In nature some things happen spontaneously (naturally), some do not. For example a hot body cools down to a temperature of the surroundings in a spon-taneous process of giving up heat. After a sufficiently long time equilibrium is reached and the body’s temperature reaches surroundings temperature. A com-pressed gas stored in a cylinder expands spontaneously into surroundings as soon as the interaction with the surroundings is made possible by opening the cylinder valve. Again after a long enough time equilibrium is reached; the gas pressure in the cylinder equals the surroundings pressure and the gas outflow from the cylinder ceases. A cigarettes smoke mixes into the air by a (spontaneous) pro-cess of molecular diffusion. All these spontaneous propro-cesses proceed naturally, in one direction, and there is no need to do any work to bring the changes about.
These spontaneous processes are irreversible. By calling a process irreversible we mean that the initial state, or any other past state, cannot be reached by any spontaneous process that would proceed in the opposite direction.
hot
Fig. 3.1: Examples of irreversible processes in closed and open systems.
Left – hot body cooling down to surroundings temperature;
Middle – compressed gas expanding to surroundings pressure (throttling);
Right – cigarette smoke diffusing into surroundings.
Consider again our three examples: the hot body being cooled down, the com-pressed gas expanding to surroundings pressure (the process is called throttling) and the cigarette smoke diffusing into the air. The first example is a case of a closed system while the second and the third examples are open systems, see Fig. 3.1. During theses spontaneous processes the system (the hot body, the compressed gas, the cigarette smoke) interacts with surroundings. The irre-versibility of open or closed processes means that it is not possible to get back to any previous states of the system and the surroundings.
After cooling the hot body to surroundings temperature, it is not possible to heat it up again using surroundings, even though doing so would not violate the first
3.1 Irreversible and Reversible Processes
law of thermodynamics. Similarly after expanding the gas to surroundings pres-sure it is not possible to compress the gas back to any previous states using only the surroundings. As soon as the cigarette smoke disperses into the air, there is no way of bringing it back to the cigarette tip using only surroundings. It does not mean that the original situation can never be restored. It can, provided that the system plus the surroundings are not kept constant but are allowed to interact with other systems. Obviously the hot body that has been cooled down to the surroundings temperature can be heated up back to its initial tempera-ture. However to accomplish this we cannot use the heat that the body has given up to surroundings but we must employ another source of energy for example a burner or an electrical heater. Similarly, we can compress the gas to its initial pressure using a compressor however, it is not possible to achieve it using energy of surroundings only.
Fig. 3.2: Irreversible process of mixing. An isolated system consisting of a box divided by a partition. The left hand side is filled with a gas whilst the right hand side is empty.
As an example of an isolated system consider a box divided by a partition into two equal parts, each of volume V , Fig. 3.2. The left hand side is filled with gas while the right hand side is empty. Suppose now that the partition is removed.
As a result of collisions with the walls and with each other, the molecules will very quickly redistribute themselves over the entire volume of the box. The final equilibrium state, where the density of molecules is uniform throughout the entire box is attained rather quickly. The question is what determines the direction of this spontaneous mixing process? It is certainly not the total energy of this isolated system since it remains constant through the whole process. However when a change within the box occurs, the energy is parcelled out (distributed) differently. This irreversible mixing process proceeds towards the greater chaotic dispersal of the total energy of the system. The irreversibility of an isolated system means that it is not possible to reach any previous states of the system and the process proceeds in one direction only. The spontaneous,
3 Equilibrium Thermodynamics
irreversible processes are always associated with a redistribution of energy into a more disordered form.
There is another observation to be made in conjunction with irreversible processes taking place in open or closed systems. The more spontaneous and violent the process is, the more difficult it is to bring the system to the initial state since the changes ("damages") to surroundings are larger. When the body is hotter or more precise when the temperature difference between the body and the surroundings is larger, the cooling process proceeds faster. To bring the body back to the initial state, we would have to use an energy source (a burner, an electrical heater) capable of heating the body to a higher temperature.
The second law of thermodynamics governs the direction of irreversible processes.
Historically there have been many formulations of the law. Probably the simplest formulation is due to Rudolf Clausius who in 1850 stated that
"Heat cannot of itself pass from a colder to a hotter body."
Other more elaborate formulation due to William Thomson (Lord Kelvin) reads:
"A transformation whose only final result is to transform into work heat extracted from a source which is at the same temperature
throughout is impossible."
Both of these formulations are not applicable in engineering practise. They are just statements indicating in which direction heat can (or rather cannot) be trans-ferred. After introducing the notion of entropy another formulation can be derived (see Section 3.3.1) which allows the second law of thermodynamics to be used in engineering.
Irreversible processes reach a different "degree of irreversibility" depending on how the processes proceed. Thus, if we alter the way of carrying out the process we may minimise the changes ("damages") to the surroundings. Perhaps we can even carry out the process in such a way that the degree of irreversibility would be zero and the process would be reversible. A reversible process is able to bring the system to any previous states, and finally to the initial state, without any changes to surroundings. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined (original and reverse) process.
Reversible processes do not occur in nature. However we mentally (conceptu-ally) construct reversible processes in order to compare real, irreversible processes against these ideal conceptual ones. Reversible processes can be approximated by actual devices but they can never be achieved. Engineers are interested in
3.1 Irreversible and Reversible Processes
reversible processes because devices producing work (engines, turbines) deliver maximum work when they realise reversible processes. Devises like pumps and compressors require minimum work when they realise reversible processes. Thus, reversible processes can be regarded as theoretical limits of the corre-sponding real processes.
Consider a closed system (of constant mass) that is at an initial state marked in Fig. 3.3 as state 1. By applying (adding) an infinite number of combinations of heat and work (see Eq. (2.56)) to the system, the matter of the system can reach an infinite number of states. Among these infinite number of thermodynamic processes the following five are typically considered:
(a) isothermal process (either expansion or compression); T = const., (b) processes proceeding under constant volume; V = const.,
(c) processes proceeding under constant pressure; p = const., (d) adiabatic processes; Q = 0,
(e) polytropic processes, pVα = const. with α = const. where α is called poly-tropic exponent.
Fig. 3.3 shows these typical processes using a p-V (work) diagram. All of them might be carried out as reversible processes if additions of the infinitesimal amounts of heat and work is carried out in a series of quasi-static steps (see below). However there are two thermodynamic processes that are highly irreversible and cannot be reversed. These are: throttling (expansion without any work) and mixing. Fig. 3.1 (middle) shows an example of throttling while Fig. 3.1 (right) and Fig. 3.2 show a highly irreversible process of mixing. Chemical reactions are highly irreversible.
Processes shown in Fig. 3.3, can be carried out reversibly if the infinitesimal amounts of heat and work are added (or removed) into the system in a process which is carried out so slowly that the system remains arbitrarily close to equi-librium at all stages of the process. Such a process is said to be "quasi-static", or "quasi-equilibrium" for the system. A quasi-static process can be viewed as a sufficiently slow process which allows the system to adjust itself internally so that properties in one part of the system do not change any faster than in another part.
3 Equilibrium Thermodynamics
p
V
V1 V2
V=const.
T=const.
p=const.
Q.
=0
1-2
L1-2 2
Q.1-2
2 2
2 1
Fig. 3.3: Typical thermodynamic processes shown using work-diagram; T = const., V = const., p = const., Q = 0, (pVα= const. not shown).
How slowly one must proceed to make sure that the process is quasi-static depends on the time (called "relaxation time") that the system requires to attain equilib-rium if it is suddenly disturbed. To be slow enough to be quasi-static implies that the process proceeds slowly compared to the relaxation time. For example, if the gas in Fig. 3.3 returns to equilibrium within a time of 10−5 seconds after the piston is suddenly moved back from position 2 to position 1, then a process wherein the piston is moved in 0.1 second can be considered quasi-static to a good approximation.
Consider a piston-cylinder, Fig. 3.4, that contains a gas at initial pressure p1. When the piston is moved suddenly to the right the molecules near the face of the piston move to the space made just available, and therefore the pressure near the piston decreases. Because of this low pressure, the system (gas) is not longer in equilibrium since a higher pressure exists at the other (left hand) end of the cylinder. Such a process is called non quasi-equilibrium. However, if the piston is moved slowly, the molecules have sufficient time to redistribute and the pressure in the entire volume of the gas remains always uniform. Since equilibrium within the gas prevails at all times, this is a quasi-equilibrium or quasi-steady-state process.
In order to carry out a reversible expansion of the compressed gas a series of infinitesimal reversible steps would have to be realised as shown in Fig. 3.4.
3.1 Irreversible and Reversible Processes
Fig. 3.4: Gas expansion process;
Left – Irreversible (non-quasi-equilibrium) expansion, Right – Reversible (quasi-equilibrium) expansion.
The work done to the system (gas) during the process is
Lrev= −
final state
Z
initial state
p · dV = Lmax (3.1)
The maximum work available from such an expansion from initial to final states is obtained when the process is carried out reversibly.
The irreversibility of a process is caused by the occurrence of the followings:
(a) friction,
(b) heat transfer through a finite temperature difference, (c) throttling (gas expansion without any work obtained), (d) non-quasi equilibrium compression or expansion, (e) electrical resistance (dissipation of heat),
(f) chemical reactions (destruction of chemical bonds).
3 Equilibrium Thermodynamics
3.2 Entropy
In Chapter 2 we have introduced the first law of thermodynamics and one of the most useful formulations of this law (see Eq. (2.52)) is:
dQt+ dL = dU (3.2)
where U stands for the internal energy and L is the absolute work done to the system. Symbol Qt (t stands for total) represents the total heat absorbed by the system:
dQt= dQf + dQ (3.3)
where Q is the amount of heat supplied from the surroundings or any other heat source, and Qf is the amount of heat generated within the system due to friction.
All these terms are expressed in joules.
We have already learned that internal energy (U) is a state variable and its dif-ferential is exact while neither heat (Q) nor work (L) are state variables and their differentials are inexact. However, T1 is an integrating factor for dQt turning it into an exact differential called entropy (S):
dS = dQt
T (3.4)
where dQtis the infinitesimal amount of heat exchanged and T is the temperature of the quasi-equilibrium, reversible process (see Example 3.1). Entropy is an extensive property so
S = m · s (3.5)
where S is the total entropy of the system in J/K, m is the mass of the system in kilograms and s is the specific entropy of the system in J/kg · K. Therefore
ds = dqt
T (3.6)
Example 3.1
The objective of this example is to demonstrate how an inexact differential can be transformed into an exact one. We wish to improve the understanding of relationships (3.4) and (3.6) which transform an inexact differential of heat into an exact differential of entropy. Consider the infinitesimal quantity:
¯
dG = α · dx + β ·x
y· dy = α · dx + β · x · d(ln y)
3.2 Entropy
Is ¯dG an exact differential? To check this, we have to calculate the second deriva-tives: the integral of ¯dG should be path dependent. Let "i" denote the initial point (1, 1) and "f " the final point (2, 2). If one then calculates the integral of ¯dG along the path i → a → f passing through the point "a" with coordinates (2, 1) one obtains:
Fig. 3.5: Integration paths from i → f
Let us calculate the same integral along the path i → b → f:
Z We can also calculate the integral along a direct path from (1, 1) to (2, 2). To this
1The cross-bar over letter ¯dindicates that the differential ¯dG is inexact. In some textbooks the inexact differentials of heat and work are marked as ¯dQand ¯dLto distinct them from the exact differentials of internal energy dU and enthalpy dH [3, 5].
3 Equilibrium Thermodynamics
end, we introduce a parametric variable t so:
x = t 1 ≤ t ≤ 2
As it can be clearly seen, we have obtained three different answers (A1), (A2) and (A3), depending on the integration path. Thus, the integrals are different, the quantity ¯dG is not an exact differential, and G is not a state function in the x, y space.
Let us make a new infinitesimal quantity dF by multiplying ¯dG by 1x so:
dF =dG¯ x = α
xdx + β y dy.
The new infinitesimal dF is an exact differential since:
∂
Integrating dF along any path between "i" (1, 1) and "f" (2, 2) should give us the same answer. To confirm this, let us calculate the integral along several paths.
Z The above integral does not have to be integrated along a straight line path. The integral can be evaluated along any path. Let us calculate the integral along a parabola: y = (x − 1)2that goes through points "i" (1, 1) and "f" (2, 2). In order
3.2 Entropy
to perform the integration we introduce a parametric variable t so that:
x = t 1 ≤ t ≤ 2 The integral Ri→fdF has been evaluated along four different paths and all the answers are identical ((B1), (B2), (B3), (B4)), as one would expect. The quantity dF is an exact differential and F is a state function in the x, y space. Since we know that dF is an exact differential, we can easily calculate the function F (x, y).
Knowing that
3 Equilibrium Thermodynamics
Thus, the function F (x, y) is of the form:
F (x, y) = α · ln x + β · ln y + const.
Knowing the function F (x, y) we can check that:
f the results of the integration along several paths (B1, B2, B3, B4).
There is another important observation to be made in conjunction with this exam-ple. The infinitesimal differential ¯dG is not exact. However, when the differential
¯
dG has been multiplied by 1x we have obtained a new differential dF that is ex-act. In mathematics the factor 1x is called the integrating factor. The integrating factor transforms an inexact differential into an exact differential. In the same way an inexact differential of heat may be transformed into an exact differential (entropy) using the integrating factor T1.
End of Example 3.1
Now our task is to derive appropriate mathematical relationships that would allow for calculating specific entropy as a function of temperature and pressure of the system s = s(T, p) and as a function of temperature and specific volume s = s(T, v). We begin with deriving the exact differential for specific entropy as a function of temperature and pressure s = s(T, p). To this end we introduce into the definition of the specific entropy (Eq. (3.6)) the first law of thermodynamics (Eq. (2.61)):
ds = dh − v · dp
T (3.7)
Introducing relationship (2.33) into the above equation we obtain:
ds = 1
3.2 Entropy
Thus, knowing the equation of state v = v(T, p) one can evaluate the derivative
∂v
∂T
p and calculate the exact differential of specific entropy as a function of tem-perature and pressure.
The procedure for deriving the exact differential of entropy as a function of tem-perature and specific volume is similar. Into the definition of specific entropy (Eq. (3.6)) we introduce the first law of thermodynamic (Eq. (2.59)) obtaining:
ds = du + p · dv
T (3.10)
and making use of Eq. (2.21) and after some algebra one obtains:
ds = cv
Eqs. (3.9) and (3.11) are the basis for calculating the entropy of any substance as a function of temperature and pressure, or temperature and specific volume, respectively. Similarly to enthalpy and internal energy we can only calculate the entropy change with respect to a certain reference state s0 = s(T0, p0) since
s(T, p) = s0+
Thus, to calculate entropy we need to know the specific heats and the equation of state of the substance. The integration can be carried out along any path from (T0, p0) to (T, p) or (T, v).
3 Equilibrium Thermodynamics
3.2.1 Entropy of Liquids and Solids Liquids and solids are incompressible and therefore
∂v
Eq. (3.12) can be then simplified to
s = s0+
and if the specific heat is independent of temperature we obtain:
s = c · ln T
T0 (3.16)
assuming that s0= 0.
3.2.2 Entropy of Ideal Gases
Invoking again the equation of state for ideal gases:
p · v = R · T
M (3.17)
from which we can easily obtain:
∂v
and inserting Eq. (3.18) into Eq. (3.12) we obtain:
s = s0+
Since cpof ideal gases is independent of pressure the above integral can be written as:
3.2 Entropy
If then cp is independent of temperature we obtain:
s = s0+ cp· ln T In the above equations s stands for specific entropy in J/kg · K, T is the temper-ature in Kelvin, cp is the specific heat at constant pressure in J/kg · K. In many text books on thermodynamics, the reader can find similar expressions for specific entropy (s) expressed in J/kmol · K written as:
s = s0+ Cp· ln T
T0 − R · ln p
p0 (3.22)
where Cp is the specific heat at constant pressure in J/kmol · K.
Using Eq. (3.13) one may derive similar expressions for calculating specific entropy as a function of temperature and specific volume:
s = s0+
and if the specific heat at constant volume is independent of temperature the following is applicable:
s = s0+ cv· ln T T0 + R
M · ln v
v0 (3.24)
3.2.3 Entropy of Phase Transition at the Transition Temperature We consider a system and its surroundings at a temperature at which two phases are in equilibrium. Such a temperature is called transition temperature and is denoted by Ttrs. For example, this temperature is 0◦C for ice in equilibrium with water at 1 bar, and 100◦C for water in equilibrium with its vapour at 1 bar.
At the transition temperature any transfer of heat between the system and its surroundings is reversible since the two phases in the system are in equilibrium.
Because our system is at a constant pressure, the change of the specific entropy of the system is:
3 Equilibrium Thermodynamics
where ∆trsh and ∆trsh are the enthalpy of the transition in J/kg and J/kmol, respectively. Standard (at p = 1 bar) transition enthalpies and transition tem-peratures are listed in the literature; Table 3.1 is an extract from reference [13].
If the phase transition is exothermic (condensation or freezing) then the entropy change is negative. This reflects the fact that during condensation or freezing the system is becoming more ordered. For endothermic transitions the entropy change is positive and the system becomes more disordered.
Table 3.1: Transition temperatures, standard enthalpies and standard entropies (p=1bar) of selected substances [13].
Substance Melting Vaporization
Tempe-rature
∆trsh ∆trss Tempe-rature
∆trsh ∆trss
K kJ/mol J/mol · K K kJ/mol J/mol · K
H2O 273.15 6.008 22.00 373.15 40.656 108.95
O2 54.36 0.444 8.17 90.18 6.820 75.63
N2 63.15 0.719 11.39 77.35 5.586 72.22
CO2 194.6 25.23 129.65
CH4 90.68 0.941 10.38 111.7 8.18 73.23
C2H6 89.85 2.86 31.83 184.6 14.7 79.63
Cl2 172.1 6.41 37.25 332.4 29.45 88.60
3.2.4 The Third Law of Thermodynamics
At T = 0 K all energy of thermal motion has been ceased and substances are in solid state. At T = 0 pure substances are arranged in regular, crystal forms in a perfect order. The third law of thermodynamics states that entropy of a pure crystalline substance at absolute zero temperature is zero2:
s(T = 0) = 0 (3.26)
3.2.5 Absolute Entropy of Pure Substances
The third law of thermodynamics specifies zero entropy at absolute zero
The third law of thermodynamics specifies zero entropy at absolute zero