joules involved in the constant pressure expansion of 3 molesof gas from an initial volume of 2.00dm3 to a final volume of 10.00dm3. The pressureremains constant at 7.00atm.

Solution

.AnsJ5674

10x00.1x00.200.10101325x00.7

VVP

dVP

PdVw

3

12

2

1

2

1

V

V

V

V

Example 2Goates, and Ott, p. 12.


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Calculate the work in J involved in the reversible compression of 10.0g of I He gasfrom an initial pressure of 1.00atm to a final pressure of 5.00atm at the constanttemperature of 27C. Assume ideal gas behaviour.

Solution

)gasideal,isothermal,reversible(V

VlnnRT

dVV

nRTw

V

nRTPIf

PdVw

2

1

V

V

V

V

2

1

2

1

Number of mol of He 50.200.4

00.10

Ratio2

1

1

2

1

2

P

P

P

nRT

P

nRT

V

V

5.00

1.00300lnx8.314x2.50

P

PnRTln

V

VnRTlnw

2

1

1

2

.J10x00.1 4

5. Heat

Heat is energy that is exchanged between a system and its surroundings because ofdifferences in temperature only. Heat added to a system is given a positive sign, andthe normal symbol for heat is q.

Heat, like work, is not a state function. One of the significant early experiments inthis area was performed by Joule in the later 1840s. He showed that the temperatureof a water bath could be raised by the friction from a paddlewheel (doing work onthe system) as an alternative to placing the container over a flame (adding heat to thesystem).


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5.1 Heat capacityThe same amount of heat applied to different systems causes different changes intemperature (a block of aluminium will increase more in temperature than a similar

block of ice for the same given amount of heat). This ability to change temperaturewith applied heat is called heat capacity, defined as the amount of heat to raise thetemperature of the system by one degree Kelvin.

There are two forms of heat capacity.

Molar heat capacityTn

qC

(units of C: JK-1 mol-1)

Specific heat capacityTm

qC

(units of C: JK-1 g-1 )

In the old calorie scale the heat capacity of water is 1 cal. K-1 g-1 . Thus the calorie isthe amount of heat needed to raise 1g of water 1C (or 1K).

(The above definitions of heat capacity are not precise - they assume heat capacitydoes not vary with T. A more precise definition of C uses infinitesimal changes in qand T.


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i.e.dT

dqC

However the above definitions will be sufficient for the cases we will be studying.)

Thus, providing there are no phase changes or chemical relations taking place,measurements of temperature can be used to calculate q.

6. Summary

The forms of energy we are concerned with are

(i) thermal energy (heat)(ii) mechanical energy (especially pressure-volume work)

In the following section we shall consider the relationships between heat and work.

7. The First Law of Thermodynamics

The First Law of Thermodynamics may be stated in a number of ways. e.g.

(i) The change in internal energy a of a system on going from an initial state A to a

final state B is determined only by the initial state A and the final state B and noton the path by which the change was effected, or

(ii) All the energy which flows in and out of a system as heat or work must beaccounted for in the change in energy content of the system.

We will be most interested in the First Law quoted in the form of the followingequation:

U = q + w (11)

where U = change in internal energy2

2 Some textbooks use E as the symbol for internal energy. U is the recommended symbol which is used in these notes. Choose either symbol and useit consistently.


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q = heat added to the systemw =work done on the system

Note that small letters, q and w, are used for changes which are dependent onpathway, a capital letter U, is used for a property that depends only on the STATEof the system. While q and w may each have many possible values, their sum (q + w)is invariable and depends only on the initial and final states.

We often wish to express the First Law in terms of a very small (infinitesimal)change. This statement is dU = dq + dw (12).

The First Law is really a restatement of the old idea of conservation of energy.However, we have replaced the vague idea of energy with the precisely defined state

function, U. Thus the First Law is equivalent to the statement internal energy is astate function.

ExampleBarrow, G.M., Physical Chemistry(McGraw-Hill, 1973), p.121.

Calculate U for the conversion, at 100C and 1atm pressure, of 1 mole of water tosteam. The latent heat of vaporization of water is 40670J mol-1 and the density ofliquid water can be taken as 1g mL-1 and water vapour can be treated as an ideal gas.

SolutionHeat energy must be supplied to supply the latent heat of vaporization to convert theliquid water to steam.

q = 40670J

Since PdVw 21

V

V (1)

and since P is constant

w = -P (V2 - V1) (2)

Volume of 18g of water


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= 18 x 10-6 m3

Volume of 1 mole of steam, at 100C and 101325 Nm-2 is:

P

nRTV

101325

373x314.8x00.1

= 0.0306 m3

Substitute in (2)

w = -101325 (0.0306 - 18 x 10-6)

= -3100J

Since U = q + w

U= 40670 - 3100

= 37570J mol-1

8. Application of the First Law under special conditions

8.1 Processes occurring at constant volumeWhen a reaction is performed in a closed vessel, no volume changes can occur, so no

pressure-volume work may be done.

i.e. PdVw 21

V

V

= 0 at constant volume

From the First Law:


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U = q + w

Since w = 0 at constant volume

U = qv

where qv is heat absorbed by the system at constant volume.

8.2 Processes occurring at constant pressureIn general, the volume of the system involved in a chemical reaction in a vessel opento the atmosphere changes as work is done on or by the surroundings. For a

macroscopic change the First Law is

U = q + w (11)

If work is restricted to pressure-volume work

PdVw 21

V

V (12)

At constant pressure

w = -P(V2 - V1) (13)

(11) becomes

U = qp - P(V2 - V1) (14)

where qp is heat absorbed at constant pressure. (14) may also be written

U2 - U1 = qp - P(V2 - VI) (15)

U2 = internal energy of the final state


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U1 = internal energy of initial state

Rearrange (15)

(U2 + PV2 ) - (U1 + PVl) = qp (l6)

At this stage it is convenient to introduce a newstate function called ENTHALPY(H) where:

H = U + PV (17)

It is easy to see that enthalpy is a state function because all of the species can theright hand side of the defining equation (17) are state functions. In terms of enthalpy,

(16) becomes

H2 - H1 = qp

or H = qp (18)

i.e. change in enthalpy = heat absorbed at constant pressure.

From an operational viewpoint, changes in properties are more significant thanabsolute values. Since

H = U + PV (17)

H= U + (PV) (19)

At constantpressure, (19) becomes

H = U + PV

At constantvolume, (19) becomes

H = U + VP


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Worked exampleG.M. Barrow, Physical Chemistry, (McGraw Hill, 1973) page 1, 140 Q.5

A chemical reaction in a gas mixture at 500C decreases the number of mole gas by0.347. If the internal energy change is +23.8kJ, what is the enthalpy change? (Assume

the gases behave ideally).

SolutionIf it is assumed that the reaction occurs at constant pressure, P, and that initiallythere are n1 mole of gas present:

Initially PV1 = n1 RT (1)

After reaction,

PV2 = (n1 - 0.347)RT (2)

Subtract (l) from (2)

P(V2 - VI) = (n1 - 0.347 - n1)RT

PV = - 0.347RT

By definition, enthalpy H is

H = U + PV

where U = internal energy.

If there is a change in enthalpy,

H = U + PV at constant pressure

i.e. H = 23.8 x 103 - 0.347 x 8.314 x 773.2

= 21.6kJ Ans.


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8.3 Relationship between qP and qVFrom the discussion above we know that:

U = qvat constant volume

and H = qp at constant pressure

but H = U + (PV) = U + PV at constant pressure

qp = qv+ PV (20)

i.e. the heat absorbed in a process taking place at constant pressure exceeds thatabsorbed in a constant volume process by PV, since no mechanical work is done inthe constant volume process.

If the reactions under discussion involve only liquids and solids, the PV term isusually negligible in comparison with the q terms. However, if gases are involvedPV is usually significant since changes in volume are often large.

ExampleWhat is the difference between qp and qvfor the following reaction carried out at25C and one atmosphere pressure?

H2(g) + O2 (g) H20(l)

(Assume hydrogen and oxygen behave as ideal gases and that the density of water is1.00g mL-1 .)

Solutionqp = qv+ PV

For a reaction involving 1.00mol of H2, 0.500mol of O2 and 1.00mol of water:

Volume of products = 18.0 x 10-6 m3


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Volume of reactants =101325

298x314.8x50.1

= 3.67 x 10-2 m3

Change in volume = V2 - V1

= 18.0 x 10-6 - 3.67 x 10-2

= - 3.67 x 10-2 m3

qp - qv = 101325 x -3.67 x 10-2

= -3.72 x 103 Jmol-1 Ans.

Where gaseous and condensed phase species both appear in a reaction it is generallysafe to consider only changes in volume of gaseous reactants.

Since PV = nRT

(PV) = (nRT)

At constant P, R and T

PV = RTnqp - qv = RTn (21)

where n = number of mole ofgaseous products

- number of mole ofgaseous reactants

The problem studied above then becomes:

qp - qv= (0 - 1.5) x 8.314 x 298


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= -3.72 x 103 Jmol-1 Ans.

9. Heat capacities at constant pressure and constant volume

The heat capacity of a system is the amount of heat required to raise the temperatureof a system by one degree on the Kelvin scale.

i.e.dT

dqC (22)

Where C = heat. capacity

dq = heat added to the system

dT = change in temperature

The heat capacity of a system is indefinite unless certain boundary conditions arespecified.

Thus at constant volume

v

vdT

dqC

(23)

and at constant pressure,

P

pdT

dqC

(24)

Since heat added to the system at constant volume is given by

dU = dqvfor an infinitesimal change


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thenV

vdT

dUC

(25)

Since heat added to the system at constant pressure is given by

dH = dqp for an infinitesimal change

thenP

PdT

dHC

(26)

For the special case of an ideal gas, internal energy is independent of volume andenthalpy is independent of pressure, so that equations (25) and (26) become,

dT

dUCv (27)

dT

dHCp (28)

If Cp and Cvare independent of T, (27) and (28) become

U = nCvT (27a)

H = nCPT (28a)

10. Relationship between Cp and Cv for an ideal gas

From the First Law of Thermodynamics, for an infinitesimal change

dq = dU - dw (29)

But dw = -PdV for reversible pressure-volume work at constant pressure (30)

From (27)

dU = CvdT (31)


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Substitute (30) and (31) into (29):

dq = CvdT + PdV (32)

For an ideal gas:

PV = RT for one mole

Differentiate with respect to T at constant pressure

RdT

dVP

PdV = RdT (33)


RdTdTv

Cdqp (34)

Divide (34) throughout by dT

Rv

CdT

dqp (35)

Rv

Cp

C (36)

wherep

C andv

C are heat capacities for one mole of substance in each case.

Note: Adiabatic processes are not treated in this course. They are considered inPhysics 1. (Adiabatic processes are those where q = 0.)


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11. Internal energy for an ideal gas

From the kinetic molecular theory the KE of an ideal gas is given by

KE = 23

RT (per mole) (37)

As this is the only energy possessed by an ideal gas

U = KE =2

3RT (38)

U =2

3RT

but U = CvT (31)

Cv=2

3R (for an ideal gas)

CP = CV+ R CP =2

5R

These approximations can be used for real gases (unless the value is quoted in theproblem). Also, since U is a function only of T

T = 0 U = 0!

(for idealgases only). It can also be shown that H = 0 alsofor T = 0. Joule(1846)


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The barrier is removed to allow the gas to expand into the evacuated half. If wedefine the systemto be the box enclosed by the insulated walls, then V = 0 (the

volume of the box doesn't change) w = 0.

Since the walls are insulated, q = 0, U = q + w = 0.

Since U = 0 T = 0.

Joule confirmed that there was no change in T (although for real gases there is asmall change due to the attractive forces between molecules; for ideal gases these areassumed to be zero. Joule's apparatus was not sensitive enough to measure thesechanges.

ExampleConsider the cyclic change carried out on one mole of an ideal gas

Calculate U, H, q and w for each pathway, and for the whole cycle.


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Step IThis is an isochoric (constant volume) step

v = 0

w = 0

q = n CVT = CVT (n = 1)

=2

3R(600 - 300) = 3741J

U = q + W = qV= 3741J

H = U + (PV) = U + VP (since V is constant) = 3741 + .02494

(200,000 - 100,000) = 6235J

Step 2

P = 0 (isobaric)

q = n C PT = 25

R (300 - 600)

= -6235J (watch sign convention!)

= H (since H = qP)

w = - PV = -200,000 (.01247 - .02494)

= 2494J

U = q + w = 2494 - 6235 = -3741J

Step 3T = 0 (isothermal)

U = 0 = H


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12.47

24.94ln300x8.314

V

VlnnRTw

1

2

= -1729J

Since q + w = U = 0

q = 1729J

for the whole cycle

U = 3741 - 3741 + 0 = 0

H = 6235 - 6235 = 0

w = 0 + 2494 - 1729 = 765 J

q = 3741 - 6235 + 1729 = -765 J

confirms U and H are state functions but q and w are not (since this is a cyclicprocess).

The formulae can be summarised in the table below (the formulae for the reversible,isothermal case apply to ideal gases only - the other two cases are more general).

V= 0 P= 0 T= 0

q nCVT nCPT nRT ln 12

V

V

w 0 -PV -nRT ln1

2

V

V

U nCVT nCPT - PV 0


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H nCVT + VP nCPT 0

Documents

Entropy and the Second Law of Thermodynamics