The First Law and Other Basic Concepts
2.5 EQUILIBRIUM AND THE THERMODYNAMIC STATE
(a) How much electrical energy is produced when 1 kg of air passes through the turbine?
(b) What is the power output of the turbine?
(c) If there is no heat transferred to the air, and if its temperature remains unchanged, what is its change in speed upon passing through the turbine?
Solution 2.2
(a) The kinetic energy of the wind on the basis of 1 kg of air is:
E K 1 = 1__
2 mu 2 = (1 kg) (9 m·s _____________−1 ) 2
2 = 40.5 kg· m 2 ·s −2 = 40.5 J Thus, the electrical energy produced per kilogram of air is 0.44 × 40.5 = 17.8 J.
(b) The power output is:
17.8 J·kg −1 × 15,000 kg· s −1 = 267,000 J·s −1 = 267 kW
(c) If the temperature and pressure of the air are unchanged, then its internal energy is unchanged. Changes in gravitational potential energy can also be neglected. Thus, with no heat transfer, the first law becomes
Δ(Energy of the system) = Δ E K = E K 2 − E K 1 = W = −17.8 J⋅ kg −1
E K 2 = 40.5 − 17.8 = 22.7 J⋅ kg −1 = 22.7 N⋅m⋅ kg −1 = 22.7 m 2 ⋅s −2
E K 2 = u__ 22
2 = 22.7 m 2 ⋅s −2 and u 2 = 6.74 m ⋅s −1 The decrease in air speed is: 9.00 − 6.74 = 2.26 m·s−1.
2.5. Equilibrium and the Thermodynamic State 31 Whether a change actually occurs in a system not at equilibrium depends on resistance as well as on driving force. Systems subject to appreciable driving forces may change at a negligible rate if the resistance to change is very large. For example, a mixture of hydrogen and oxygen at ordinary conditions is not in chemical equilibrium, because of the large driving force for the formation of water. This reaction (hydrogen combustion) would occur rapidly and violently if initiated by a spark. However, if chemical reaction is not initiated, this system may exist in long-term thermal and mechanical equilibrium, and purely physical processes can be analyzed without regard to possible chemical reaction.
Likewise, living organisms are inherently far from overall thermodynamic equilibrium.
They are constantly undergoing dynamic changes governed by the rates of competing bio- chemical reactions, which are outside the scope of thermodynamic analysis. Nonetheless, many local equilibria within organisms are amenable to thermodynamic analysis. Examples include the denaturing (unfolding) of proteins and the binding of enzymes to their substrates.
The systems most commonly found in chemical technology are fluids, for which the primary characteristics (properties) are temperature T, pressure P, specific or molar volume V, and composition. Such systems are known as PVT systems. They exist at internal equilibrium when their properties are uniform throughout the system, and conform to the following axiom:
Axiom 3: The macroscopic properties of a homogeneous PVT system at internal equilibrium can be expressed as a function of its temperature, pressure, and composition.
This axiom prescribes an idealization, a model that excludes the influence of fields (e.g., electric, magnetic, and gravitational) as well as surface effects and other less common effects.
It is entirely satisfactory in a multitude of practical applications.
A concept associated with internal equilibrium is a thermodynamic state for which a PVT system has a set of identifiable and reproducible properties, including not only P, V, and T, but also internal energy and other properties yet to be introduced. However, the notation of Eqs. (2.3) through (2.6) suggests that the internal energy terms on the left are different in kind from the quantities on the right. Those on the left reflect changes in the thermodynamic state of the system as reflected by its properties. For a homogeneous pure substance we know from experience that fixing two of these properties also fixes all the others, and thus determines its thermodynamic state. For example, nitrogen gas at a temperature of 300 K and a pressure of 105 Pa (1 bar) has a fixed specific volume or density and a fixed molar internal energy. Indeed, it has a complete set of intensive thermodynamic properties. If this gas is heated or cooled, compressed or expanded, and then returned to its initial temperature and pressure, its intensive properties are restored to their initial values. They do not depend on the past history of the substance nor on the means by which it reaches a given state. They depend only on present conditions, however reached. Such quantities are known as state functions. For a homoge- neous pure substance, if two state functions are held at fixed values the thermodynamic state of the substance is fully determined.5 This means that a state function, such as specific internal energy, is a property that always has a value; it can therefore be expressed mathematically as a function of coordinates such as temperature and pressure, or temperature and density, and its values can be identified with points on a graph.
5For systems of greater complexity, the number of state functions that must be specified in order to define the state of the system may be different from two. The method of determining this number is found in Sec. 3.1.
On the other hand, the terms on the right sides of Eqs. (2.3) through (2.6), representing heat and work quantities, are not properties; they account for the energy changes that occur in the sur- roundings. They depend on the nature of the process, and they may be associated with areas rather than points on a graph, as suggested by Fig. 1.3. Although time is not a thermodynamic coordinate, the passage of time is inevitable whenever heat is transferred or work is accomplished.
The differential of a state function represents an infinitesimal change in its value. Inte- gration of such a differential results in a finite difference between two of its values, e.g.:
∫ V1 V2dV = V 2 − V 1 = ΔV and
∫ UU1 2dU = U 2 − U 1 = ΔU
The differentials of heat and work are not changes, but are infinitesimal amounts. When inte- grated, these differentials give not finite changes, but finite amounts. Thus,
∫ dQ = Q and
∫ dW = W
For a closed system undergoing the same change in state by several processes, experiment shows that the amounts of heat and work required differ for different processes, but that the sum Q + W [Eqs. (2.3) and (2.5)] is the same for all processes.
This is the basis for the identification of internal energy as a state function. The same value of Δ U t is given by Eq. (2.3) regardless of the process, provided only that the change in the system is between the same initial and final states.
Example 2.3
A gas is confined in a cylinder by a piston. The initial pressure of the gas is 7 bar, and the volume is 0.10 m3. The piston is held in place by latches.
(a) The whole apparatus is placed in a total vacuum. What is the energy change of the apparatus if the restraining latches are removed so that the gas suddenly expands to double its initial volume, the piston striking other latches at the end of the process?
(b) The process described in (a) is repeated, but in air at 101.3 kPa, rather than in a vacuum. What is the energy change of the apparatus? Assume the rate of heat exchange between the apparatus and the surrounding air is slow com- pared with the rate at which the process occurs.
Solution 2.3
Because the question concerns the entire apparatus, the system is taken as the gas, piston, and cylinder.
2.5. Equilibrium and the Thermodynamic State 33
(a) No work is done during the process, because no force external to the system moves, and no heat is transferred through the vacuum surrounding the apparatus. Hence Q and W are zero, and the total energy of the system does not change. Without further information we can say nothing about the distribution of energy among the parts of the system. This may be different than the initial distribution.
(b) Here, work is done by the system in pushing back the atmosphere. It is evaluated as the product of the force of atmospheric pressure on the back side of the piston, F = PatmA, and the displacement of the piston, Δl = ΔVt/A, where A is the area of the piston and ΔVt is the volume change of the gas. This is work done by the system on the surroundings, and is a negative quantity; thus,
W = − F Δl = − P atm ΔV t = − (101.3) (0.2 − 0.1) kPa⋅ m 3 = − 10.13 ___kN m 2 ⋅m 3 or
W = −10.13 kN·m = − 10.13 kJ
Heat transfer between the system and surroundings is also possible in this case, but the problem is worked for the instant after the process has occurred and before appreciable heat transfer has had time to take place. Thus Q is assumed to be zero in Eq. (2.2), giving:
Δ(Energy of the system) = Q + W = 0 − 10.13 = −10.13 kJ
The total energy of the system has decreased by an amount equal to the work done on the surroundings.
Cylinder Piston
Latch
Gas under pressure
Latch
Before After
Solution 2.4
Assume that the system changes only in its internal energy and thus that Eq. (2.3) is applicable. For path acb, and thus for any path leading from a to b,
Δ Uabt = Q acb + W acb = 100 − 40 = 60 J (a) For path aeb,
Δ Uabt = 60 = Q aeb + W aeb = Q aeb− 20 and Q aeb = 80 J (b) For path bda,
Δ Ubat = − Δ Uabt = − 60 = Q bda + W bda = Q bda + 30 and
Q bda = − 60 − 30 = − 90 J
Heat is therefore transferred from the system to the surroundings.
Example 2.4
When a system is taken from state a to state b in the accompanying figure along path acb, 100 J of heat flows into the system and the system does 40 J of work.
(a) How much heat flows into the system along path aeb if the work done by the system is 20 J?
(b) The system returns from b to a along path bda. If the work done on the system is 30 J, does the system absorb or liberate heat? How much?