Home / AP® Exam / AP Physics 2 Revision Resources / AP Physics 2- 9.6 Entropy and the Second Law of Thermodynamics- Study Notes

AP Physics 2- 9.6 Entropy and the Second Law of Thermodynamics- Study Notes- New Syllabus

AP Physics 2- 9.6 Entropy and the Second Law of Thermodynamics – Study Notes

AP Physics 2- 9.6 Entropy and the Second Law of Thermodynamics – Study Notes – per latest Syllabus.

Key Concepts:

  • Entropy and Second Law of Thermodynamics 

AP Physics 2-Concise Summary Notes- All Topics

Entropy:

Entropy measures the tendency of energy to spread out or the unavailability of energy to do work.

  • Localized energy will naturally disperse over time.
  • Entropy is a state function: it depends only on the current state or configuration of the system, not on the path taken.
  • Maximum entropy occurs when a system reaches thermodynamic equilibrium.
  • The change in a system’s entropy depends on interactions with its surroundings.

Mathematical Expression:

\( \Delta S = \displaystyle \int \dfrac{dQ_{rev}}{T} \)

  • \( \Delta S \): Change in entropy
  • \( dQ_{rev} \): Infinitesimal heat absorbed reversibly
  • \( T \): Absolute temperature at which heat is transferred

Essential conclusions:

  • Entropy is a measure of energy dispersal and disorder.
  • Isolated systems always move toward maximum entropy (thermodynamic equilibrium).
  • Reversible processes maintain total entropy, irreversible processes increase it.
  • Entropy changes in a system are influenced by energy exchanges with surroundings.

Second Law of Thermodynamics:

The total entropy of an isolated system can never decrease. It remains constant only when all processes are reversible.

  • Isolated systems spontaneously evolve toward thermodynamic equilibrium, the state of maximum entropy.
  • In a closed system, entropy can decrease because energy can be exchanged with the surroundings.

Example :

One mole of an ideal gas expands reversibly and isothermally at 300 K from 5 L to 10 L. Find the entropy change.

▶️ Answer/Explanation

Entropy change for isothermal expansion: \( \Delta S = n R \ln \dfrac{V_f}{V_i} \)

\( \Delta S = 1 \times 8.314 \ln \dfrac{10}{5} = 8.314 \ln 2 \approx 5.76 \, J/K \)

Answer: Entropy change ≈ 5.76 J/K.

Example :

Heat of 400 J is removed reversibly from a system at 500 K. Find the change in entropy.

▶️ Answer/Explanation

\( \Delta S = \dfrac{Q_{rev}}{T} = \dfrac{-400}{500} = -0.8 \, J/K \)

Answer: Change in entropy = -0.8 J/K.

Scroll to Top