Entropy and the Second Law of Thermodynamics
Why time runs one way, why a shattered egg never reassembles, why no engine can be perfect, and why air never spontaneously crowds into one corner of the room — all written in a single quantity that only ever grows
- Why irreversible processes run one way only, and that their direction is set by the entropy change, not by energy.
- The definition of entropy change, \(\Delta S = \int_{i}^{f} dQ/T\), that entropy is a state function, and for an ideal gas \(\Delta S = nR\ln(V_{f}/V_{i}) + nC_{V}\ln(T_{f}/T_{i})\).
- The second law in a closed system: \(\Delta S \ge 0\) (= for reversible, > for irreversible).
- How a Carnot engine works and its efficiency ceiling \(\varepsilon_{C} = 1 - T_{L}/T_{H}\), with \(\varepsilon = W/|Q_{H}|\) for any engine.
- Why no perfect engine or perfect refrigerator can exist, and the coefficient of performance \(K_{C} = T_{L}/(T_{H}-T_{L})\).
- The statistical meaning of entropy through microstates and multiplicity, \(W = N!/(n_{1}!\,n_{2}!)\), and Boltzmann's \(S = k\ln W\).
What Is Physics?
Time has a direction — the direction in which we age. We live among one-way processes: an egg dropped on the floor, a pizza baked, a car driven into a lamppost, a beach eroded by waves. These are irreversible — they cannot be undone by some small tweak of their surroundings. One goal of physics is to understand why time has a direction and why such processes never run backward.
This question may seem far from practical life, yet it sits at the very heart of every engine, because it sets the ultimate limit on how well an engine can run. The key to it all is a single quantity: entropy.
Irreversible Processes and Entropy
The one-way character of nature is so familiar we rarely notice it — but it is deeply strange. If you wrapped your hands around a hot mug and your hands grew colder while the coffee grew hotter, you would be astonished. Yet energy would be perfectly conserved either way. Pop a balloon and the helium spreads through the room; it never spontaneously regathers into the balloon's shape — though energy conservation would permit it.
So energy conservation does not pick the direction of an irreversible process. Something else does — the change in entropy \(S\).
Unlike energy, entropy is not conserved. For irreversible processes it grows, which is why the change in entropy is called the arrow of time. An exploding popcorn kernel marks the forward direction of time; the reverse — fragments reassembling into the kernel — would lower entropy, and so it never happens.
Change in Entropy
We define the entropy change of a system going from state \(i\) to state \(f\) in terms of the heat exchanged and the temperature at which it flows:
There is a catch. During an irreversible process like a free expansion, the gas has no well-defined pressure or temperature midway through — there is no path to integrate along. The way out rests on entropy being a state function: the change \(\Delta S\) depends only on the endpoints, not the route. So we replace the messy irreversible process with any convenient reversible process linking the same two states, and compute \(\Delta S\) for that.
For an ideal gas taken reversibly from \((V_{i},T_{i})\) to \((V_{f},T_{f})\), integrating the first law gives a result that depends only on the endpoints:
The Second Law of Thermodynamics
A puzzle: reverse a reversible isothermal expansion and the gas alone loses entropy (\(Q < 0\)). Does that break the entropy postulate? No — that postulate applies only to closed systems undergoing irreversible processes. Include the reservoir, and the reservoir gains exactly the entropy the gas loses, so for the enlarged closed system \(\Delta S = 0\). This widens the postulate into the full law:
The equals sign applies to reversible processes, the greater-than sign to irreversible ones. Entropy may fall in part of a closed system, but there is always an equal or larger increase elsewhere, so the total never decreases. Real processes — with friction and turbulence — are always irreversible to some degree, so the entropy of a real closed system always rises. Constant-entropy processes are idealizations.
Entropy in the Real World: Engines
A heat engine extracts energy as heat and does useful work. At its core is a working substance (water in a steam engine, a fuel-air mix in a car) that must operate in a cycle, returning to each state again and again. Just as the ideal gas illuminates real gases, an ideal engine illuminates real ones. The benchmark is the Carnot engine — the most efficient engine possible, conceived by Sadi Carnot in 1824, before either the first law or entropy was known.
Each cycle, the Carnot engine absorbs \(|Q_{H}|\) from a hot reservoir at \(T_{H}\), dumps \(|Q_{L}|\) to a cold reservoir at \(T_{L}\), and delivers work \(W\). Its cycle is two isothermal steps (the only steps with heat transfer) joined by two adiabatic steps. On a p–V diagram the enclosed area is the net work; on a temperature–entropy diagram the cycle is a clean rectangle.
The Efficiency of a Carnot Engine
An engine's thermal efficiency is "what we get" (work) over "what we pay for" (heat absorbed). Combining that with the cycle relations above gives the famous Carnot limit:
The temperatures are in kelvins. Because \(T_{L} > 0\) and \(T_{H}\) is finite, \(\varepsilon_{C} < 1\) always — no engine can be 100% efficient. A perfect engine (\(|Q_{L}| = 0\), all heat turned to work) would need \(T_{L} = 0\) or \(T_{H} \to \infty\), both impossible.
In short: there are no perfect engines. A real car engine that could reach the Carnot limit between its temperatures might hit ~55%; in practice it manages ~25%. A nuclear plant's Carnot ceiling might be ~40%, with ~30% achieved. No design trick beats the limit set by \(\varepsilon_{C}\).
Real Engines and Refrigerators
One can prove no real engine beats a Carnot engine between the same temperatures. Suppose an engine X claimed \(\varepsilon_{X} > \varepsilon_{C}\). Couple it to drive a Carnot refrigerator; the combination would transfer heat from cold to hot with no net work — a perfect refrigerator, which the second law forbids. The only flawed assumption is the claim itself, so \(\varepsilon_{X} \le \varepsilon_{C}\).
A refrigerator runs the cycle in reverse: work \(|W|\) is done to pump heat \(|Q_{L}|\) from cold to hot. Its figure of merit is the coefficient of performance — "what we want" (heat removed) over "what we pay for" (work):
A perfect refrigerator running with no work input would give the cold reservoir entropy \(-|Q|/T_{L}\) and the hot one \(+|Q|/T_{H}\), for a net \(\Delta S < 0\) — a violation. In short: there are no perfect refrigerators. If you want yours to run, you must plug it in.
A Statistical View of Entropy
There is a second, deeper definition of entropy — by counting. Imagine \(N\) indistinguishable molecules split between the two halves of a box, \(n_{1}\) on the left and \(n_{2}\) on the right. A configuration is a split like (4, 2); a microstate is one specific arrangement of the actual molecules. The number of microstates in a configuration is its multiplicity:
Entropy is the logarithm of the number of microstates, scaled by \(k = 1.38\times10^{-23}\,\mathrm{J/K}\). The logarithm is natural: entropies of two systems add while their probabilities multiply, and \(\ln ab = \ln a + \ln b\). This formula — engraved on Boltzmann's tombstone — and the thermodynamic \(\int dQ/T\) give the same entropy change. For huge factorials, use Stirling's approximation \(\ln N! \approx N\ln N - N\).
Putting It to Work
Problem. Two identical copper blocks (\(m = 1.5\,\mathrm{kg}\), \(c = 386\,\mathrm{J/kg\cdot K}\)), one at 60 °C and one at 20 °C, are placed in contact in an insulated box and reach 40 °C. Find the net entropy change.
Solution. Replace the irreversible mixing with two reversible steps, each \(\Delta S = mc\ln(T_{f}/T_{i})\), in kelvins (333 K, 313 K, 293 K).
The hot block loses less entropy than the cold block gains, because the same heat is more "valuable" (lower \(T\)) where it arrives. The net is positive — exactly as the second law demands for an irreversible process.
Problem. One mole of nitrogen (ideal) is confined to the left half of a container; the stopcock opens and the volume doubles freely. Find \(\Delta S\).
Solution. A free expansion is irreversible, so substitute a reversible isothermal expansion (same \(T_{i} = T_{f}\)): \(\Delta S = nR\ln(V_{f}/V_{i})\).
Positive, as required. Section 8 derives the very same \(nR\ln 2\) by counting microstates with \(S = k\ln W\) — two utterly different routes to one answer.
Problem. A Carnot engine runs between 850 K and 300 K, doing 1200 J of work per 0.25 s cycle. Find (a) efficiency, (b) power, (c) \(|Q_{H}|\), (d) \(|Q_{L}|\), and (e) the entropy changes of the working substance.
Solution. Use \(\varepsilon_{C} = 1 - T_{L}/T_{H}\), then \(P = W/t\), \(|Q_{H}| = W/\varepsilon\), \(|Q_{L}| = |Q_{H}| - W\), and \(\Delta S = Q/T\).
The two entropy changes cancel — the net per cycle is zero, exactly as a state function over a closed cycle must be.
Problem. An inventor claims an engine that is 75% efficient operating between the boiling and freezing points of water. Possible?
Solution. Compare with the Carnot ceiling between \(T_{H} = 373\,\mathrm{K}\) and \(T_{L} = 273\,\mathrm{K}\).
The best possible efficiency between those temperatures is about 27%. A real engine must do worse still, so a claimed 75% is impossible — the inventor is mistaken.
Problem. For 100 indistinguishable molecules in a box, compare the multiplicities of the (50, 50) and (100, 0) configurations.
Solution. Apply \(W = N!/(n_{1}!\,n_{2}!)\) to each.
The even split is more likely than the all-on-one-side split by a factor of about \(10^{29}\). And 100 is a tiny sample — for a mole the lopsided arrangements are effectively impossible, which is why gases sit at equilibrium.
Problem. Show that doubling the volume of \(n\) moles of ideal gas in a free expansion gives \(\Delta S = nR\ln 2\) — using statistical mechanics.
Solution. Initially all \(N\) molecules are on one side (\(W_{i} = 1\)); finally they spread over both (\(W_{f} = N!/[(N/2)!]^{2}\)). Apply \(S = k\ln W\) with Stirling's approximation.
Identical to the thermodynamic result of Example 2 (using \(Nk = nR\)). The microscopic counting and the macroscopic \(\int dQ/T\) are two faces of one quantity.
Chapter Summary
One-way processes set time's arrow. Their direction is fixed by entropy, not energy.
\(\Delta S = \int_{i}^{f} dQ/T\); isothermal \(\Delta S = Q/T\). A state function.
\(\Delta S = nR\ln(V_{f}/V_{i}) + nC_{V}\ln(T_{f}/T_{i})\) — depends only on endpoints.
\(\Delta S \ge 0\) for a closed system (= reversible, > irreversible).
\(\varepsilon = W/|Q_{H}|\); Carnot \(\varepsilon_{C} = 1 - T_{L}/T_{H}\). No perfect engine.
\(K = |Q_{L}|/|W|\); Carnot \(K_{C} = T_{L}/(T_{H}-T_{L})\). No perfect fridge.
Two isotherms + two adiabats; \(W = |Q_{H}| - |Q_{L}|\), \(|Q_{H}|/T_{H} = |Q_{L}|/T_{L}\).
\(W = N!/(n_{1}!\,n_{2}!)\); all microstates equally probable.
\(S = k\ln W\); \(\ln N! \approx N\ln N - N\) (Stirling).
Problems
Take \(R = 8.31\,\mathrm{J/mol\cdot K}\), \(k = 1.38\times10^{-23}\,\mathrm{J/K}\), and water's specific heat \(4187\,\mathrm{J/kg\cdot K}\) (copper 386, lead 128). Always work in kelvins. For irreversible processes, build a reversible path between the same endpoints; for engines and fridges, lean on \(\varepsilon_{C} = 1 - T_{L}/T_{H}\) and the first law over a cycle.
- Suppose 4.00 mol of an ideal gas expands reversibly and isothermally from \(V_{1}\) to \(2.00V_{1}\) at 400 K. Find (a) the work done and (b) the entropy change. (c) If the expansion were reversible and adiabatic instead, what is \(\Delta S\)?
- A 2.50 mol sample of ideal gas expands reversibly and isothermally at 360 K until its volume doubles. What is the entropy increase?
- Find (a) the heat absorbed and (b) the entropy change of a 2.00 kg copper block whose temperature rises reversibly from 25.0 °C to 100 °C.
- What is the entropy change of a 12.0 g ice cube that melts completely in water just above the freezing point? (Heat of fusion 333 kJ/kg.)
- A 50.0 g copper block at 400 K is placed with a 100 g lead block at 200 K in an insulated box. Find (a) the equilibrium temperature and (b) the entropy change of the two-block system.
- In an insulated container, 200 g of aluminum (\(c = 900\,\mathrm{J/kg\cdot K}\)) at 100 °C is mixed with 50.0 g of water at 20.0 °C. Find (a) the equilibrium temperature and (b) the net entropy change.
- A Carnot engine absorbs 52 kJ and exhausts 36 kJ as heat each cycle. Find (a) its efficiency and (b) the work done per cycle.
- A Carnot engine has an efficiency of 22.0% and operates between reservoirs differing by 75.0 C°. Find the (a) lower and (b) higher reservoir temperatures.
- A Carnot engine whose cold reservoir is at 17 °C has an efficiency of 40%. By how much must the hot-reservoir temperature be raised to push the efficiency to 50%?
- How much work must a Carnot refrigerator do to move 1.0 J as heat (a) from 7.0 °C to 27 °C, and (b) from −73 °C to 27 °C?
- A heat pump (a Carnot engine in reverse) keeps a building at 22 °C while it is −25.0 °C outside, delivering 7.54 MJ per hour. At what rate must work be done to run it?
- A Carnot air conditioner moves heat from a 70 °F room to 96 °F outdoors. For each joule of electrical work, how many joules are removed from the room?
- Construct a table like the six-molecule table for a box of eight indistinguishable molecules: list the configurations, their multiplicities, and the total number of microstates.
- A box holds \(N = 50\) molecules. Find (a) the multiplicity of the central (25, 25) configuration, (b) the total number of microstates, and (c) the percentage of time the system spends in the central configuration.