Table of contents |
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Introduction |
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Work and Reversibility |
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ΔSsys for an Isothermal Expansion (or Compression) |
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The Relationship between Internal Energy and Entropy |
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Solved Example |
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Figure 13.4.1: Reversible vs. Irreversible Expansions and Compressions.
Definition: Reversible Changes
A reversible change is one carried out in such as way that, when undone, both the system and surroundings (that is, the world) remain unchanged.
Figure 13.4.2: Note that the reversible condition implies wmax and qmin. The impossibility of extracting all of the internal energy as work is essentially a statement of the Second Law.
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy.
Entropy is still described, particularly in older textbooks, as a measure of disorder. In a narrow technical sense this is correct, since the spreading and sharing of thermal energy does have the effect of randomizing the disposition of thermal energy within a system. But to simply equate entropy with “disorder” without further qualification is extremely misleading because it is far too easy to forget that entropy (and thermodynamics in general) applies only to molecular-level systems capable of exchanging thermal energy with the surroundings. Carrying these concepts over to macro systems may yield compelling analogies, but it is no longer science. It is far better to avoid the term “disorder” altogether in discussing entropy.
Note: If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.
Because the pressure of an ideal gas is inversely proportional to its volume, i.e.,
P = nRT/V (13.4.7)
we can easily alter Equation 13.4.6 to express the entropy change associated with a change in the pressure of an ideal gas: (13.4.8)
Also the concentration c = n/V for an ideal gas is proportional to pressure
P = cRT (13.4.9)
we can expressing the entropy change directly in concentrations, we have the similar relation
ΔS = ln (c1/c2)(13.4.10)
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture.
Because the quantity of heat transferred (qrev) is directly proportional to the absolute temperature of an object (T) (qrev ∝ T), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (ΔS ∝ qrev). Combining these relationships for any reversible process,
qrev = TΔS (13.4.11)
and
ΔS = qrev/T (13.4.12)
Because the numerator (qrev) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is
wrev = −PΔV, (13.4.13)
we can express Equation 13.4.3 as follows:
ΔU = qrev + wrev
= TΔS − PΔV (13.4.14), (13.4.15)
Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the PV work done.
To illustrate the use of Equation 13.4.12 and Equation 13.4.15, we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its T constant (Figure 13.4.5). The internal energy of the gas does not change because the temperature of the gas does not change; that is, ΔU = 0 and qrev = −wrev. During expansion, ΔV > 0, so the gas performs work on its surroundings:
wrev = −PΔV < 0. (13.4.16)
According to Equation 13.4.15, this means that qrev must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore(13.4.17)
and the entropy change of the surroundings is
ΔSsurr = −qrev/T. (13.4.18)
The corresponding change in entropy of the universe is then as follows:
(13.4.19) (13.4.20) (13.4.21)
Thus no change in ΔSuniv has occurred.
Figure 13.4.5: Expansion of Gas at Constant Temperature. (CC BY-SA-NC; Anonymous by request)
(13.4.22)
Figure 13.4.6: Thermograms Showing That Heat Is Absorbed from the Surroundings When Ice Melts at 0°C
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case,
(13.4.23) (13.4.24) (13.4.25)
The amount of heat lost by the surroundings is the same as the amount gained by the ice, so
(13.4.26) (13.4.27) (13.4.28)
Once again, we see that the entropy of the universe does not change:
(13.4.29)(13.4.30)(13.4.31)
In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
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Entropy Changes in Reversible Processes
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The entropy of the universe increases during a spontaneous process. It also increases during an observable non-spontaneous process.
Figure 13.4.7: Spontaneous Transfer of Heat from a Hot Substance to a Cold Substance
When molten lava flows into cold ocean water, so much heat is spontaneously transferred to the water that steam is produced.
Example: Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
(a) What is ΔS for this process?
(b) Which is the more highly ordered form of tin—white or gray?
Given: ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy: Use Equation 13.4.12 to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Ans: We know from Equation 13.4.12 that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔU are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where qrev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
(b) The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.
Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C) to facilitate tin pest.
During a spontaneous process, the entropy of the universe increases.
ΔS = qrev/T
A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases.
1. What is the concept of reversibility in thermodynamics? | ![]() |
2. How is entropy related to internal energy in thermodynamics? | ![]() |
3. Can you provide an example of an isothermal expansion or compression process? | ![]() |
4. How does the entropy of a system change in a reversible process? | ![]() |
5. What are some applications of entropy changes in reversible processes? | ![]() |