An Introduction to Thermodynamics phần 2 potx

A completely general proof that the reversible work is a minimum for the isothermal 7 reversible process cannot be given here, but this important result can be illustrated by considering

This scale was formerly called “centigrade” in English-speaking countries, a name that 5

can be confused with 1/100 degree in other languages

temperature on the Celsius scale Temperature differences are the same for the two scales With the appropriate numerical substitutions,

∆E = 2 mol x 21 J/mol·K x (323 - 298) K

= 1.05 kJ/mol

This is the amount of (thermal) energy that must be added to the gas to raise its temperature

25 C.o

IDEAL-GAS EXPANSIONS Now let 2 mol of hydrogen, at 3 atm pressure, expand at a constant temperature of 50 C through a pinhole (to maintain a slow expansion) and against a piston o Assume the external pressure acting on the piston is 1 atm, as in Figure 1 The work done on the gas is the product of the force exerted on the gas (by the piston) and the distance through which the piston moves, with a negative sign because the pressure (of the gas) acts opposite to the

pressure acting on the gas The force, on the gas, times the displacement, is f dx = - (PA) dx = -

P (Adx) = - P dV Because of the unusual arrangement (effusion controlled), the pressure acting

on the gas at the point of displacement is the external pressure, P Therefore the work done on ext

the gas is

(Effusion controlled) W = I f dx = - I P dV (9) ext

Because P is maintained constant, it may be removed from the integral ext

W = - P ext I dV = - P ∆V (9a) ext

3 5

Pa 10

x 013 1

K 323

K x J/mol 314

8

x mol 2

=

⋅

=

V

3 5

Pa 10

x 1.013

x 3

K 323

K x J/mol 314 8

x mol 2

=

⋅

=

V

K mol

atm ml 06 82 K mol

cal 987 1 K mol

J 314 8

⋅

⋅

=

⋅

=

⋅

=

R

The term “ideal” means simply that the substance obeys a certain equation An ideal gas 6

obeys the equation PV = nRT; in later chapters we will encounter the “ideal solution”, which

obeys an equation known as Raoult’s law The ideal-gas equation combines Boyle’s law and Charles’, or Gay-Lussac’s, law into a single, more convenient expression The temperature must

be on an absolute scale, which we will always take as the Kelvin scale; n is the number of moles of gas; and R is a universal constant, whose value depends on the units chosen for pressure and

volume It should be noted that the product of pressure and volume has the dimensions and units

of energy (although PV is not a measure of energy) Such gases as He, H , O , and N closely2 2 2

follow the ideal-gas equation at room temperature; such easily condensable gases as CO or H O2 2 vapor follow the equation less closely

The initial and final volumes can be calculated from the ideal-gas equation , PV = nRT.

Thus the work is

W = - P ∆V = - P (V - V ) ext ext f i

= - 1.013 x 10 Pa (0.0530 - 0.01767) m = - 3.58 kJ5 3

The necessary constants are given in Table 1

Table 1 GAS CONSTANT AND CONVERSION FACTORS

1 joule = 1 newton·meter = 0.239 cal = 1 pascal·meter3

1 cal (thermochemical) = 4.1840 joule = 41.3 ml·atm*

1 ml·atm = 0.1013 joule = 0.0242 cal

1 atm = 1.01325 x 10 N/m5 2

* Various experimentally defined values of the calorie appear in the literature (including the dietician's Calorie = 1 kcal) Only the thermochemical calorie is defined exactly

In order that the temperature may remain constant it is necessary to supply thermal energy to the gas to compensate for the energy expended in doing work From the first-law equation,

Q = ∆E - W

For the special case of an ideal gas, the energy depends only on the temperature, not on the

pressure or volume At constant temperature, therefore, ∆E = 0 and the thermal energy that must

be supplied is

Q = - W = 3.58 kJ

A more important example than the expansion against a constant external pressure is the

“reversible” and isothermal (constant temperature) expansion A process is said to be

thermodynamically reversible if it can be reversed at any stage by an infinitesimal increase in the opposing force or an infinitesimal decrease in the driving force It should be clear that such a reversible process is an example of a limiting case, which is only approximately achieved in practice Not only does it require a frictionless mechanism, but if the unbalance of forces is infinitesimal, the rate will be infinitesimal and the time required will be infinite The question of reversibility will be considered in more detail later

From the assumption that the operation is isothermal and the gas is ideal,

∆E = 0

and

Q = - W = I P dV

Because the expansion is to be reversible, the pressure on the piston differs only infinitesimally

from the equilibrium pressure of the gas, P = nRT/V The work done on the gas is then

(T, I.G., rev.) W = - I nRT dV/V = - nRT ln(V /V ) (10) f i

A completely general proof that the reversible work is a minimum for the isothermal 7

reversible process cannot be given here, but this important result can be illustrated by considering

the expansion or compression of a gas When the piston moves away from the gas molecules, the

molecules do not strike the piston as hard as when the piston is stationary The pressure exerted

by the gas on the piston, and hence by the piston on the gas, is therefore less than the equilibrium pressure of the gas Hence IP dV < I P dV; the work done by the gas is a maximum for a eff

reversible expansion Changing the sign, to obtain the work done on the gas, changes the

direction of the inequality; the work done on the gas, W, is a minimum when the effective pressure

is equal to the equilibrium pressure, for expansion or compression

One may see elsewhere the expression W = - I P dV applied when P , the external ext ext

pressure, differs from the pressure of the system (= the gas) Except in unusual circumstances

(e.g., an effusion-controlled expansion), this is not even a good approximation See Am J Phys.

37, 675-679 (1969)

For example, if 2 mol of hydrogen at 3 atm and 50 C is expanded isothermally and reversibly

to a final pressure of 1 atm,

Q = - W = 2 mol x 8.314 J/mol·K x 323 K x ln V /V f i Because the temperature is constant, V /V = P /P = 3 f i i f

Q = - W = 5.37 kJ ln 3 = 5.90 kJ

Notice that the amount of thermal energy that must be supplied is greater for the reversible expansion than in the previous example of expansion against a constant external pressure A

reversible expansion (or compression) must always give a maximum value for Q, and accordingly

a minimum value for W.7

An integral may be represented by the area under a curve (see Appendix) and the three

examples considered above may be compared graphically When the gas was warmed, at constant

volume, the pressure increased by the factor T /T = 323/297 = 1.09 In a plot of pressure against f i

volume, this is represented by a vertical line (Figure 2a) and the area under such a line is zero

No work is performed

The expansion against a constant opposing force is represented by the plot in Figure 2b The gas, in escaping through the pinhole, changes its pressure from 3 atm to 1 atm, but there is no

force exerted on moving surroundings during this escape When the gas is through the pinhole,

however, it has the pressure of 1 atm and the volume increases, at this pressure, while the gas pushes against the piston (the surroundings), until all of the gas has reached the same pressure

and occupies the total volume V The work done by the gas, - W, is the area under the vertical f

line, which is zero, plus the area under the horizontal line, which is 1 atm times the volume

change

When the pressure of the gas changes smoothly, as in Figure 2c, the area under the curve

(-W) is a maximum for the given temperature The pressure follows the curve PV = constant

(where the constant is nRT) which is part of a hyperbola.

The condition to be satisfied in order that an expansion or compression of any fluid should be thermodynamically reversible, and thus that the work done on the fluid should be - IP dV, is

simply that the fluid have a well-defined, uniform pressure throughout In other words, there must be no “leaks”, allowing different pressures at different points, and the motion of the piston confining the fluid must be slow compared to the “relaxation time” of the fluid, or the time for pressures to equilibrate in the fluid For a gas, the expansion or compression must be very slow compared to the speed of sound in the gas Unless there is such a well-defined pressure, or in certain situations (see Figure 1) two or more well-defined pressures, for the system,

thermodynamics cannot be applied It is then necessary to apply the more difficult methods of non-equilibrium, or irreversible, thermodynamics That is, equilibrium thermodynamics is

sufficient for reversible processes and certain types of irreversible processes, whereas “irreversible thermodynamics” deals with the time-dependent equations that become necessary in other

problems involving irreversible phenomena, such as diffusion rates and shock-wave propagation The existence throughout the process of a well-defined “state” (an equilibrium state, for which pressure is defined) for the system is sufficient to ensure that the expansion or compression will be reversible, but there may well be other irreversibilities in the total process Some of the work done by the gas may be against frictional forces (in the surroundings), or work done on a massive piston may appear as kinetic energy of the piston, subsequently converted to thermal energy by collision of the piston with mechanical stops (in the surroundings) Or there may be conduction of thermal energy to or from the surroundings at a lower or higher temperature Thus the total process may consist of a sum of parts, some of which are reversible and some

irreversible It is particularly important in such circumstances to define carefully the system and surroundings and the exact process to be considered Interchanging “system” and “surroundings” labels may be helpful in analyzing the total process

PHASE CHANGES Let 5 g of ice melt at 0 C under 1 atm pressure Experimental measurementso have shown that the amount of thermal energy that must be supplied, called the “heat of fusion”,

is 334 J/g The work done is -IP dV = - 1 atm x ∆V The volumes of ice and water are 1.09

cm /g and 1.00 cm /g at 0 C The energy change for the ice can be calculated from the first-law3 3 o equation

∆E = Q + W = 5 g x 334 J/g - 1 atm(1.00 - 1.09)ml/g x 5 g

Conversion of the work term from ml·atm to J gives

∆E = 1.67 kJ + 0.046 J = 1.67 kJ

Although the total energy absorbed by the ice, ∆E, is slightly greater than the heat of fusion, Q,

the work done on the ice-water system by the atmosphere (0.05 J) is negligible for nearly all purposes

If 5 g of water is vaporized at 100 C and 1 atm pressure, the energy change can be calculatedo

This heat of vaporization is required to separate the molecules from each other — that is, 8

it represents an increase in the potential energy of the molecules Because the temperature does not change, the kinetic energy remains constant

in the same manner The heat of vaporization is 2.258 kJ/g, and for calculation of the work done

it is sufficient to assume that the water vapor is an ideal gas

∆E = Q + W

= 2.258 kJ/g x 5 g - 1 atm[ 5/18 mol x (8.314 J/mol·K)/1 atm x 373.15 K - 5 x 10 m ]-6 3

= 11.290 kJ - 0.861 kJ = 10.43 kJ

In problems of this type, however, an important shortcut is often satisfactory The volume of the liquid is sufficiently small compared to that of the vapor that it may be neglected Then the work term becomes

W = - P(V - V ) = - PV = - PnRT/P =- nRT f i f The value of R in J/mol·K may be inserted into this expression to avoid completely the units of

volume and the conversion from atm to Pa The entire problem can thus be written

∆E = Q + W = 5 g x 2258 J/g - (5/18) mol x 8.314 J/mol·K x 373 K

= 11.290 kJ - 0.861 kJ = 10.43 kJ

CHEMICAL REACTIONS An example of a gas-phase chemical reaction is the combustion of hydrogen at 100 C and 1 atm pressure (or slightly less) to give water vapor.o

2 H + O 2 2 &6 2 H O2

The heat of reaction (the amount of thermal energy absorbed by the system) is - 242.5

kJ/mol(H O) That is, a negative amount of thermal energy is absorbed by the reacting system, or2

a positive amount of thermal energy is given up by the reacting system to the surroundings The work done on the system is

W = - IP dV = P(V - V ) = - P(n RT/P - n RT/P) = - (n - n )RT f i f i f i

= - (∆n)RT (11)

and the energy change of the reacting system is

∆E = Q + W

= 2 mol(- 242.5 kJ/mol) - (- 1 mol) x 8.314 J/mol·K x 373 K = - 481.90 kJ + 3.10 J = - 481.90 kJ/(2 mol H O)2

The same reaction, at 100 C and 1 atm pressure (or slightly more) but producing liquid watero rather than water vapor, will have a heat of reaction that differs from that of the preceding

Later we will consider work done in other ways, especially by electrical fields We are 9

explicitly excluding such work terms for the present because they are unnecessary here

problem by the heat of vaporization of water

Q = - 481.90 kJ - 2 mol x 18 g/mol x 2.258 kJ/g = - 481.90 kJ - 81.2 kJ

= - 563.19 kJ/(2 mol H O)2

The work differs, also, because ∆V is different Neglecting the volume of the liquid, ∆V =

(∆n)RT/P = (- 3)RT/P.

W = - P ∆V = 3 RT = 3 mol x 8.314 J/mol·K x 373K = 9.30 kJ

Then the energy change of the system is

∆E = Q + W = - 563.19 kJ + 9.30 kJ

= - 553.89 kJ/(2 mol H O)2

Enthalpy

We have covered, in brief outline, the methods of finding the change of energy of any system,

by measuring or calculating the amount of thermal energy transfer, Q, and the amount of work done, W But already, we have seen that complications arise, such as having to find work done,

against the atmosphere, when that work term is itself of little interest

The way to avoid these annoying correction terms, in general, is to define a new quality that focuses on what we do want to know Our second step, therefore, in understanding an applying thermodynamics, is to learn how to define and calculate new variables that will simplify our measurements and calculations

Chemical reactions and phase changes are more often carried out at constant pressure than at

constant volume At constant volume (no work done) the energy change is equal to Q, the

amount of thermal energy transferred to the system, but under constant pressure a correction must be made for the work performed on the system by the external pressure Consider a

completely general constant-pressure process in which the only work done is because of the volume change.9

(P) ∆E = Q + W = Q - P(V - V )2 1

which can be written

∆E = Q - (P V - P V )2 2 1 1

where P = P Set ∆E = E - E and rearrange.1 2 2 1

Q = E - E + P V - P V2 1 2 2 1 1

As you might guess, H was chosen as a symbol because ∆H = Q, the “heat” (i.e., the

10

amount of thermal energy transfer) One must, however, avoid any attempt to associate H with the (total) thermal energy Warning: PV has the dimensions, and units, of energy but, as shown explicitly in equation 13, PV is not energy.

= E + P V - (E + P V )2 2 2 1 1 1

or

(P) Q = ∆(E + PV) (12)

This last equation will be encountered so often that it is a great convenience to introduce a new

symbol for the quantity E + PV.

Definition of Enthalpy H = E + PV (13) Then, if pressure is constant and the only form of work is W =- IP dV, equation 12 becomes

(P) ∆H = Q (14)

The function H is called the “enthalpy”.10

Note that the enthalpy is actually defined by equation 13, which contains no restrictions on pressure, temperature, or volume Nevertheless, the enthalpy will be found to be most convenient for problems in which pressure is constant

A comparison between energy and enthalpy, E and H, can be made by calculating the

enthalpy changes for the same processes for which energy changes were previously found

TEMPERATURE CHANGES AND HEAT CAPACITY, C P We found that for 2 mol of H warmed at2

constant volume from 25 C to 50 C, ∆E = 1.05 kJ For the same process, the enthalpy change iso o

∆H = ∆(E + PV) = ∆E + ∆(PV) = ∆E + V ∆P

= ∆E + V(nRT /V - nRT /V) = ∆E + nR∆T f i = ∆E + 2 mol x 8.3144 J/mol·K x 25 K

= 1.05 kJ + 416 J = 1.47 kJ The enthalpy change is greater than the energy change because the increase in temperature causes

an increase in P and thus in the product PV.

If the gas is warmed at constant pressure, rather than at constant volume, the enthalpy

change can be calculated in much the same way

∆H = ∆E + ∆(PV) = ∆E + P ∆V = ∆E + nR ∆T

= 1.47 kJ

The enthalpy, like the energy, depends only on the temperature, for an ideal gas It is for this

reason that we find the same ∆H (as well as the same ∆E) when the hydrogen gas is warmed by

25 C whether the process is at constant volume or constant pressure, or under other conditions.o

q dT

dH

=

(15)

P P C T H =       ∂ ∂ ∫  =∫      ∂ ∂ = ∆ dT C dT T H H P P ( ) ( ) C R dT RT d dT dE dT PV d dT dE dT dH C P = = + = + = V + It should be observed that in the constant-pressure process the correction term to obtain the enthalpy change from the energy change is a work term, -W = P ∆V But in the constant-volume process the correction term, though of equal magnitude, is not a work term, having instead the form V ∆P (You will learn very quickly that I V dP … W.) The constant-pressure warming process can be treated in a somewhat different manner Employing equation 14, for constant pressure, (P) dH = q Then, dividing by dT, (P)

or For a diatomic gas near room temperature, C is about 7/2 R = 29 J/mol·K Inserting this P (approximate) value for H ,2 = 2 mol x 29 J/mol·K x 25 K = 1.45 kJ

The heat capacity at constant pressure, C , is greater than, or occasionally equal to, the heat P capacity at constant volume, C The difference is small for solids and liquids (typically about 1.7 V

J/mol·K), but for an ideal gas the difference is appreciable and is easily calculated The constant-pressure and constant-volume restrictions can be dropped from the derivatives of equations 6a and 15 for the special case of an ideal gas, because both the energy and enthalpy of an ideal gas are independent of pressure and volume Taking 1 mol of gas,

or

(I.G.) C = C + R (16) P V

IDEAL-GAS EXPANSIONS. The work performed on an ideal gas as it expands isothermally was calculated previously for two important special cases (equations 9 and 10) Because the energy

of the ideal gas is independent of pressure and volume, the energy change must be zero in an

isothermal expansion Thus we conclude that Q + W = 0; the work done on the gas is equal and

An exception is that heat of combustion is given as a positive number (although the

11

amount of thermal energy absorbed by the system is negative) and is therefore - ∆H If there is

any question concerning the convention followed, look for a reaction such as hydrogen plus

oxygen to give water, where we know energy is given off, so ∆Hreaction is negative

opposite to Q, the thermal energy absorbed by the gas in the expansion.

At constant temperature, the product PV is also constant for an ideal gas Therefore, at constant temperature, ∆H = ∆(E + PV) = ∆E = 0 Enthalpy change is also zero for any

isothermal expansion (or compression) of an ideal gas

One often sees the generalization that a gas cools on expansion That is generally not true

except when the gas does work on its surroundings (thus giving up energy to the surroundings) with no compensating transfer of thermal energy to the gas (or for certain situations involving non-ideal gases) Obviously there is no cooling in an isothermal expansion

PHASE CHANGES. The heat of fusion is the amount of thermal energy absorbed by a solid when it melts (fuses) under constant pressure From equation 14 this is identically ∆H for the melting process Similarly, the heat of vaporization is ∆Hvap Although ∆Efusion is nearly the same as

∆Hfusion, because of the small volume changes involved, the difference between ∆Evap and ∆Hvap

can be appreciable, as was shown above Tabulated values are invariably the enthalpy changes

CHEMICAL REACTIONS. The heat of reaction, as tabulated, assumes constant pressure and is therefore identical with the ∆H of reaction That is, it is assumed that the pressures associated11 with each reactant and each product are the same before and after the reaction, so the total

pressure remains constant during the reaction

The assumption of constant individual pressures is much less important when considering heats of reaction than in some later applications Keep in mind, however, in interpreting

thermodynamic processes, that we are not considering a process that starts with reactants, only, and concludes with products, only Rather, we assume the only change is the chemical reaction

Thus we begin with a mixture of products and reactants and allow the reaction to proceed

without change in the physical conditions (temperature, pressure, or concentrations) of the

substances One way of achieving this experimentally would be to have the total amount of material large, and allow only a small part of the reactants to combine (The measured values are then calculated per mole of a reactant or a product.) We will find, in Chapter 4, an alternative method is often to measure a reaction in an electrochemical cell, under equilibrium conditions

We will return to this assumption of fixed conditions when we explicitly consider chemical equilibria

State Functions

We have seen that those properties — such as energy, enthalpy, temperature, pressure, and

volume — that depend on the state of a system are called state functions If measurements

carried out in Boston, Brooklyn, Birmingham, and Berkeley are to be compared, we must be