Entropy

The Carnot limit of the previous lesson looks like a statement about engines, but hidden inside it is a quantity that depends only on the state of a substance, and finding that quantity is the whole business of this lesson.

From Carnot to a state function

Recall the result of the Carnot analysis: a reversible engine working between reservoirs at absolute temperatures $T_{H}$ and $T_{C}$ exchanges heat in the fixed ratio $| Q_{H} | / | Q_{C} | = T_{H} / T_{C}$ . That was derived without reference to any working substance, so it holds for steam, for air, for a rubber band. It is a statement about temperature itself.

Now write it with signs taken carefully. Adopt the convention that $Q$ is positive when heat enters the system, so for an engine $Q_{H} > 0$ (absorbed from the hot reservoir) and $Q_{C} < 0$ (rejected to the cold one). Then $| Q_{C} | = - Q_{C}$ , and the Carnot ratio $Q_{H} / T_{H} = - Q_{C} / T_{C}$ rearranges into something strikingly tidy:

\frac{Q_{H}}{T_{H}} + \frac{Q_{C}}{T_{C}} = 0

The heats themselves do not cancel round the cycle, because the engine produces net work and $Q_{H} + Q_{C} = W > 0$ . It is the heats divided by the temperatures at which they are exchanged that sum to zero. Nothing in the derivation made that cancellation obvious, and it is the clue worth chasing.

Chase it by generalising. Take any reversible cycle at all, traced as a closed loop on a $p$ against $V$ diagram. Cover the interior with a fine mesh of adiabats and isotherms. Each cell of that mesh is bounded by two adiabats and two isotherms, which is exactly a miniature Carnot cycle, so each contributes zero to the sum of $δ Q / T$ . Add up every cell. Every interior boundary is traversed twice in opposite directions and cancels in pairs, so what survives is a zigzag approximation to the original loop. Refine the mesh without limit and the zigzag converges on the loop itself, leaving

\oint \frac{δ Q_{r e v}}{T} = 0

for every reversible cycle whatsoever. A quantity whose integral vanishes round every closed path cannot depend on the path: the integral from state 1 to state 2 is the same along any reversible route, because two different routes joined head to tail form a closed loop that integrates to zero. That is precisely the defining property of a state function, in the way that a conservative force field has a potential. Rudolf Clausius reached this conclusion in the 1850s and in 1865 named the new quantity entropy, from the Greek for transformation, deliberately shaping the word to resemble energy because he thought the two belonged together.

The definition and how to use it

The new state function is defined by

d S = \frac{δ Q_{r e v}}{T}

with $T$ the absolute temperature of the system at the moment the heat $δ Q_{r e v}$ crosses its boundary. The units follow immediately: joules divided by kelvin, J/K. Substances are usually tabulated as specific entropy in J/(kg K) or molar entropy in J/(mol K), which is why the property tables of the previous lesson list $s$ alongside $u$ , $h$ and $v$ .

The subscript on $δ Q_{r e v}$ carries an enormous amount of weight, and misreading it is the commonest error in the subject. It does not mean that entropy is defined only for reversible processes. It means that the recipe for computing an entropy change uses reversible heat. Since $S$ is a state function, $Δ S = S_{2} - S_{1}$ depends on the two end states and on nothing else: not on how fast the change happened, not on how much friction was involved, not on whether the process was violent enough that the system had no definable temperature partway through.

The practical consequence is a licence. To find $Δ S$ for an irreversible process, discard the actual process entirely, invent any convenient reversible path linking the same two states, and integrate $δ Q_{r e v} / T$ along that. The invented path need not resemble the real one and usually does not. A gas bursting into a vacuum does no work and exchanges no heat, so the real $\int δ Q / T$ is zero; the entropy change is not, because the correct calculation replaces the burst with a slow isothermal expansion between the same two states, which does absorb heat.

The Clausius inequality

What if the cycle is not reversible? Consider a system taken round any cycle, exchanging heat with its surroundings at various temperatures. Insert a thought device: instead of letting each parcel of heat $δ Q$ come directly from the surroundings, supply it through a small reversible Carnot engine drawing from a single reservoir held at $T_{0}$ . That engine delivers $δ Q$ to the system at the system's boundary temperature $T$ , and to do so it must draw $δ Q_{0} = T_{0} δ Q / T$ from the reservoir.

Now look at the composite of system plus auxiliary engines over one complete cycle. The system returns to its initial state and the engines are cyclic, so no internal energy has changed anywhere. The only net effects are that heat $Q_{0} = T_{0} \oint δ Q / T$ has left the single reservoir and some net work has been produced. That is a device which, working in a cycle, takes heat from one reservoir and delivers work with no other effect, and the Kelvin-Planck statement of the Second Law forbids it. The only escape is that the work is not positive, which requires

\oint \frac{δ Q}{T} \leq 0

This is the Clausius inequality. Equality holds when every step is reversible, recovering the earlier result, and the inequality is strict whenever any irreversibility is present. Note that $T$ here is the temperature at the boundary where the heat crosses, which for an irreversible process may differ sharply from the temperature deep inside the system.

The entropy balance

Split an irreversible cycle into an irreversible path from state 1 to state 2 and a reversible return from 2 to 1. The Clausius inequality applied to the whole loop gives $\int_{1}^{2} δ Q / T + \int_{2}^{1} δ Q_{r e v} / T \leq 0$ , and the second integral is $S_{1} - S_{2}$ by definition. Rearranging,

Δ S \geq \int \frac{δ Q}{T}

which becomes an equation once the deficit is given a name:

Δ S = \int \frac{δ Q}{T} + S_{g e n}, S_{g e n} \geq 0

Read this as an accounting statement and its meaning is clear. Entropy enters or leaves a system with heat, one joule per kelvin for every joule crossing at temperature $T$ : that is the transfer term. On top of that, entropy is manufactured inside the system by every irreversibility present, by friction, by unrestrained expansion, by mixing, by heat flowing across a finite temperature difference: that is $S_{g e n}$ , the generation. Energy has no such term. Entropy is not conserved, and the Second Law is exactly the assertion that $S_{g e n}$ can be zero or positive but never negative.

For an isolated system the transfer term vanishes because no heat crosses the boundary, leaving $Δ S \geq 0$ . That is the famous statement, and it is worth being blunt about how narrow it is. "Entropy always increases" is true for an isolated system and false as a general claim about anything else. A refrigerator lowers the entropy of its contents every day and breaks no law, because the contents are not isolated: the machine dumps more entropy into the kitchen than it removes from the food. The same is true of a freezing pond and a growing tree.

Example. A pipe leaks 60 kJ of heat from a reservoir at 600 K straight into a reservoir at 300 K. Each reservoir is large enough that its temperature does not move. What is the entropy change of each reservoir, and of the two together?

Each reservoir exchanges heat at a single temperature, so the transfer term is just $Q / T$ and nothing is generated inside a reservoir. The hot one loses heat: $Δ S_{h o t} = - 60000 / 600 = - 100$ J/K. The cold one gains it: $Δ S_{c o l d} = + 60000 / 300 = + 200$ J/K. Together they form an isolated pair, and the total is $- 100 + 200 = + 100$ J/K. Positive, as the Second Law demands: the 60 kJ was divided by the larger temperature on the way out and the smaller one on the way in.

Now you. A different pipe leaks 45 kJ from a reservoir at 450 K into a reservoir at 300 K. Find the entropy change of each reservoir and show that the total is positive.

Answer

$Δ S_{h o t} = - 45000 / 450 = - 100$ J/K and $Δ S_{c o l d} = + 45000 / 300 = + 150$ J/K, so the total is $- 100 + 150 = + 50$ J/K, positive as required.

Computing entropy changes

For an ideal gas the recipe follows in three lines. The First Law for a reversible step is $d U = δ Q_{r e v} - p d V$ , and for an ideal gas $d U = n C_{V} d T$ , so $δ Q_{r e v} = n C_{V} d T + p d V$ . Divide by $T$ and use $p / T = n R / V$ :

d S = n C_{V} \frac{d T}{T} + n R \frac{d V}{V}

Integrating with constant $C_{V}$ gives $Δ S = n C_{V} \ln (T_{2} / T_{1}) + n R \ln (V_{2} / V_{1})$ . Both terms vanish when the state does not change, as they must. For a phase change at constant temperature the integral is trivial, since $T$ comes out of it: $Δ S = Q / T = m L / T$ , with $L$ the latent heat. Melting and boiling therefore raise entropy sharply, and condensation lowers it.

Example. 2.0 mol of a monatomic ideal gas, $C_{V} = \frac{3}{2} R = 12.471$ J/(mol K), is taken from 300 K to 450 K while its volume doubles. What is $Δ S$ ?

Both terms contribute:

Δ S = 2.0 \times 12.471 \ln \frac{450}{300} + 2.0 \times 8.314 \ln 2 = 10.11 + 11.53 = + 21.64 J/K

The heating and the expansion each raise the entropy, and because $S$ is a state function this answer holds however the gas actually got from one state to the other.

Now you. 1.5 mol of the same gas goes from 250 K to 500 K while its volume triples. Find $Δ S$ .

Answer

Δ S = 1.5 \times 12.471 \ln \frac{500}{250} + 1.5 \times 8.314 \ln 3 = 12.97 + 13.70 = + 26.67 J/K

Combining the same First Law step with $δ Q_{r e v} = T d S$ gives the fundamental relation

d U = T d S - p d V

It is derived from a reversible process, yet it holds for any process at all, because every symbol in it is a property. $U$ , $T$ , $S$ , $p$ and $V$ are all state functions, so a relation among their differentials is a relation among the states, not among the paths. The reversible process was only the scaffolding used to find it. For an irreversible change $T d S$ is no longer the heat and $p d V$ is no longer the work, but their difference is still $d U$ .

A block of ice melting in a warm room

Take one kilogram of ice at $0$ °C in a room held at $25$ °C, and let it melt completely. The latent heat of fusion of water is $L = 334$ kJ/kg, so the ice absorbs $Q = 334$ kJ. It stays at $273.15$ K throughout the melt, so this is a constant-temperature heat addition and

Δ S_{i c e} = \frac{334000}{273.15} = + 1222.8 J/K

The room supplies that heat and is large enough that its temperature does not move, so it undergoes a constant-temperature heat removal at $298.15$ K:

Δ S_{r o o m} = \frac{- 334000}{298.15} = - 1120.2 J/K

The room's entropy falls, which is permitted, because the room is not isolated. Take the two together and the pair is isolated, so the total is the number that must obey the Second Law:

Δ S_{t o t a l} = 1222.8 - 1120.2 = + 102.6 J/K

Positive, as required. The origin of the surplus is visible in the arithmetic: the same 334 kJ was divided by the smaller number on the way in and the larger number on the way out. Heat crossing a finite temperature drop always generates entropy, and here $S_{g e n} = 102.6$ J/K appears in the thin layer of air and water where the drop actually occurs. Had the room been at $0.001$ °C the melt would have taken forever and the surplus would have been negligible, which is the reversible limit.

Example. 0.75 kg of water boils away completely at 100 °C, where the latent heat of vaporisation is $L = 2257$ kJ/kg. What is the entropy change of the water?

Boiling is a constant-temperature heat addition at $373.15$ K, so the same recipe applies:

Δ S = \frac{m L}{T} = \frac{0.75 \times 2257000}{373.15} = + 4536.4 J/K

Now you. 0.20 kg of ice melts completely at 0 °C, with $L = 334$ kJ/kg. Find the entropy change of the ice.

Answer

Δ S = \frac{0.20 \times 334000}{273.15} = + 244.6 J/K

Two blocks of metal in contact

Now a case with no phase change. Take two identical one-kilogram blocks of copper, specific heat $c = 385$ J/(kg K), one at $300$ K and one at $400$ K, and clamp them together inside perfect insulation.

Energy is conserved and no work is done, so the heat lost by the hot block equals the heat gained by the cold one: $m c (400 - T_{f}) = m c (T_{f} - 300)$ . The masses and specific heats cancel because the blocks are identical, giving $T_{f} = (300 + 400) / 2 = 350$ K, the plain arithmetic mean.

Each block is heated or cooled at constant volume, so $Δ S = \int m c d T / T = m c \ln (T_{f} / T_{i})$ , and the reversible path is easy to imagine: bring each block through a sequence of reservoirs differing infinitesimally in temperature. The total is

Δ S = m c [\ln \frac{350}{300} + \ln \frac{350}{400}] = 385 (0.15415 - 0.13353) = + 7.94 J/K

The cold block gains more entropy than the hot one loses, for a structural reason. As a single logarithm the sum is $Δ S = m c \ln (T_{f}^{2} / (T_{1} T_{2}))$ , where $T_{f}$ is the arithmetic mean of $T_{1}$ and $T_{2}$ and $\sqrt{T_{1} T_{2}}$ is their geometric mean. The arithmetic mean of two unequal positive numbers always exceeds the geometric mean, so the argument of the logarithm always exceeds one and $Δ S > 0$ for any starting difference, vanishing only when the blocks start equal. This is heat flowing downhill, costed: the process runs one way and not the other, and entropy is the ledger that says which.

Isentropic processes and isentropic efficiency

A process that is both adiabatic and reversible has $δ Q = 0$ and $S_{g e n} = 0$ , so $Δ S = 0$ and the entropy is constant. Such a process is called isentropic, and it is the reference case for every device meant to exchange work rapidly and not heat: turbines, compressors, pumps and nozzles all run fast enough that heat loss through the casing is small compared with the work crossing the shaft.

The real device is not reversible. Friction in the bearings and turbulence in the blade passages generate entropy, so a real adiabatic turbine leaves the fluid at higher entropy than it entered with, and at a higher exit enthalpy than the isentropic case for the same exit pressure. Since the work delivered is the enthalpy drop, the real turbine delivers less. The isentropic efficiency compares the two directly: the ratio of actual work to ideal work for a turbine, and the ratio of ideal to actual for a compressor, where irreversibility makes the real machine demand more input than the ideal one.

Typical values are worth knowing: a large steam turbine reaches around $0.90$ , an axial compressor around $0.85$ , a small pump considerably less. That single number lets an engineer take an isentropic result, computable from property tables alone, and turn it into a prediction about a machine that exists. It is the workhorse calculation of the field, and the Applied Thermodynamics course that follows this one develops it in full for cycles, plant and refrigeration.

What has and has not been achieved

Entropy is now on the same footing as internal energy and enthalpy. It has a definition, $d S = δ Q_{r e v} / T$ ; it has units; it is tabulated for real substances; it can be computed for gases, for phase changes and for irreversible processes by substituting a reversible path. It supplies a criterion for which of two directions a process will run in, and a number, $S_{g e n}$ , for how much capacity to do work was thrown away in the running.

That is a great deal, and it is also entirely operational. Every statement above defines entropy by what it does, not by what it is. Nothing here explains why the quantity should exist at all, why it takes the particular form $δ Q / T$ , or why nature should care about it. Energy at least has an intuitive story behind it. Entropy so far has only a rule.

The gap closes when the question is asked at the scale of the molecules rather than the vessel. Counting the microscopic arrangements available to a system produces a quantity with exactly the properties derived here, along with an explanation of why the Second Law has the direction it does and what happens as temperature approaches absolute zero. That counting is the subject of the next lesson.