Introduction to Entropy熵入门

Big Idea · Energy (ENE)AP Science Practice 6 · ArgumentationAP link · Unit 9~40 min

Learning goals

  • Explain entropy in terms of how dispersed matter and energy are (the number of possible arrangements)
  • Predict the sign of ΔS from changes of state, numbers of gas molecules, dissolving, temperature and volume
  • Explain that the total entropy increases in a spontaneous process, so when a system's entropy decreases, the surroundings' entropy increases by more

Before you start

  • Kinetic molecular theory (U3.5)
  • Endothermic and exothermic processes (U6.1)
  • Energy of phase changes (U6.5)

Entropy: how spread out are matter and energy?

Put a drop of perfume in one corner of a room, and soon it can be smelled everywhere; leave a hot block of metal on a table, and its heat spreads into the air. In each case matter or energy goes from concentrated to spread out. Entropy熵 S measures this dispersal: the more possible arrangements a system’s particles and energy have, the higher its entropy.

Take two connected containers. While the particles are shut in the left side, there is only one arrangement; once the valve opens, each particle can be on the left or the right, and the number of arrangements jumps. The more particles there are, the less likely it is that all of them happen to be back on the left:

Try it: how particles spread between two halves of a container

That is why a gas spontaneously fills its whole container but never spontaneously squeezes back into a corner. For an isolated system, a spontaneous process always increases the total entropy; this is one statement of the second law of thermodynamics热力学第二定律.

Predicting the sign of ΔS

The entropy change is ΔS = S(final) − S(initial). In these cases the system’s entropy usually increases (ΔS > 0):

  • Change of state: solid → liquid → gas; the particles become freer at each step, so entropy increases.
  • More gas molecules: if the total gas coefficients increase in a reaction, ΔS > 0; if they decrease, ΔS < 0. Gases matter far more than solids and liquids.
  • Dissolving: when a solid dissolves in water, its particles spread through the whole solution, so ΔS is usually positive.
  • Raising the temperature: particles move more vigorously, so there are more ways to distribute the energy.
  • A larger gas volume: the particles have more room, so there are more arrangements.

A system’s entropy can decrease spontaneously

At −5 °C, water freezes spontaneously, and the system’s entropy decreases. This doesn’t break the second law: the heat released by freezing passes to the surroundings and raises their entropy by more, so the total entropy of system plus surroundings still increases. Whether a process is spontaneous can’t be judged from the system’s entropy alone (U9.3 uses ΔG to account for both).

Example 1: the sign of ΔS

Is ΔS of the system positive or negative for each process? (a) Dry ice sublimes: CO₂(s) → CO₂(g). (b) 2NO(g) + O₂(g) → 2NO₂(g). (c) A sample of argon is compressed from 5.0 L to 1.0 L at constant temperature. (d) Water vapor condenses into droplets on a cold window.

Step-by-step solution
  1. Identify: what decides the sign of ΔS. Entropy reflects how many arrangements the particles can have. Typical cases with more arrangements (ΔS > 0):

    • solid → liquid → gas
    • more moles of gas
    • more volume for a gas to move in

    The reverse gives ΔS < 0.

  2. (a) Dry ice sublimes. A solid becomes a gas, and the particles have far more room: ΔS > 0.

  3. (b) 2NO(g) + O₂(g) → 2NO₂(g). Compare the gas coefficients on each side:

    2+1=3→22 + 1 = 3 \to 2

    Fewer gas molecules: ΔS < 0.

  4. (c) Argon compressed at constant temperature. From 5.0 L to 1.0 L, the gas has less room and fewer arrangements: ΔS < 0.

  5. (d) Water vapor condenses. A gas becomes a liquid: ΔS < 0.

  6. Think about it. Condensation has ΔS < 0 yet happens on its own, because the heat it releases raises the entropy of the surroundings by more than the system’s entropy falls.

Example 2: counting arrangements

Two glass bulbs of equal volume are joined by a narrow tube and contain 4 gas molecules; each molecule is equally likely to be in either bulb. (1) How many arrangements are there? (2) What is the probability that all 4 molecules are in the left bulb? (3) What is the most likely distribution? (4) With 20 molecules, what is the probability that all of them are in the left bulb?

Step-by-step solution
  1. Count all the arrangements. Each molecule has 2 choices, left or right, independently of the others:

    24=162^4 = \boxed{16}
  2. All 4 on the left. Only 1 arrangement:

    P=116≈6.3%P = \frac{1}{16} \approx \boxed{6.3\%}
  3. Most likely distribution: compare them all. The number of arrangements with k molecules on the left is C(4, k):

    • 4 : 0 → 1
    • 3 : 1 → 4
    • 2 : 2 → 6
    • 1 : 3 → 4
    • 0 : 4 → 1

    The total is 16, as it should be. Two in each bulb is most likely:

    P=616=37.5%P = \frac{6}{16} = 37.5\%
  4. All 20 molecules on the left.

    P=1220≈9.5×10−7P = \frac{1}{2^{20}} \approx \boxed{9.5 \times 10^{-7}}
  5. Connect to entropy. The more molecules there are, the less likely “all on one side” becomes and the more completely the spread-out states dominate; for 10²³ molecules, a gas filling its container evenly is essentially certain. That is what an entropy increase means statistically.

Practice: is ΔS of the system positive or negative?
  1. Liquid bromine evaporates: Br₂(l) → Br₂(g)
  2. Water freezes
  3. 2SO₂(g) + O₂(g) → 2SO₃(g)
  4. 2KClO₃(s) → 2KCl(s) + 3O₂(g)
  5. A cup of hot water cools to room temperature (the water only)
  6. Sucrose dissolves in water
  7. Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
  8. Helium expands from 1.0 L to 3.0 L at constant temperature

Common mistakes

Mistake 1: “Entropy is disorder, so anything neat-looking has zero entropy.” Entropy measures how many ways matter and energy can be arranged. Only a perfect crystal at absolute zero has zero entropy; at ordinary temperatures every substance has entropy greater than zero.

Mistake 2: “If the system’s entropy decreases, the process can’t be spontaneous.” The second law is about the total entropy of system plus surroundings. When water freezes at −5 °C, the system’s entropy decreases, but the heat released raises the surroundings’ entropy by more.

Mistake 3: “Exothermic processes always lower entropy and endothermic ones always raise it.” The signs of ΔH and ΔS aren’t linked. Dissolving a solid, for example, can be endothermic or exothermic, but its ΔS is usually positive.

Quick check

Ammonia and hydrogen chloride gases meet and form white smoke: NH₃(g) + HCl(g) → NH₄Cl(s). Which statement is correct?

Two identical connected containers hold 4 gas molecules. What is the probability that all 4 are in one particular container?

Which change increases the entropy of a fixed amount of gas?

A hot iron block is dropped into cold water, and heat flows from the iron to the water. The best explanation is:

Water freezes spontaneously at −5 °C even though the system’s entropy decreases. Why is this consistent with the second law?

Carbon burns in limited oxygen: 2C(s) + O₂(g) → 2CO(g). Which gives the correct sign of ΔS and the correct reason?

AP-style free response

AP-style free response

A little ethanol is poured into an open evaporating dish. After a few minutes all the ethanol has evaporated, and the dish feels cool.

(a) Is ΔS positive or negative for C₂H₅OH(l) → C₂H₅OH(g)? Justify your answer at the particle level.

(b) From the fact that the dish feels cool, decide whether ΔH for the evaporation of ethanol is positive or negative, and explain.

(c) Evaporation of ethanol is endothermic, yet it happens spontaneously at room temperature. Use the idea of entropy to explain why.

(d) A glass of ice water is placed in ethanol vapor, and droplets form on the glass. Give the sign of ΔS for the system in this process, and explain why the process can still occur.

(e) If the ethanol is kept in a sealed bottle, the amount of liquid decreases for a while and then stops changing. Explain at the particle level.

Model answer and scoring

(a) ΔS > 0. As a gas, the molecules can move freely through a much larger space, so the number of possible arrangements increases greatly.

(b) ΔH > 0. Evaporating ethanol absorbs heat from the dish and the surroundings, so the dish feels cool.

(c) Evaporation greatly increases the entropy of the system (gas molecules have far more possible arrangements). At room temperature, the effect of the entropy increase outweighs the unfavorable heat absorption, so the process is spontaneous.

(d) ΔS < 0 (a gas becomes a liquid). Condensation releases heat to the cold glass and the surroundings, raising their entropy by more than the system’s entropy decreases, so the total entropy still increases.

(e) As vapor builds up in the bottle, the rate at which vapor molecules return to the liquid increases. When the rates of evaporation and condensation become equal, dynamic equilibrium is reached and the amount of liquid stops changing (U7.1).

Scoring: (a) sign with particle reasoning; (b) sign with evidence; (c) role of the entropy increase; (d) sign and total-entropy explanation; (e) equal rates.

Summary

  • Entropy measures how dispersed matter and energy are: the more possible arrangements, the higher the entropy.
  • Solid → liquid → gas, more gas molecules, dissolving, heating, and expanding a gas usually give ΔS > 0.
  • In a spontaneous process the total entropy of system plus surroundings increases; the system’s entropy can decrease as long as the surroundings’ entropy increases by more.

AP tip: To predict the sign of ΔS, first look for gases and how the gas coefficients change, and justify it with “the particles are more dispersed / there are more possible arrangements,” not just “more disorder.”

Practice this chapter: Unit 9 · Thermodynamics and Electrochemistry · 12 questionsPractice this topic: U9.1 · Introduction to Entropy · 5 questions

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