An Introduction to Thermodynamics phần 2 potx

This scale was formerly called “centigrade” in English-speaking countries, a name that 5 can be confused with 1/100 degree in other languages. 7/10/07 1- 14 temperature on the Celsius scale. Temperature differences are the same for the two scales. With 5 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 I dV = - P ∆V (9a) ext ext 3 5 1 m 0.0530 Pa 10 x 013 . 1 K 323K x J/mol .3148 x mol 2 = ⋅ =V 3 5 2 m 0.01767 Pa 10 x 1.013 x 3 K 323K x J/mol .3148 x mol 2 = ⋅ =V Kmol atmml 06.82 Kmol cal 987.1 Kmol 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 closely 2 2 2 follow the ideal-gas equation at room temperature; such easily condensable gases as CO or H O 2 2 vapor follow the equation less closely. 7/10/07 1- 15 The initial and final volumes can be calculated from the ideal-gas equation , PV = nRT. 6 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 kJ 5 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·meter 3 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/m 5 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, 7/10/07 1- 16 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). 7/10/07 1- 17 For example, if 2 mol of hydrogen at 3 atm and 50 C is expanded isothermally and reversibly o 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 7/10/07 1- 18 (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 measurements o 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-law 3 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 calculated o 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. 7/10/07 1- 19 in the same manner. The heat of vaporization is 2.258 kJ/g, and for calculation of the work done 8 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 &6 2 H O 2 2 2 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, or 2 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 water o 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. 7/10/07 1- 20 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 V 2 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. 7/10/07 1- 21 = 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 . We found that for 2 mol of H warmed at P 2 constant volume from 25 C to 50 C, ∆E = 1.05 kJ. For the same process, the enthalpy change is o 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 dT q dT dH = (15) P P C T H =       ∂ ∂ ∫ ∫ =       ∂ ∂ =∆ dTCdT T H H P P ( ) ( ) RC dT RTd dT dE dT PVd dT dE dT dH C VP +=+=+== 7/10/07 1- 22 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 ∆H is negative. reaction 7/10/07 1- 23 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 ∆H . Although ∆E is nearly the same as vap fusion ∆H , because of the small volume changes involved, the difference between ∆E and ∆H fusion vap vap 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 associated 11 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 [...]... initial state, E 1, and the energy of the final state, E2, to find ∆E Integrands of this type are called exact differentials Most integrals we will encounter are well defined in themselves For example, ∫ x2 x1 ∫ 1 2 1 2 x2 − x1 2 2 xdx = x2 , y2 x1 , y1 (xdy + ydx ) = ∫ x2 , y2 x1 , y1 d ( xy ) = x 2 y 2 − x1 y 1 Some, however, are not defined For example, ∫ x 2 ,y x1 , y1 2 ydx cannot be evaluated... because P changes with V), so P dV is an inexact differential W is an amount, not a change in work or any other property, so it is at best misleading to describe W as a differential (exact or inexact) Careful terminology can be an aid to understanding Careless terminology is a hindrance to understanding EQUATION OF STATE Knowledge of two properties, or state functions, is generally sufficient to know,... of x, then we can make a plot of y vs x, as in Figure 3 Then, for example, if y = x2 + 3, ∫ x2 , y2 x1 , y1 7/10/07 ydx = ∫ (x x2 x1 2 ) + 3 dx = 1 3 1  x 2 + 3 x 2 −  x 13 + 3 x 1  3 3  1- 25 We can say that Iy dx has been integrated along the line shown in Figure 3 This is called a line integral.14 Line integrals appear in thermodynamics when it is necessary to evaluate a quantity that depends... discussion to equilibrium states, so that there can be no strains in solids, or other residual effects It follows, therefore, that the change in such an equilibrium property, in going from one state to another, can depend only on the initial and final states Mathematically, this means that the change, described by an integral, depends only on the limits of the integral For example, ∆E = E 2 − E1 = ∫ E2 E1... with substances in substantially identical states An important question is how much information we must have about any system in order to determine its state, and hence to fix completely its physical and chemical properties EXTENSIVE AND INTENSIVE PROPERTIES If the amount of the system is important, it must be specified, either by mass or by equivalent measure such as the number of moles 12 Some properties,... temperature and pressure are known the volume can be calculated (assuming the necessary measurements have been previously made) whether the substance is solid, liquid, or gaseous, and from this the density, viscosity, and refractive index follow To define the state of a gas it would be equally satisfactory to specify the pressure and volume, or the volume and temperature On the other hand, knowing... 12 A mole is a certain number of molecules or other particles (Avogadro’s number, equal to 6. 02 x 1 023 ) Thus it is necessary to indicate what kind of molecules are being counted (e.g., H is not the same as H2) In practice, the number of moles is determined from the mass and the molar mass (or so-called “molecular weight”) with units of g/mol understood At one time, quantities such as “gram mole” and... Nevertheless, a weight equal to the molar mass number in pounds may appear under the title of “pound mole” 13 “Specific” quantities are generally defined as dimensionless ratios of the (extensive) properties of a substance to the (extensive) properties of another substance Thus specific gravity is the ratio of the density of a substance to the density of water (or, for gases, the ratio to the density of air)... that equilibrium state, if there are no “extraneous” variables such as electric or magnetic fields) For example, if the pressure and temperature of n moles of an ideal gas are known, the volume can be found from the “equation of state”, V = nRT/P, or PV = nRT Any substance, whether an ideal gas or not, follows an equation of state relating the pressure, volume, and temperature, but the true equation of... very complicated function and, indeed, may not be known accurately Several functions have been proposed as good approximations to the equations of state for real gases The best-known is that suggested by van der Waals:  n2a   P + 2 (V − nb ) = nRT  V    In this equation, a and b are constants that depend on the particular gas to be described When the volume is large and the temperature is high, . a ∫ =−=∆ 2 1 12 E E dEEEE 2 1 2 2 2 1 2 1 2 1 xxxdx x x −= ∫ ( ) ( ) 1 122 , , , , 22 11 22 11 yxyxxydydxxdy yx yx yx yx −==+ ∫∫ ∫ 22 11 , , yx yx ydx ( )       +−+=+= ∫∫ 1 3 12 3 2 2 , , 3 3 1 3 3 1 . 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, . substitutions, ∆E = 2 mol x 21 J/mol·K x ( 323 - 29 8) 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