Entropy

A quantity hidden in the Carnot cycle

Look again at the Carnot cycle of Lesson 3. It took in \(Q_H = 2880\,\text{J}\) at \(500\,\text{K}\) and gave out \(Q_C = 1728\,\text{J}\) at \(300\,\text{K}\). The heats differ, but divide each by its temperature: \(2880/500 = 5.76\,\text{J/K}\) and \(1728/300 = 5.76\,\text{J/K}\). They are equal. This is the rule \(Q_C/Q_H = T_C/T_H\) written another way, \(Q_H/T_H = Q_C/T_C\).

So the gas takes in \(5.76\,\text{J/K}\) of "heat over temperature" at the source, and gives out exactly as much at the sink. In 1865 Clausius saw that this ratio measures a property of the gas itself. He called the property entropy, \(S\). This lesson uses \(R = 8.31\,\text{J/(mol K)}\), \(c_\text{water} = 4200\,\text{J/(kg K)}\), \(L_f = 3.34 \times 10^5\,\text{J/kg}\) and \(L_v = 2.26 \times 10^6\,\text{J/kg}\).

The unit of entropy is the joule per kelvin. The sign follows the heat: a system that takes in heat gains entropy, and one that gives out heat loses it. A system that takes in \(1000\,\text{J}\) at a steady \(500\,\text{K}\) gains \(1000/500 = 2\,\text{J/K}\). The reservoir that supplied it loses \(2\,\text{J/K}\).

The temperature in the denominator is the heart of the idea. The same \(1000\,\text{J}\) taken in at \(250\,\text{K}\) raises the entropy by \(4\,\text{J/K}\), twice as much. Heat given to a cold body changes it more than the same heat given to a hot one. A spoon of warm water barely changes a bucket of hot water, but it melts some of a block of ice.

The same change on every path

Heat is not a property of a state. The chapter on the First Law showed that two paths between the same states take in different heats. Entropy is different. The change \(\int dQ_\text{rev}/T\) is the same on every reversible path between two states. So entropy is a state function, like internal energy. Fig 5.1 tests this on three paths.

The three paths in Fig 5.1 take in very different heats. The isotherm takes in \(1728\,\text{J}\). Cooling first and then expanding takes in \(1247\,\text{J}\), and expanding first and then cooling takes in \(2493\,\text{J}\). Yet the sum of \(dQ/T\) ends at \(5.76\,\text{J/K}\) every time. On the second path the sum even goes negative while the gas gives out heat, and then climbs back.

Two results follow at once. First, after a complete cycle a system is back in its starting state, so its entropy change round any cycle is zero. Second, to find \(\Delta S\) for any process, even an irreversible one, choose any reversible path between the same two states and work along that. Lesson 6 uses this trick for free expansion and for heat flow.

Entropy of melting and boiling

A change of state happens at a fixed temperature, so the heat is taken in at one \(T\) and \(\Delta S = mL/T\). One kilogram of ice melting at \(273\,\text{K}\) gains \(3.34 \times 10^5/273 = 1223\,\text{J/K}\). One kilogram of water boiling at \(373\,\text{K}\) gains \(2.26 \times 10^6/373 = 6059\,\text{J/K}\), about five times as much.

Why is boiling so much larger? The latent heat of boiling is almost seven times that of melting, while the temperature is only \(37\,\%\) higher. Loosening a solid into a liquid changes the arrangement of the molecules a little. Spreading a liquid out into a gas, more than a thousand times larger in volume, changes it enormously. Entropy grows with that freedom.

Entropy of heating

When a body is warmed, its temperature changes as the heat goes in, so \(Q/T\) cannot be used whole. Break the warming into small steps. Each step takes in \(dQ = mc\,dT\) at its own temperature \(T\), so \(dS = mc\,dT/T\). Adding the steps is an integral, and the gap derivation below does it.

A \(1\,\text{kg}\) copper block (\(c = 400\,\text{J/(kg K)}\)) warmed from \(300\,\text{K}\) to \(600\,\text{K}\) gains \(400\ln 2 = 277\,\text{J/K}\). One kilogram of water warmed from \(273\,\text{K}\) to \(373\,\text{K}\) gains \(4200\ln(373/273) = 1311\,\text{J/K}\). Cooling runs the formula backwards: \(T_2 < T_1\) makes the logarithm negative, and the body loses entropy.

The curve in the sim bends over. At \(300\,\text{K}\) one more kelvin adds \(C_p/T = 0.069\,\text{J/K}\). At \(600\,\text{K}\) it adds only \(0.035\,\text{J/K}\), and at \(900\,\text{K}\) only \(0.023\,\text{J/K}\). The heat for each kelvin is the same, but it is taken in at a higher temperature. So \(\Delta S\) reaches \(14.4\,\text{J/K}\) at \(600\,\text{K}\) and only \(22.8\,\text{J/K}\) at \(900\,\text{K}\).

Entropy of an ideal gas

For a gas the first law gives the reversible heat of a small step: \(dQ = dU + p\,dV\). The internal energy of an ideal gas depends on the temperature alone, so \(dU = nC_v\,dT\). The pressure is \(p = nRT/V\). Dividing by \(T\) separates the two parts neatly.

The formula contains every simple process. On an isotherm the first term vanishes, so \(\Delta S = nR\ln(V_2/V_1)\). One mole doubling its volume gains \(8.31\ln 2 = 5.76\,\text{J/K}\), as in Fig 5.1. At constant volume only the first term is left. At constant pressure \(V\) grows in proportion to \(T\), and the two terms combine into \(nC_p\ln(T_2/T_1)\), since \(C_v + R = C_p\).

Heat one mole of helium from \(300\,\text{K}\) to \(600\,\text{K}\). At constant volume, \(\Delta S = \tfrac32 R\ln 2 = 8.64\,\text{J/K}\). At constant pressure, \(\Delta S = \tfrac52 R\ln 2 = 14.4\,\text{J/K}\). The second is larger because the gas also expands, and it must take in more heat for the same rise in temperature.

On a reversible adiabat no heat flows at any step, so \(dS = 0\) and the entropy stays constant. Such a process is isentropic. The formula agrees. On leg B to C of the Carnot cycle the gas cooled from \(500\,\text{K}\) to \(300\,\text{K}\) while its volume grew from \(20\,\text{L}\) to \(71.7\,\text{L}\). Then \(20.775\ln(0.6) + 8.31\ln(3.586)\), which is \(-10.61 + 10.61 = 0\). The cooling lowers the entropy exactly as much as the expansion raises it.