Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

Theme B — The Particulate Nature of Matter

B.4.4 Calculating Changes in Entropy — IB Physics HL Revision Notes & Topical Questions

5 min read 9 parts Lesson 36 of 53 1 quiz

📚 IB Physics HL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
0%read
Lesson summary

B.4.4 Calculating Changes in Entropy — IB PHYSICS HL notes: Calculate entropy changes using ΔS = Q/T and ΔS = kB ln(Ω2/Ω1) for heat transfers, changes of state and gas expansions.Equation — ΔS = ΔQ/TValid when heat is transferred reversibly at a constant absolute temperatureSymbo…

  • Link to microstates
  • Gas expanding from V to 2V
  • Equivalent form
  • Change of state
  • Isothermal gas expansion
  • Doubling the volume

Section 1 of 9Macroscopic definition

Equation
  • ΔS = ΔQ/T
  • Valid when heat is transferred reversibly at a constant absolute temperature
Symbols
  • ΔS in J K−1, ΔQ in J, T in K
Sign
  • Sign heat gained gives ΔS > 0 and heat lost gives ΔS < 0
  • A reversible process returning to the original state has ΔS = 0
Link to microstates
  • Link to microstates at low temperature a small amount of heat changes the available arrangements by a larger fraction so ΔQ/T is larger

Section 2 of 9Microscopic definition

Microscopic definition
Equation
  • S = kB ln Ω and ΔS = kB ln(Ω2/Ω1)
Gas expanding from V to 2V
  • Gas expanding from V to 2V the number of microstates rises by 2N
  • ΔS = NkB ln 2
Equivalent form
  • ΔS = NkB ln(V2/V1) = nR ln(V2/V1) because nR = NkB

Section 3 of 9Constant temperature cases

Constant temperature cases
Reservoir
  • Reservoir a body so large that its temperature does not change when it gains or loses heat
Change of state
  • Change of state melting or boiling occurs at constant T with Q = mL
  • ΔS = mL/T
Isothermal gas expansion
  • ΔU = 0 so Q = W = nRT ln(V2/V1)
  • ΔS = Q/T = nR ln(V2/V1)
Doubling the volume
  • ΔS = nR ln 2 = 5.76 J K−1 per mole
State function
  • State function entropy depends only on the initial and final states so a free expansion between the same states gives the same ΔS

Section 4 of 9Heat flow between two reservoirs

1Hot reservoir
  • Hot reservoir loses heat Q so its entropy change is −Q/Th
2Cold reservoir
  • Cold reservoir gains heat Q so its entropy change is +Q/Tc
3Total
  • ΔStotal = Q/Tc − Q/Th = Q(Th − Tc)/(ThTc)
4Result
  • Th > Tc gives a positive total so spontaneous heat flow from hot to cold increases total entropy
5Limit
  • Th = Tc gives zero, the reversible limit

Section 5 of 9Method for entropy problems

1
Identify every body or reservoir that gains or loses heat.
2
Convert each temperature to kelvin.
3
Give heat gained a positive sign and heat lost a negative sign.
4
Calculate ΔS = Q/T for each body.
5
Add the changes for the system and surroundings to find the total.

Section 6 of 9Interpreting the total

1Positive total
  • Positive total the process is possible but irreversible
2Zero total
  • Zero total the process is possible and reversible
3Negative total
  • Negative total the process is impossible
4Degree of irreversibility
  • Degree of irreversibility a larger total entropy production means a more irreversible process

Section 7 of 9More from the specification

1
Thermodynamic Cycles and Heat Engine Efficiency
  • In a heat engine operating through a closed thermodynamic cycle, thermal energy absorbed from a hot reservoir is converted into mechanical work, with maximum theoretical efficiency strictly determined by second law of thermodynamics: η = 1 - (TC / TH).
2
Internal Energy of an Ideal Gas
  • The internal energy U of a monatomic ideal gas depends solely on its absolute thermodynamic temperature (U = 3/2 n R T),
  • as intermolecular forces between gas particles are negligible.
3
Entropy and Reversible Heat Transfer
  • When thermal energy Q is transferred to a system at constant absolute temperature T, entropy changes by ΔS = ΔQ / T;
  • while local entropy of a system decreases during cooling, total entropy of the universe always increases in irreversible processes.
4
Work Done During Gas Expansion
  • During an isobaric expansion against external pressure P, the work done by expanding gas particles is given by W = P ΔV, linking mechanical work and heat exchange to changes in internal energy through the first law of thermodynamics (Q = ΔU + W).
5
Thermodynamic Cycles and Heat Engine Efficiency
  • In a heat engine operating through a closed thermodynamic cycle, thermal energy absorbed from a hot reservoir is converted into mechanical work, with maximum theoretical efficiency strictly determined by second law of thermodynamics: η = 1 - (TC / TH).
6
Internal Energy of an Ideal Gas
  • The internal energy U of a monatomic ideal gas depends solely on its absolute thermodynamic temperature (U = 3/2 n R T),
  • as intermolecular forces between gas particles are negligible.
7
Entropy and Reversible Heat Transfer
  • When thermal energy Q is transferred to a system at constant absolute temperature T, entropy changes by ΔS = ΔQ / T;
  • while local entropy of a system decreases during cooling, total entropy of the universe always increases in irreversible processes.
8
Work Done During Gas Expansion
  • During an isobaric expansion against external pressure P, the work done by expanding gas particles is given by W = P ΔV, linking mechanical work and heat exchange to changes in internal energy through the first law of thermodynamics (Q = ΔU + W).
9
Carnot cycle
  • A reversible cycle of two isothermal and two adiabatic processes between two reservoirs.
10
Isolated system
  • System with no external interaction.
11
Atmospheric pressure
  • Pressure of the air, about 1.01 × 105 Pa at sea level.
12
Pressure
  • Force per unit area, P = F/A.
    What the symbols mean: p = pressure (Pa = N m−2) · F = force acting at right angles to the surface (N) · A = area (m2)
13
Efficiency
  • Useful output divided by total input.
14
Adiabatic
  • With no thermal energy transfer
15
Compression
  • Decrease in length under a squashing force.
16
Piston
  • A moving part in a cylinder that applies or receives pressure.
17
Molecule
  • Tiny particle of the fluid, too small to be seen.
18
Internal energy
  • Total kinetic and potential energy of the particles in a system
19
Thermal energy
  • Energy of random molecular motion, increased by friction.

Section 8 of 9Key terms

TermMeaning
ΔS = ΔQ/TThe entropy change when heat ΔQ enters a system reversibly at constant absolute temperature T.
ReservoirA body so large that its temperature does not change when it gives or takes heat.
State functionA quantity that depends only on the state of the system and not on how it was reached.
Reversible processAn idealised process in which the total entropy of the universe stays constant.
Latent heatEnergy transferred during a change of state at constant temperature.

Section 9 of 9Exam technique

1
Using temperature in °C in ΔS = ΔQ/T — always use kelvin.
2
Ignoring the sign — heat leaving a body gives a negative entropy change for that body.
3
Using ΔS = ΔQ/T when the temperature changes a lot — the equation needs a constant temperature.
4
Considering only the system — include the entropy change of the surroundings.
5
Saying an irreversible process has ΔS = 0 for the universe — it has ΔStotal > 0.

Last updated:

\n