An Introduction to Thermodynamics phần 5 docx

It was also shown that a necessary and sufficient condition for equilibrium at constant temperature and pressure is that the free energy of the system be a minimum, and hence that the fr

Similarly, to the body absorbing a certain amount of thermal energy at the temperature T , it can2

make little if any difference to the state, or entropy, of the body, where the energy came from At the same time, we must recognize we have introduced an approximation by assuming the energy transfer was thermal energy transfer under equilibrium conditions If, for example, the energy is transferred by radiation, higher-temperature radiation has a greater chance of knocking electrons free as photoelectrons, so the state of the receptor may depend, at least to some extent, on the temperature of the higher-temperature radiator The closer the experimental situation approaches true equilibrium, the more assurance we can have that the answers are adequate For example, in the paragraph preceding equation 6, in which thermal energy is transferred to the system (ice) from the surroundings (a thermostat) at a slightly higher temperature, there is no reason to doubt the conclusions

Problems

1 Figure 5 shows a capillary tube dipping into a dish of water The tubing is bent, and water dripping from the suspended open end turns a paddle wheel before returning to the dish By covering the apparatus, water can be retained and the energy required is returned to the dish by thermal conduction from the surroundings What is wrong?

2 Figure 6 shows a wheel on which are mounted small bar magnets The force between the fixed magnet and the individual magnets attached to the wheel may be taken as

proportional to 1/d, where d is the distance separating the

magnets Show that the force between the fixed magnet and the nearest magnet is smaller than the sum of the forces

pulling the magnets along the right-hand side That is, 1/d1

< 1/d + 1/d + ··· A magnetic shield (readily available) is2 3

placed as shown The wheel then rotates counterclockwise What is wrong?

−

=

∆

i

i

i N N

nR

3 Find ∆S when 2 mol of an ideal gas at 25 C is expanded into vacuum to 10 times its original

volume

4 A vessel containing 5 mol of an ideal gas, A, is connected to an identical vessel containing 5 mol of another ideal gas, B, and the two are allowed to reach equilibrium For this process, at

27 C, findo

a ∆S for the gas A.

b ∆S for the gas B.

c ∆S for the entire system (∆S(A) + ∆S(B)).

d What would ∆S be if A and B were the same gas?

5 If a gas at pressure P and volume V is mixed isothermally with another gas at pressure P and1 volume V to give a mixture at pressure P and volume V = V + V , the mole fractions are N =2 1 2 1

V /(V + V ) and N = V /(V + V ) Show that the “entropy of mixing” of ideal gases in this1 1 2 2 2 1 2

fashion is

where n = Gn is the total number of moles and N = n /n is the mole fraction of the i gas i i i

th

6 A vessel containing 5 mol of an ideal gas, A, is connected to an identical vessel containing 10 mol of an ideal gas, B, and the two are allowed to reach equilibrium For the process, at 27 C,o find

a ∆S, assuming A and B are different gases.

b ∆S, assuming A and B are the same Explain the distinction between these two cases and

between these and problem 4

7 Which of the following have the higher value of S?

a CO at 25 C, 1 atm2 o or “dry ice” at 1 atm

b a coiled spring or the spring “relaxed”

c 1 g of liquid water at 25 C o or 1 g of water vapor, at the vapor pressure of water at

25 C.o

d silica glass or quartz (crystalline silica)

8 When a gas escapes through a pinhole into an evacuated chamber, the first gas to escape is

later compressed by the following gas Explain why W = 0, despite this compression.

9 Calculate ∆G for the processes described in problems 3, 4, and 6.

10 The heat of vaporization of cyclohexane, C H , is 30.46 kJ/mol at its normal boiling point,6 12 80.7 C.o

a Calculate ∆S for the vaporization of 1 mol of cyclohexane at 80.7 C and 1 atm Compareo with the Trouton constant

b Calculate ∆G for the same process.

11 Calculate the free energy change when 3 mol of Ar is compressed isothermally from 2 to 6 atm at 50 C.o

12 The standard free energies of formation of ethylene (C H ) and of ethane (C H ) at 25 C and2 4 2 6 o

1 atm are 68.4 and -32.0 kJ/mol The standard enthalpies of formation are 52.4 and -84.0 kJ/mol For the reduction of ethylene with H to give ethane,2

a find ∆G

b find ∆S o

13 Calculate ∆S , the entropy change under standard conditions at 25 C, for the reaction ofo o carbon (graphite) with fluorine gas to give the gas CF , using values as necessary from Table 1.4

14 For each of the following reactions at 25 C calculate (using Table 1)o

4 FeO + O 2 6 2 Fe O2 3

2 Fe O + ½ O 3 4 2 6 3 Fe O2 3

3 FeO + ½ O 2 6 Fe O3 4

Which form of iron oxide — FeO, Fe O , or Fe O — is most stable in the presence of an oxygen3 4 2 3 atmosphere at this temperature? Explain

15 Would Fe O (see problem 14) become more or less stable, with respect to FeO and oxygen,3 4

as temperature is increased? Explain

16 For the reaction at 25 C, with each gas at 1 atm,o

3 H + N 2 2 6 2 NH3

a Find ∆H b Find ∆E.

c Find ∆G d Find ∆S.

e Is the reaction as written, at constant T and P, spontaneous?

f Does the reaction as written take up or give off thermal energy?

g Find the pressure, P, of NH for which ∆G = 0.3

17 In the following reactions, would you expect ∆S to be positive or negative? Explain.

a Ag O (s) 2 6 2 Ag (s) + ½ O (g)2

b 2 CO (g) + O (g) 2 6 2 CO (g)2

c 2 C (s) + O (g) 2 6 2 CO (g)

d CaH (s) + 2 H O (l) 2 2 6 Ca(OH) (s) + 2 H (g)2 2

e H (g) + I (g) 2 2 6 2 HI (g)

f 2 H (g) + O (g) 2 2 6 2 H O (l)2

g n C H (g) 2 4 6 — (C H ) — (polyethylene)2 4 n

18 A Dewar flask contains 500 ml of water at 25 C To the flask is added 100 g of ice at - 6 C o o For the process which then occurs, find

a ∆H

b ∆S

c the final temperature

19 Find

for the process of subliming 6 g of ice at - 10 C and warming the vapor to 200 C and 4 atm.o o

20 One kilowatt hour of energy passes from a heater, at a temperature of 1000 K, to a room at

27 C Find ∆S.o

21 For the freezing of 5 g of supercooled water, at - 10 C, to ice at - 10 C, calculateo o

(36) ln

ln adiabatic) rev,

, (I.G.

2

1

1

2

V

V R T

T

(37) ln

ln adiabatic) rev,

(I.G.,

1 2

1

2

P

P R T

T

22 The standard free energies (1 atm) of NO and N O at 25 C are 51.3 and 99.8 kJ/mol For2 2 4 the reaction

2 NO (1 atm) - N O (P atm)2 2 4

a find ∆G if P = 1.

b find the value of P (i.e., the pressure of N O ) that would make ∆G = 0.2 4

23 In a reversible, adiabatic expansion, q = 0 Show, by setting dE = n C dT and w = - P dV V that, if P = nRT/V, the equation w = dE can be integrated (after dividing through by T to separate

variables) to give

24 Show that, by replacing dV by d(nRT/P), the equation for a reversible, adiabatic expansion

integrates (after separating variables) to give

(Note that alternatively equation 37 may be obtained from equation 36 by replacing V by nRT2 2 /P and V by nRT /P )2 1 1 1

25 Show that equations 36 and 37 for an adiabatic, reversible expansion of an ideal gas can be put into the following forms:

a TV R/C V = constant, or T /T = (V /V ) R/C V

b PV = constant, or P /P = (V /V ) , where γ = C /C γ γ

26 An increase of entropy may correspond to an increase of randomness of velocities, or momenta, rather than of positions For example, when a gas is heated at constant volume there can be no change in the spatial disorder but there is more of a spread in the molecular speeds What conclusions can be drawn concerning the change in temperature when an isolated

supersaturated solution spontaneously separates into two phases by precipitation of solute?

3 Physical Equilibria

The power of thermodynamics lies in its applicability to all substances in all states It is not limited to such abstractions as ideal gases, ideal solutions, or perfect crystals, although certain of the equations take on especially simple forms for such special cases In the following discussions

of physical equilibria and chemical equilibria the exact thermodynamic equations will be derived wherever possible This will enable a rational choice of approximations for each application and a clearer understanding of when, and what, approximations are being introduced

General Conditions for Equilibrium

The fundamental requirement for a system to be at equilibrium, with respect to a certain process, is that under the existing conditions the process should be thermodynamically reversible The requirement is, from the second law, that the entropy of the system and the surroundings should remain unchanged for small changes in the state of the system This is a basic statement, from which other conditions can be derived It is not necessarily, however, the most useful way

of stating the condition of equilibrium In this chapter the conditions for equilibrium not involving chemical change will be put into a variety of forms Depending upon the process considered, one

or another of these will be found most convenient

CONDITIONS OF TEMPERATURE, PRESSURE, AND FREE ENERGY It was shown in the previous chapter that the second law requires that two bodies in equilibrium should have the same

temperature and the same pressure It will therefore be assumed in the following discussion that these conditions are satisfied We will be interested in the question of what that temperature and pressure must be, and how these parameters are affected by each other and by changes in

chemical composition

It was also shown that a necessary and sufficient condition for equilibrium at constant

temperature and pressure is that the free energy of the system be a minimum, and hence that the free-energy change of the system be zero for small changes in the system The condition will form the basis of the derivations in this chapter

CLAPEYRON EQUATION Pure water freezes at 0 C and boils at 100 C, but only under “normal”o o conditions of 1 atm pressure An increase of pressure will lower the freezing point but will raise the boiling point Both effects are easily predicted by the Clapeyron equation, which is to be developed

Let some substance exist in two phases, A and B, in equilibrium at some temperature and

pressure This might be the equilibrium between ice and liquid water, or liquid water and its vapor, or some other solid-solid, solid-liquid, solid-vapor, or liquid-vapor equilibrium Then the condition for equilibrium is

∆G = G - G = 0 (1) B A

(3)

V S V V S S dT dP A B A B ∆ ∆ = − − = (4)

V T H dT dP ∆ ∆ = C 0.14 K 334 10 x 1.013 x 10 x 0.09035 x 273 x 19 J/m 1 Pa 1 Pa 10 x 013 1 atm 1 x g / m 10 x 9.035 K x 273 J/g 334 -T atm 19 o 5 6 -3 5 3 8 -− = − = ∆ = ∆ ∆ = ∆ = ∆ ∆ T V T H T P The substitution of ∆P/∆T for dP/dT is justified because the change in temperature is 1 small More generally, one could carry out the integration to obtain IdP = ∆P = ∆H/∆V IdT/T = ∆H/∆V ln T /T However, ln T /T = ln (1 + (T - T )/T ) and for small values of x, ln(1 + x) = x 2 1 2 1 2 1 1 Therefore, for (T - T )/T small, ∆P = ∆H/∆V ∆T/T or ∆P/∆T = ∆H/T∆V.2 1 1 1 We change the temperature and simultaneously change the pressure in such a way that equilibrium is maintained This requires that G and G change by the same amount A B dG = dG (2) A B From equations 21, 22, and 23 in Chapter 2, the change in free energy with temperature and pressure is dG = V dP- S dT B B B dG = V dP- S dT A A A and therefore V dP - S dT = V dP - S dT B B A A This can be rearranged to give (Equil between two phases)

Because equilibrium is maintained, ∆G = 0 and ∆S = ∆H/T This gives the Clapeyron equation in its usual form: (Equil between two phases)

From this equation the change in a freezing point, or other phase transition, with change of

applied pressure can be calculated

For example, the freezing point of water under 1 atm of air is exactly 0 C The freezing pointo under 20 atm pressure can be calculated as follows.1

where we have inserted the value 0.917 g/cm for the density of ice and 1 atm = 1.01325 x 10 Pa3 5 (1 Pa = 1 pascal = 1 N/m = 1 J/m ).2 3

Escaping Tendency

The thermodynamic picture of equilibrium is static, with no tendency for change in the total amounts of material in the two phases The molecular picture of equilibrium is dynamic, with molecules continually passing back and forth from one phase to the other; but this dynamic model

is entirely consistent with the static concept of equilibrium because the number of molecules passing in one direction is just equal, at equilibrium, to the number of molecules passing in the other direction

Employing either the macroscopic and static picture or the dynamic molecular view, we may say that at equilibrium the tendency for molecules, or material, to “escape” from one phase to the other is the same as the tendency for molecules, or material, to “escape” in the reverse direction

In this language, the condition for equilibrium is that the escaping tendency for any substance must be the same in all phases in equilibrium.

The free energy is a measure of this escaping tendency The higher the free energy, the

greater the escaping tendency and the lower the stability of a phase The minimum free energy, or minimum escaping tendency, represents the most stable, or equilibrium, condition However, for many problems the free energy itself is not convenient First, we cannot know an absolute value for the free energy We know only values relative to arbitrary standard states Second, whatever

the arbitrary value assigned for a standard state, the value of G will become negative and infinite

as the pressure of a gas or concentration of a solute approach zero Gases at low pressures, and dilute solutions, are too important to permit such awkward behavior of the function chosen to describe them We turn, therefore, for this purpose to the vapor pressure or to an idealization of the vapor pressure

VAPOR PRESSURE AND CLAUSIUS-CLAPEYRON EQUATION At room temperature (25 C) watero molecules leave the surface of the liquid at a rate equal to that at which molecules of the vapor strike the surface, if the vapor is also at room temperature and at a pressure of 23.756 torr ; that2

is, at 100% relative humidity If the temperature is raised the molecules in the liquid have higher speeds and more of them will escape from the surface The molecules in the vapor will also have higher speeds and they will strike the surface more frequently Is this enough to maintain

equilibrium? It is not, because as so often happens, we encounter an exponential dependence relating energy and temperature

The number of molecules escaping from the surface depends on the number having speeds above some limiting value The number of molecules having speeds above a limiting energy value depends exponentially on the temperature, whereas the increase in average speed in the vapor depends only on the square root of temperature Therefore, in order to maintain equilibrium, the density of molecules, or the pressure, must be increased, and in fact it must increase

exponentially

An approximate equation describing the change of vapor pressure with temperature can be derived from the Clapeyron equation by assuming the vapor to be an ideal gas The volume of

P T

H dT

dP=∆

(5)

ln 2 RT H dT P d PdT dP = = ∆ % 5 3 035 0 K (373) K x J/mol 314 8 K 1 x J/mol 0,657 4 2 2 = = ⋅ = ∆ P P (6)

ln

2 1 1

2

T RT

T H P

torr 1 28 0.0369

x 760

0369 0 ln 30 3 373

x 298

x 314 8

x(-75) 18

x 2258 760

ln 2

2

=

=

=

−

=

=

P P

one mole of the vapor is RT/P and the volume of the liquid is negligible in comparison, so RT/P may be taken as ∆V Thus

or

(Ideal vapor in equil with liquid or solid)

This is called the Clausius-Clapeyron equation It serves for calculations of changes of boiling

point, or sublimation point, with change of pressure, which is equivalent to saying that it gives the variation of the vapor pressure of a solid or liquid with change in temperature Vapors at their condensation points are not ideal gases, so it is only an approximation, but it is often adequate

The percentage change in vapor pressure, ∆P/P, for a one-degree change in temperature for

water at its normal boiling point, can be found from the Clausius-Clapeyron equation The heat

of vaporization of water is 2.258 kJ/g or 40,657 J/mol, at 100 C Therefore,o

For larger ranges of temperature the differential form of the Clausius-Clapeyron equation

(equation 5) must be integrated to give the equation

This form assumes that the heat of vaporization (or heat of sublimation) is a constant over the

temperature range ∆T = T - T It should be recognized that ∆H does depend on temperature,2 1 because the heat capacities of condensed phase and gas phase are different The change in ∆H is, however, usually small compared to ∆H itself if ∆T is small, and the error can be further reduced

by substitution of an average value for ∆H into equation 6.

From equation 6 the vapor pressure of water at room temperature would be calculated as follows:

Substitution of an average value for ∆H of about 2.34 kJ/g would change the final answer to P =2

25.2 torr The experimental value is 23.756 torr Over a smaller temperature range better

agreement would be expected

The pressure appearing in the Clausius-Clapeyron equation is the pressure of the (ideal) pure vapor in equilibrium with the liquid It is not necessarily the same as the total pressure applied to

ln 1 2 f f RT G= ∆ (8)

1 Lim 0 = → P f P the liquid For example, the pressure on water in the laboratory is 1 atm, applied by the air, but the pressure of the water vapor at equilibrium is only 23.756 torr, or about 0.03 atm The pressure of a single component is often called the “partial pressure,” to distinguish it from the total pressure It is the equilibrium partial pressure, or vapor pressure, that is a measure of the escaping tendency of the liquid The escaping tendency of the liquid is affected, but only slightly, by the total applied pressure FUGACITY When the vapor is ideal there is a simple, exact relationship between vapor pressure and the thermodynamic properties, especially the free energy, G (equation 28, Chapter 2) It would be a great help to have an exact relationship for real gases and the escaping tendencies of liquids and solids that could be employed for thermodynamic calculations The logical requirement for this new function, f, is that it should coincide with the vapor pressure of a solid or liquid, and with the partial pressure of a vapor, whenever the vapor is ideal, but it should retain the same functional relationship to G even when the vapor is non-ideal This new measure of escaping tendency may be considered a more practical vapor pressure We define the function, f, by the two equations and An alternative form of the first of these equations is to write G = RT ln f + B(T) (7a)

with the unknown function B(T), depending on temperature, added so that absolute values can be assigned to f even though absolute values are not known for G The function f is called fugacity, from the same Latin root (fugere) as “fugitive” The value to be inserted for G when the substance is not pure is an effective value, to be discussed later From the defining equations (7 and 8) a value for the fugacity can be found experimentally for any substance At a sufficiently low pressure of the vapor, fugacity will equal pressure The change in free energy with pressure can be found from equation 22, Chapter 2, ∆G = IV dP (9)

If the volume is known for each pressure, this integral can be evaluated (by numerical integration

if necessary) and the fugacity at the higher pressure can be found from the change in free energy:

(10)

ln ln 1 2 1 2 2 1 P f RT f f RT VdP G P P = = = ∆ ∫ α + = P RT V (10a)

ln 2 1 1 2 ∫ + = ∆ P P dP P P RT G α (11)

i

P

P P

f ≈

In this equation, P is chosen to be sufficiently small so that P = f Alternatively, the volume1 1 1

may be written

where α is a correction term that varies with pressure and is zero for zero pressure Then, for any

P and P ,1 2

In practice these somewhat tedious integrations are seldom necessary

Fugacity tables are available for gases When the actual volume of a gas is known at a certain temperature and pressure, the fugacity may be approximated by means of the equation

f P V/RT2

or, letting P = RT/V, i

This equation can be derived from van der Waals’ equation or other similar equations of state by neglecting second-order correction terms Note that the fugacity differs from the true pressure of

a non-ideal gas in the opposite direction from that of the “ideal” pressure, P = RT/V i

In many applications we need not know the absolute value of the fugacity but only a relative

value compared to some standard state The ratio, f /f , where f is the fugacity in the particularo o

standard state chosen, is called the activity The activity is a dimensionless ratio; the value of the

ratio depends on the choice of standard state as well as the state of the substance being described Different standard states are frequently employed in different problems, or often even in the same problem, so different values of the activity are obtained for a given state of any substance

In the discussions that follow the reader may, without appreciable error, substitute the term

“vapor pressure” for fugacity, or else the term “partial pressure” if the substance is itself a vapor The equations are exactly true as given and are approximately true if vapor pressures (or partial pressures) are chosen as approximations to the fugacity (Note, however, that the total applied pressure may be very different from the vapor pressure, or fugacity, of a solid or liquid.)

EXACT CLAUSIUS-CLAPEYRON EQUATION The Clapeyron equation (equation 4) is an exact