Let one drop of ink fall into a still glass of water. Slowly it spreads, in curling threads, until the whole glass is a pale blue. Wait as long as you like: the colour never gathers itself back into a drop. Nothing is lost and no energy is created, yet the change runs one way only.
Heat flowing across a gap
Let heat \(Q\) flow from a hot reservoir at \(T_H\) to a cold reservoir at \(T_C\), as it does by itself. The hot reservoir loses entropy \(Q/T_H\). The cold one gains \(Q/T_C\). Since \(T_C\) is smaller, the gain is larger than the loss. The total change is \(Q(1/T_C - 1/T_H)\), which is positive. This lesson uses \(R = 8.31\,\text{J/(mol K)}\) and \(c_\text{water} = 4200\,\text{J/(kg K)}\).
Take \(1000\,\text{J}\) flowing from \(400\,\text{K}\) to \(300\,\text{K}\). The hot body changes by \(-1000/400 = -2.50\,\text{J/K}\), and the cold body by \(+1000/300 = +3.33\,\text{J/K}\). The total is \(+0.83\,\text{J/K}\). If heat flowed the other way, from cold to hot, the total would fall by the same amount. That never happens.
A sealed tin of \(1\,\text{kg}\) of water at \(80\,°\text{C}\) is pressed against a sealed tin of \(1\,\text{kg}\) of water at \(20\,°\text{C}\). Heat flows until both are at \(50\,°\text{C}\). What happens to the total entropy of the two?
In the figure above the heat that passes is \(4200 \times 30 = 126\,\text{kJ}\). The hot water cools from \(353\,\text{K}\) to \(323\,\text{K}\), so its entropy changes by \(4200\ln(323/353) = -373\,\text{J/K}\). The cold water warms from \(293\,\text{K}\) to \(323\,\text{K}\) and gains \(4200\ln(323/293) = +409\,\text{J/K}\). The total, \(+36.4\,\text{J/K}\), grows all the time and stops growing only when the flow stops.
Free expansion and mixing
In a free expansion a gas rushes into a vacuum. No heat flows, so it is tempting to write \(\Delta S = 0\). But the process is irreversible, and \(Q/T\) along it is not the entropy change. Entropy is a state function, so use a reversible path between the same states. An ideal gas keeps its temperature in a free expansion, so the isotherm will do.
Along that isotherm \(\Delta S = nR\ln(V_2/V_1)\). One mole rushing into a vacuum of equal size gains \(8.31\ln 2 = 5.76\,\text{J/K}\), though no heat moved at all. The surroundings are untouched, so the universe gains \(5.76\,\text{J/K}\). Two different gases mixing do the same: each spreads into the whole vessel, and each gains \(nR\ln 2\) if the halves were equal.
What if both halves hold the same gas at the same temperature and pressure? Removing the partition then changes nothing that can be measured, so the entropy does not change. Only the spreading of one kind of molecule among another raises it. The drop of ink spreading through the water is this kind of change.
The entropy of an isolated system never decreases. It stays constant in a reversible process and grows in an irreversible one: \[\Delta S_\text{universe} = \Delta S_\text{system} + \Delta S_\text{surroundings} \ge 0\]
The entropy of one part can fall. Two kilograms of water cooling from \(373\,\text{K}\) to \(273\,\text{K}\) lose \(2622\,\text{J/K}\), as in Lesson 5. But the heat they give out goes to cooler surroundings, and there it carries more entropy. The total for water and surroundings still grows. The law is about the whole, never about one part.
Ludwig Boltzmann found what entropy counts. A state of a gas can be built from its molecules in \(W\) different ways, and \(S = k\ln W\), where \(k = 1.38 \times 10^{-23}\,\text{J/K}\). Let \(N\) molecules double their volume: each has twice the room, so \(W\) grows by \(2^N\) and \(\Delta S = Nk\ln 2\). For one mole, \(N_A k\ln 2 = 5.76\,\text{J/K}\), the free expansion again. A spread-out state has far more ways, so it is what we see.
\(1000\,\text{J}\) of heat flows from a reservoir at \(500\,\text{K}\) to a reservoir at \(250\,\text{K}\). What is the change in the entropy of the universe, in J/K?
One mole of ideal gas expands freely into a vacuum until its volume is doubled. No heat flows. What is its change in entropy?
A hot body at \(400\,\text{K}\) gives out \(600\,\text{J}\) of heat to its surroundings at \(300\,\text{K}\). Both are large enough to keep their temperatures. What is the total change in entropy, in J/K?
The Clausius inequality
Follow a real engine round one cycle and add up \(dQ/T\) for the heat it takes in and gives out. Here \(T\) is the temperature of the reservoir at each step. For a Carnot engine the sum is zero: \(Q_H/T_H - Q_C/T_C = 0\). A real engine rejects more heat than a Carnot engine would. So the sum comes out negative.
Take an engine that takes \(1000\,\text{J}\) at \(500\,\text{K}\) and gives out \(700\,\text{J}\) at \(300\,\text{K}\). The sum is \(1000/500 - 700/300\), which is \(2 - 2.33 = -0.33\,\text{J/K}\). The engine itself returns to its state, so its own entropy is unchanged. The two reservoirs gain \(-2 + 2.33 = +0.33\,\text{J/K}\). A Carnot engine would give out only \(600\,\text{J}\), with a sum of zero.
The same bookkeeping explains the Clausius statement. A refrigerator that moves \(Q_C\) from \(T_C\) and gives \(Q_H\) to \(T_H\) changes the universe by \(Q_H/T_H - Q_C/T_C\), which cannot be negative. So \(Q_H/T_H \ge Q_C/T_C\). With no work, \(Q_H = Q_C\), and this would need \(T_H \le T_C\), which is false. So the work cannot be zero.
The Clausius inequality is beyond the Class 12 syllabus, but it is the second law in its most general form. From it, for any process from a state 1 to a state 2, \(\Delta S \ge \int dQ/T\), with equality only when the process is reversible. For an isolated system \(dQ = 0\) at every step, so \(\Delta S \ge 0\): its entropy can never fall.
An irreversible change wastes a chance to do work. Heat of \(1000\,\text{J}\) flowing straight from \(400\,\text{K}\) to \(300\,\text{K}\) does no work and creates \(0.83\,\text{J/K}\). A Carnot engine between the same bodies could have delivered \(1000 \times (1 - 300/400) = 250\,\text{J}\). That lost work equals \(T_C\) times the entropy created: \(300 \times 0.833 = 250\,\text{J}\).
An engine takes \(1000\,\text{J}\) from a reservoir at \(500\,\text{K}\) and gives \(500\,\text{J}\) to a reservoir at \(300\,\text{K}\) in each cycle. What does the Clausius inequality say about it?
Temperature against entropy
A p-V diagram shows work as an area, because \(W = \int p\,dV\). The definition of entropy gives heat the same treatment. For a reversible step \(dQ = T\,dS\), so the heat taken in along a path is \(Q = \int T\,dS\). This is the area under the path on a graph of \(T\) against \(S\). This is a T-S diagram.
Two processes become straight lines on it. An isotherm has a fixed \(T\), so it is horizontal. A reversible adiabat has a fixed \(S\), so it is vertical. A path moving to the right takes in heat, and a path moving to the left gives it out. Round a cycle, the area enclosed is the net heat, and so the net work.
In Fig 6.1 the curved loop on the left becomes a rectangle on the right. Leg A to B, heat in at \(500\,\text{K}\), is the top edge, running right by \(5.76\,\text{J/K}\). Leg B to C, adiabatic, is the right edge, falling to \(300\,\text{K}\). Leg C to D, heat out at \(300\,\text{K}\), is the bottom edge, running left. Leg D to A is the left edge. The rectangle's area is \(200 \times 5.76 = 1152\,\text{J}\), the net work.
A cycle on a T-S diagram runs along a straight line from X (\(300\,\text{K}\), \(1\,\text{J/K}\)) to Y (\(500\,\text{K}\), \(3\,\text{J/K}\)). Then it runs straight down to Z (\(300\,\text{K}\), \(3\,\text{J/K}\)), and back along \(T = 300\,\text{K}\) to X. Find the heat taken in, the heat given out, the net work and the efficiency.
X to Y moves right, so heat is taken in. The area under it is a trapezium: \(\tfrac12(300 + 500) \times 2 = 800\,\text{J}\).
Y to Z is vertical, a reversible adiabat, with no heat. Z to X moves left at \(300\,\text{K}\), giving out \(300 \times 2 = 600\,\text{J}\).
Net work: \(800 - 600 = 200\,\text{J}\), the area of the triangle. Efficiency: \(200/800 = 0.25\), less than the Carnot \(1 - 300/500 = 0.40\), because part of its heat is taken in below \(500\,\text{K}\).
On a T-S diagram an ideal gas goes along a horizontal line at \(400\,\text{K}\) from \(S = 1\,\text{J/K}\) to \(S = 3\,\text{J/K}\). How much work does the gas do, in joules?
On the T-S diagram of a Carnot engine, start at the top-left corner and go round clockwise. Put the four legs in the order they are met.
The Carnot efficiency in two lines
Lesson 3 needed seven steps and an ideal gas to find the Carnot efficiency. Entropy does it in two lines, for any working substance. Over one cycle the entropy of the working substance returns to its first value. It gains \(Q_H/T_H\) on the hot isotherm, loses \(Q_C/T_C\) on the cold one, and changes nothing on the adiabats.
\[ \dfrac{Q_H}{T_H} - \dfrac{Q_C}{T_C} = 0 \]
Why. entropy is a state function, so round a cycle \(\Delta S = 0\); only the two isotherms change it.
\[ \eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H} \]
Read it. step 1 gives \(Q_C/Q_H = T_C/T_H\); put it into the definition of efficiency.
The rectangle of Fig 6.1 says the same thing as a picture. Its height is \(T_H - T_C\) and its width is \(\Delta S\), so \(W = (T_H - T_C)\Delta S\). The heat taken in is the whole area under the top edge, \(Q_H = T_H\,\Delta S\). Dividing, \(\eta = (T_H - T_C)/T_H = 1 - T_C/T_H\). With \(500\,\text{K}\), \(300\,\text{K}\) and \(\Delta S = 5.76\,\text{J/K}\): \(1152/2880 = 0.40\).
The entropy law also proves Carnot's theorem again. Any engine returns to its state, so the universe changes by \(Q_C/T_C - Q_H/T_H\), which cannot be negative. So \(Q_C/Q_H \ge T_C/T_H\), and \(\eta = 1 - Q_C/Q_H \le 1 - T_C/T_H\). The equal sign belongs to reversible engines alone.
On a T-S diagram, a Carnot cycle is a rectangle between \(500\,\text{K}\) and \(300\,\text{K}\), \(2\,\text{J/K}\) wide. How much heat does it take in per cycle, in joules?
Use the entropy change of the working substance over one Carnot cycle to show that \(Q_H/T_H = Q_C/T_C\).
Entropy is a state function, and after a full cycle the working substance is back in its starting state, so its total entropy change is zero. On the hot isotherm it takes in \(Q_H\) at the fixed temperature \(T_H\), gaining \(Q_H/T_H\). On the cold isotherm it gives out \(Q_C\) at \(T_C\), losing \(Q_C/T_C\). On the two reversible adiabats no heat flows, so its entropy does not change. So \(Q_H/T_H - Q_C/T_C = 0\), which gives \(Q_H/T_H = Q_C/T_C\).
Now return to the drop of ink. As it spreads, the ink molecules reach far more arrangements, and the entropy of the glass grows. Gathering back into a drop would lower the entropy of an isolated glass, which the second law forbids. Cooling chai, a braking car and a working engine all obey the same rule: in every real change the entropy of the universe grows.
In 1877 the Austrian physicist Ludwig Boltzmann showed that entropy counts the number of ways the molecules of a system can be arranged. Many scientists of his time doubted that atoms existed, and they attacked his ideas for years. Experiments soon after his death in 1906 settled the question in his favour. His gravestone in Vienna carries his formula, \(S = k\log W\).
- Writing \(\Delta S = 0\) for a free expansion because no heat flows.\(Q/T\) gives \(\Delta S\) only along a reversible path. Use the isotherm: \(\Delta S = nR\ln(V_2/V_1)\).
- Thinking that the entropy of every system must rise.The law is about an isolated system, or the universe. A part can lose entropy if its surroundings gain more.
- Reading heat from a T-S diagram as the area to the axis of \(T\).Heat is the area between the path and the \(S\) axis, under the path, \(Q = \int T\,dS\).