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10.3The second law of thermodynamics
A glass of water at 10 Â°C left in a room at 20 Â°C will warm up until it reaches the temperature of the room. But the reverse never happensâif it did, the cold water would lose thermal energy, making it colder, and the room would become warmer.
What law of physics prevents this reverse process from happening? It is not the first law of thermodynamics; the reverse process is consistent with the first law.
It is forbidden by a special law of physicsâthe second law of thermodynamics. This law involves a new concept, that of entropy. We loosely refer to entropy as a measure of the disorder of a system.
But what is disorder?
âA gas is more disordered than a solid at the same temperature because its atoms move about randomly whereas those of the solid are regularly arrangedâwe know less about the position of the gas atoms.
âBlue paint mixed together in a container with red paint is more disordered than a layer of red paint on top of a layer of blue paint.
âA Rubikâs cube is ordered when each face has the same colour but quickly becomes disordered when faces are turned and colours scramble.
Systems, left on their own, evolve so as to increase disorder. If we plant rows of plants with fixed distances from plant to plant we will have an ordered system. After a few years, seeds produced by the plants will fall to the ground and create new plants. The new plants will be in random positions, with no rows and no fixed distances from plant to plant. The system will evolve into a disordered state. As we will see in the next section, this is because a disordered state is more likely than an ordered state.
Even though we will not define entropy precisely we will define the change in entropy, ÎS, to be:
ÎS=QT
where Q is a quantity of heat given to or removed from the system at a temperature T (in kelvin). The unit of entropy is J Kâ1.
When heat is given to the system, Q > 0 and entropy increases. When heat is removed, Q < 0 and entropy decreases. For a reversible process that returns the system to its original state, ÎS = 0. One such reversible process is the isothermal expansion of a gas and the subsequent isothermal compression back to the initial state. Since the expansion and compression are isothermal, leaving the temperature constant, we may write
ÎS1=+|Q|T
during expansion and
ÎS2=â|Q|T
during compression. (The gas receives thermal energy upon expanding and discards thermal energy upon compressing.) The net entropy change is thus zero.
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EXAM TIP
The formula ÎS=QT must be used with great care. Strictly, it applies when a quantity of heat enters or leaves a system without appreciably changing the temperature. If temperature changes, calculus is required to evaluate ÎS=â«dQT, which is not part of this course.
Let us apply the expression for ÎS given above to the case of the flow of heat between a hot body at temperature Th and a cold body at temperature Tc (Figure 10.16a). If a quantity of heat Q flows from the hot to the cold body (as we know will happen), the total entropy change of the Universe is:
ÎS=âQTh+QTc=Q(1Tcâ1Th)âÎS>0
(the temperature of each body is assumed unchanged during this small exchange of thermal energy). The entropy of the system has increased, and so this is what we expect to happen. The reverse process (Figure 10.16b) does not happen. It corresponds to a net decrease in the entropy of the Universe:
ÎS=QThâQTc=Q(1Thâ1Tc)âÎS<0

Figure 10.16: a When thermal energy flows from a hot to a cold body, the entropy of the universe increases. b If the reverse were to happen without any performance of work, the entropy would decrease.

Similarly, when energy is given to a solid at its melting temperature, the solid will use that energy to turn into a liquid at the same temperature. The entropy formula again shows that the entropy increases as the solid absorbs the latent heat of fusion.
This allows us to state the second law of thermodynamics in its general form involving entropy:
KEY POINT
The entropy of an isolated system never decreases. Entropy increases in realistic irreversible processes and stays the same in theoretical, idealized, reversible processes.
There are other equivalent formulations of the second law:
One, proposed by Rudolf Clausius, states that it is impossible for heat to flow from a cold to a warmer body without the performance of work.
And another, proposed by William Thomson (first Baron Kelvin), states that it impossible for a device working in a cycle to change all the heat extracted from a hot reservoir into mechanical work.
CHECK YOURSELF 5
An ideal gas expands adiabatically, doubling its volume.
During this expansion did the entropy of the gas increase, decrease or stay the same?
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WORKED EXAMPLE 10.8
An ideal gas in state A can reach state B at constant volume and state C at constant pressure, Figure 10.17. The temperatures at B and C are the same. Determine which process results in the greatest entropy change.

Figure 10.17: For worked example 10.8.

Answer
Since the temperatures at B and C are the same the change in internal energy for both paths is the same. The path to B involves zero work done, but the path to C involves positive work done by the gas. Hence Q is greater for the path to C and so is the entropy change.
WORKED EXAMPLE 10.9
A metallic sphere of mass 0.50 kg, specific heat capacity 480 J kgâ1 Kâ1 and temperature 90 Â°C is dropped into a very large quantity of water at 10 Â°C. After some time the temperature of the sphere becomes 10 Â°C. The temperature of the water does not change appreciably because the mass of the water is large. The entropy of the sphere decreases by 60 J Kâ1. Explain how this observation is consistent with the second law of thermodynamics.
Answer
The heat that enters the water is ÎQ = mcÎT = 0.50 Ã 480 Ã 80 = 1.92 Ã 104 J. This happens at a constant temperature of 283 K and so the entropy change of the water is an increase of ÎQT=1.92Ã104283=68Â J Kâ1. The overall entropy of the Universe therefore increased by 8 J Kâ1, consistently with the second law.
A statistical view of entropyâa simple model
Suppose we want to arrange four coins along a line. Each coin can show heads (H) or tails (T) on its top face. One arrangement is that all coins show heads. Another is that two show heads and two show tails and so on.
The possibilities are summarised in Table 10.3.
Number of H
Possible arrangements
Number of arrangements
0
TTTT
1
1
HTTT, THTT, TTHT, TTTH
4
2
HHTT, TTHH, HTHT, THTH, HTTH, THHT
6
3
HHHT, HHTH, HTHH, THHH
4
4
HHHH
1
Total
24 = 16
Table 10.3: The possible arrangements of four coins.
Our system is the set of coins. The system can be described by the number of heads that appear.
Since this number can be anything from 0 to 4 we say that our system is described by five macrostates.
To each macrostate correspond microstates, the number of different arrangements of coins which give the same macrostate.
We denote by the expression Î©(H) the number of microstates that correspond to a macrostate H.
So, Î©(1) = 4, Î©(2) = 6 and so on. What happens when the number of coins increases?
Table 10.4 shows the situation for 10 coins.
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Number of H
Number of arrangements
0
1
1
10
2
45
3
120
4
210
5
252
6
210
7
120
8
45
9
10
10
1
Total
210 = 1024
Table 10.4: The number of possible arrangements of 10 coins.
Here Î©(2) = 45, Î©(5) = 252, and so on.
The mathematicians among you should have recognised that the numbers of microstates are binomial coefficients. For example, Î©(2)=10C2=10!2!Ã8!=45.
Suppose we arrange the 10 coins randomly and ask for the probability of obtaining the state H = 1.
The answer is 101024â0.0098.
The probability for the state H = 5 is 2521024â0.25 and so it is far more likely than the state with H = 1.
What happens if we increase the number of coins even further? Table 10.5 shows the situation with 100 coins.
Number of H
Number of arrangements
0
1
1
100
2
4950
â®
â®
40
1.4 Ã 1028
45
6.1 Ã 1028
50
1.0 Ã 1029
55
6.1 Ã 1028
60
1.4 Ã 1028
â®
â®
98
4950
99
100
100
1
Total
2100 â 1.27 Ã 1030
Table 10.5: The number of possible arrangements of 100 coins.
From Figure 10.18 we see that the probability for the state H = 1 is negligibly small. The most likely outcomes of a random throw of 100 coins would be about equal numbers of H and T. It is possible that the outcome might be H = 1 but it is highly unlikely.
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Figure 10.18: The variation of the number of microstates with the number of heads for 10 and 100 coins.

This situation becomes much more pronounced as we let the number of coins increase and become, say 1023, a number close to Avogadroâs number (the number of particles in a mole).
We define a quantity called the entropy, S, of a given macrostate of the system as:
S=kBlnÎ©
where kB is the Boltzmann constant and Î© is the number of possible microstates of the system.
We see that the states which are likely to occur have large entropy. We can now see the connection between entropy and disorder. The likely states are also the more disordered ones. The system of 10 coins with 5H and 5T is more disordered than the system where all 10 coins show H. There are many ways (252 ways) to get the system with 5H and 5T but only one to get it with 10H. Entropy is a measure of disorder.
The change in entropy in going from a state with H = 45 to the most likely state H = 50 is ÎS = 1.38 Ã 10â23 Ã ln(1.0 Ã 1029) â 1.38 Ã 10â23 Ã ln(6.1 Ã 1028) â 7 Ã 10â24 J Kâ1. There is an increase in entropy. Going the reverse way from H = 50 to H = 45 would imply a decrease in entropy.
Real systems
The ideas developed in the last section also apply to actual physical systems, for example a gas. Here, too, we have macrostates (determined for example by the total energy of the gas), and for each macrostate we have many microstates. (To actually calculate the number of microstates for a real system such as a gas is beyond the level of this course.) If the number of microstates is Î© we again define the entropy of the system to be S = kB ln Î©.
CHECK YOURSELF 6
What is the largest entropy of a system of six coins?
WORKED EXAMPLE 10.10
Refer to Table 10.5.
aWhat is the probability of obtaining the state with H = 1?
bSuppose you throw 100 coins every second. After how much time would you expect to get the outcome of H = 1?
Answer
aThe probability is p=1001.27Ã1030â7.9Ã10â29.
bTo get H = 1 we must throw the 100 coins N times such that Np = 1. Thus, N=1p=1.27Ã1028. Since it takes 1 s to throw the coins, the time is 1.27 Ã 1028 s or about 1020 years. (This is a long time! The age of the universe is about 1017 s.)
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THEORY OF KNOWLEDGE
The entropy of the universe increases, and we are led from ordered to less ordered and more chaotic situations. But the world around us is full of examples of systems that evolve from highly disordered to very ordered forms (and hence forms with less entropy). Life itself evolved from simple microorganisms to the complex forms we see today. A small quantity of water freezing leads to the highly structured, ordered (and beautiful) snowflakes. For this to happen, energy has been exchanged with the surroundings in such a way that the systems evolved into highly ordered, complex forms. However, the decrease in entropy in these systems is accompanied by a larger increase in the entropy of the surroundings, leading to an overall increase in the entropy of the universe.
The nature of time has always mystified scientist and nonscientists alike. The fact that there appears to be a connection between the direction from past to future (the arrow of time) and thermodynamics is a fascinating and not well understood aspect of the second law.
The laws of mechanics do not distinguish the arrow of time: a film of billiard balls colliding with each other does not look strange if run backwards. So why do the same laws when applied to very many particles in a gas, say, show a preference for one direction of time, the one leading to increased entropy and more disorder? The answer has to do with the fact that the system is led to a more disordered, higher entropy state because this state is the vastly more likely state; the result of the very many random collisions among the particles of the gas.
Entropy in thermodynamics may be defined as S = k In Î© where k is Boltzmannâs constant and Î© is the number of ways in which a particular macroscopic state of a system may be realised. It is quite extraordinary that this formula of thermodynamics has featured extensively in Claude Shannonâs theory of information, a crucial part of modern telecommunications theory.
TEST YOUR UNDERSTANDING
25Explain why in an isothermal compression of an ideal gas the entropy of the gas decreases. Why does this decrease in entropy of the gas not violate the second law of thermodynamics?
26In an isothermal expansion of an ideal gas all the thermal energy supplied is used to perform work. Doesnât this violate the second law of thermodynamics?
27In an isothermal expansion of an ideal gas the entropy of the gas increases. How do we understand this increase in entropy in terms of microstates?
28In an adiabatic expansion you would expect the entropy to increase since the volume increases, thus increasing disorder. What accounts for the fact that the entropy change is zero?
29Could you cool down your kitchen by keeping the refrigerator door open?
30Calculate the change in entropy of 25 g of ice at 0 Â°C that melts into water at 0 Â°C. Take the specific latent heat of fusion of ice to be 334 kJ kgâ1.
31Two very large bodies at 0 Â°C and 100 Â°C are brought into thermal contact. Calculate the overall change in entropy when an amount of heat of 50 J leaves the hot body and enters the cold body.
32A 15 g ice cube at 0 Â°C is dropped into a very large container of water at 20 Â°C. Take the specific latent heat of fusion of ice to be 334 kJ kgâ1 and the specific heat capacity of water to be 4200 J kgâ1 Kâ1.
aCalculate the change in entropy of the ice after all the ice has turned into water at 0 Â°C.
bThe 15 g of water at 0 Â°C will eventually reach 20 Â°C. During this process the entropy of the 15 g of water increases by 4.5 J Kâ1. Calculate the overall change in entropy of the ice and the water after equilibrium is reached.
