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IB Physics HL/Notes/B.4 Thermodynamics

IB Physics HLB.4 ThermodynamicsNotes

Track First law of thermodynamics

The first law is an accounting equation. Energy entering the gas by heating either increases the internal energy of the gas or leaves the gas as work done on the surroundings. If a problem uses the opposite sign convention for work, state it clearly and keep it consistent.

The first law of thermodynamics is conservation of energy for a thermodynamic system.
Using the IB convention here, Q is positive when energy is transferred into the gas by heating.
W is positive when the gas does work on the surroundings, such as during expansion.
The equation is Q = ΔU + W, equivalently ΔU = Q - W.
All three quantities are energy transfers or energy changes measured in joules.

Build the first-law energy equation with the IB sign convention.

Formula
Target formula Q = ΔU + W; ΔU = Q - W
ΔU
change in internal energy of the gas
J
Q
thermal energy transferred into the gas by heating
J
W
work done by the gas on surroundings
J
1Choose the closed gas system.system = gas
2Take heat transferred into the gas as positive.Q > 0 into system
3Take work done by the gas as positive.W > 0 for expansion
4Apply conservation of energy.Q = ΔU + W, so ΔU = Q - W

State the first law of thermodynamics for a closed gas system and define the sign convention for Q and W.

Common mark losses are omitting the sign convention, mixing work done by and on the gas, or treating Q as a state variable.

State the first law of thermodynamics for a closed gas system and define the sign convention for Q and W.

Choose

Use Gas work

Gas work is mechanical energy transfer due to a moving boundary such as a piston. The simple equation W = PΔV applies directly for constant pressure. For a changing pressure process, use the area under the p-V curve, with the sign determined by the direction of volume change.

Work done by a gas occurs when its volume changes against an external pressure.
For constant pressure, work done by the gas is W = PΔV.
On a p-V diagram, work done by the gas is the area under the process curve.
Expansion has ΔV > 0, so W is positive for work done by the gas.
Compression has ΔV < 0, so W is negative for work done by the gas, meaning work is done on the gas.

Build the gas-work relationship from a p-V process.

Formula
Target formula W = PΔV at constant pressure; W = signed area under p-V curve
W
work done by the gas
J
P
pressure
Pa
ΔV
change in volume
m^3
1Find the change in volume.ΔV = V_final - V_initial
2Check whether pressure is constant.constant P?
3For constant pressure, multiply pressure by volume change.W = PΔV
4For changing pressure, use the p-V graph area.W = area under curve

Explain how to determine work done by a gas from a p-V diagram and state the formula for constant pressure.

Common mark losses are using PΔV for non-constant pressure without considering area, or using the wrong sign for compression.

Explain how to determine work done by a gas from a p-V diagram and state the formula for constant pressure.

Choose

Internal energy change

There are two common routes to internal energy change. In energy-transfer problems, use the first law and the sign convention. In monatomic ideal-gas temperature problems, use the temperature dependence of internal energy. For an ideal gas, a volume change does not by itself determine ΔU unless it also changes temperature.

Internal energy U is a state function, so ΔU depends on the initial and final states, not the path.
From the first law with W as work done by the gas, ΔU = Q - W.
For a monatomic ideal gas, U = (3/2)nRT, so ΔU = (3/2)nRΔT.
Equivalently, ΔU = (3/2)Nk_BΔT when particle number is used.
For an isothermal process of an ideal gas, ΔT = 0 and therefore ΔU = 0.

Choose the correct route for finding change in internal energy.

Formula
Target formula ΔU = Q - W; for monatomic ideal gas ΔU = (3/2)nRΔT = (3/2)Nk_BΔT
ΔU
change in internal energy
J
Q
heat transferred into the gas
J
W
work done by the gas
J
n
amount of gas
mol
ΔT
temperature change
K
1Check whether Q and W are given or whether temperature change is given.energy route or temperature route
2For heat and work data, use the first law with sign convention.ΔU = Q - W
3For a monatomic ideal gas, use the temperature dependence.ΔU = (3/2)nRΔT
4If temperature is constant for an ideal gas, internal energy does not change.ΔT = 0 ⇒ ΔU = 0

A monatomic ideal gas changes temperature by ΔT. State how to calculate ΔU and explain how this connects to the first law.

Common mark losses are using the monatomic formula without checking the gas model, or forgetting the sign convention in ΔU = Q - W.

A monatomic ideal gas changes temperature by ΔT. State how to calculate ΔU and explain how this connects to the first law.

Choose

Track Entropy and disorder

Entropy gives a statistical direction to thermodynamics. A gas spreading out, heat flowing from hot to cold, or energy becoming dispersed are more likely because there are many more microscopic ways to be spread out than concentrated. The word disorder is useful only when tied to this microstate idea.

Entropy S is a state quantity linked to the number of accessible microstates of a system.
A microstate is one specific microscopic arrangement of particles and energy consistent with the same macroscopic state.
More accessible microstates means higher entropy and greater microscopic disorder.
Spontaneous processes tend to move toward states with greater total entropy for the system plus surroundings.
Entropy is often associated with energy becoming more spread out and less available to do useful work.

Sort each situation by its entropy cue.

Sort
Unsorted
5
higher entropy cue
0
lower entropy cue
0
not enough information
0

Explain entropy in terms of microstates and use this to describe why a gas spreading out is spontaneous.

Common mark losses are saying only “entropy is disorder” without mentioning microstates or the system plus surroundings.

Explain entropy in terms of microstates and use this to describe why a gas spreading out is spontaneous.

Choose

Track Entropy equations

Practice

The equation ΔS = Q_rev/T is used for reversible heat transfer or for a thermal reservoir at constant temperature. The statistical equation S = k_B ln(Ω) connects entropy to microstates. Both forms point to the same idea: entropy increases when energy or particles can be arranged in more microscopic ways.

For reversible thermal energy transfer at temperature T, entropy change is ΔS = Q_rev/T.
Q_rev is positive when heat is added to the system and negative when heat leaves it.
Temperature must be absolute temperature in kelvin; entropy has unit J K^-1.
Statistical entropy is S = k_B ln(Ω), where Ω is the number of accessible microstates.
A larger number of microstates means larger entropy.

Build the appropriate entropy equation.

Formula
Target formula ΔS = Q_rev/T; S = k_B lnΩ
ΔS
entropy change
J K^-1
Q_rev
reversible heat transfer
J
T
absolute temperature
K
S
entropy
J K^-1
k_B
Boltzmann constant
J K^-1
Ω
number of accessible microstates
1Decide whether the problem gives heat transfer or microstate count.Q_rev/T or k_B lnΩ
2For reversible heat transfer at temperature T, divide heat by absolute temperature.ΔS = Q_rev/T
3For number of accessible microstates, use Boltzmann’s formula.S = k_B lnΩ
4Check units and signs.T in K, ΔS in J K^-1

State two equations used to calculate entropy or entropy change, defining all symbols and conditions.

Common mark losses are using Celsius temperature, omitting the reversible heat-transfer condition, or failing to define Ω.

State two equations used to calculate entropy or entropy change, defining all symbols and conditions.

Choose

Track Second law of thermodynamics

Practice

The second law is a direction rule, not a failure of energy conservation. The first law says energy is conserved; the second law says which energy transfers are possible spontaneously. In real processes, total entropy of system plus surroundings increases, or stays constant only for an ideal reversible process.

Entropy form: the total entropy of an isolated system cannot decrease.
Clausius form: heat cannot spontaneously flow from a colder body to a hotter body without work being done.
Kelvin form: no cyclic heat engine can convert all heat absorbed from a reservoir into work with no other effect.
The second law explains the direction of real thermodynamic processes.
A subsystem may decrease in entropy if the surroundings increase by at least as much.

Repair the second-law misconceptions.

Spot Errors

State two forms of the second law of thermodynamics and explain what they imply about real processes.

Common mark losses are saying entropy of every subsystem must increase, or claiming a 100% efficient cyclic heat engine is possible.

State two forms of the second law of thermodynamics and explain what they imply about real processes.

Choose

Irreversibility

Irreversibility is the practical directionality of thermodynamics. Friction converts ordered mechanical energy into dispersed thermal energy, heat flows hot to cold, and gases spread into available volume. These processes are allowed by the first law, but the second law explains why they do not naturally reverse.

An irreversible process cannot be reversed without leaving some change in the system or surroundings.
Real spontaneous processes are irreversible because they increase total entropy.
Examples include friction, heat conduction through a finite temperature difference, free expansion, and mixing of gases.
A reversible process is an ideal limiting process that can be reversed by an infinitesimal change and has no net entropy production.
Energy conservation can still hold in irreversible processes; irreversibility is about entropy and direction.

Sort each process by whether it is irreversible or an ideal reversible limit.

Sort
Unsorted
5
irreversible real process
0
ideal reversible limit
0
needs external work to reverse
0

Track Local entropy decrease

A local decrease in entropy is not a loophole in the second law. It means the chosen subsystem is not isolated. A refrigerator lowers entropy inside its cold compartment, but it requires work and releases heat to the room, increasing the entropy of the surroundings by a larger amount.

The second law applies to the total entropy of an isolated system, not necessarily to every part individually.
A local subsystem can decrease in entropy if it exchanges energy with the surroundings.
The process is allowed when ΔS_system + ΔS_surroundings ≥ 0.
Examples include refrigerators, freezing water, crystal formation, and living systems maintaining order using energy input.
The key exam move is to define the system boundary and include the surroundings.

Decide whether a local entropy decrease violates the second law.

Decision
Water freezes while releasing heat to surroundings.
A refrigerator cools its interior using electrical work.
An isolated system is claimed to have a net entropy decrease.

Use Gas processes

Each process name tells you what is constrained. Use that constraint to choose the equation, p-V graph shape, and first-law simplification. This is the habit that prevents mixing up a vertical isovolumetric line with a zero-pressure or zero-work process.

Isobaric means constant pressure; it appears as a horizontal line on a p-V diagram and W = PΔV.
Isovolumetric or isochoric means constant volume; it appears as a vertical line and W = 0.
Isothermal means constant temperature; for an ideal gas ΔU = 0 and pV is constant.
Adiabatic means no heat exchange with surroundings, so Q = 0.
On a p-V diagram, an adiabatic curve is steeper than an isotherm for an ideal gas.

Match each gas process to its fixed condition and p-V consequence.

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Describe the p-V diagram shape and key energy condition for isobaric, isovolumetric, isothermal, and adiabatic processes.

Common mark losses are mixing up isothermal and adiabatic curves, or forgetting W = 0 for constant volume.

Describe the p-V diagram shape and key energy condition for isobaric, isovolumetric, isothermal, and adiabatic processes.

Choose

Use Adiabatic ideal gas

Adiabatic processes are thermally isolated or fast enough that heat transfer is negligible. For a monatomic ideal gas the pressure-volume relation uses γ = 5/3. Combine this with PV = nRT when you need temperature at an initial or final state.

Adiabatic means there is no heat exchange with the surroundings: Q = 0.
For a monatomic ideal gas, γ = 5/3 and P V^γ = constant.
The calculation form is P1V1^(5/3) = P2V2^(5/3).
During adiabatic expansion, the gas does work and internal energy decreases, so temperature falls.
On a p-V diagram, an adiabatic curve is steeper than an isothermal curve through the same point.

Build the adiabatic monatomic ideal-gas relation.

Formula
Target formula P1V1^(5/3) = P2V2^(5/3); Q = 0
P1
initial pressure
Pa
V1
initial volume
m^3
P2
final pressure
Pa
V2
final volume
m^3
γ
adiabatic index, 5/3 for monatomic ideal gas
1Check that no heat is exchanged.Q = 0
2Check that the gas is monatomic and ideal.γ = 5/3
3Use the adiabatic pressure-volume relation.PV^γ = constant
4Write the two-state form for calculation.P1V1^(5/3) = P2V2^(5/3)

A monatomic ideal gas undergoes an adiabatic expansion. State the pressure-volume relation and explain the temperature change.

Common mark losses are confusing adiabatic with isothermal, forgetting Q = 0, or using 5/3 for a non-monatomic gas.

A monatomic ideal gas undergoes an adiabatic expansion. State the pressure-volume relation and explain the temperature change.

Choose

Track Heat engine cycles

Heat engines repeat a set of thermodynamic processes so the working gas returns to its starting state. The useful output is net work per cycle, read from the enclosed p-V area. The second law requires some heat to be rejected; not all heat input can become work in a cyclic engine.

A cyclic process returns the gas to its original pressure, volume, and temperature.
Because internal energy is a state function, ΔU = 0 over one complete cycle.
On a p-V diagram, the net work done by the gas per cycle equals the area enclosed by the loop.
A clockwise cycle represents net work done by the gas; an anticlockwise cycle represents net work done on the gas.
A heat engine absorbs heat from a hot reservoir, does useful work, and rejects some heat to a colder reservoir.

Interpret the p-V cycle for a heat engine.

Graph

p-V cycle diagram

1identify whether the path is a closed cycle
2use enclosed area to determine net work per cycle
3use direction of the loop to decide whether work is done by or on the gas
4state that ΔU = 0 for a complete cycle

For a heat engine shown as a closed loop on a p-V diagram, explain how to determine net work and ΔU over one cycle.

Common mark losses are using the area under only one path, forgetting the loop direction, or giving a non-zero ΔU for a complete cycle.

For a heat engine shown as a closed loop on a p-V diagram, explain how to determine net work and ΔU over one cycle.

Choose

Compare Useful and Input Energy

Efficiency compares what the engine usefully delivers with what it receives. The second law requires some heat rejection in a cyclic engine, so Q_C is not zero for a real engine. Use magnitudes for Q_H and Q_C unless the question defines a signed convention.

Thermal efficiency η is useful work output divided by heat input from the hot reservoir.
For a heat engine, η = W_out/Q_H.
Over a complete cycle, ΔU = 0, so W_out = Q_H - Q_C.
Therefore η = 1 - Q_C/Q_H, where Q_C is heat rejected to the cold reservoir.
Efficiency is dimensionless and is often expressed as a percentage at the end.

Build the heat-engine efficiency equation.

Formula
Target formula η = W_out/Q_H = 1 - Q_C/Q_H
η
thermal efficiency
W_out
useful work output per cycle
J
Q_H
heat input from hot reservoir per cycle
J
Q_C
heat rejected to cold reservoir per cycle
J
1Identify heat input from the hot reservoir.Q_H
2Identify useful work output per cycle.W_out
3Use ΔU = 0 over a full cycle.W_out = Q_H - Q_C
4Divide useful output by input.η = W_out/Q_H = 1 - Q_C/Q_H

A heat engine absorbs Q_H from a hot reservoir and rejects Q_C to a cold reservoir each cycle. State the efficiency equation and explain why η is less than 1 for a real engine.

Common mark losses are using the wrong denominator, mixing signed and magnitude quantities, or claiming all heat input can become work.

A heat engine absorbs Q_H from a hot reservoir and rejects Q_C to a cold reservoir each cycle. State the efficiency equation and explain why η is less than 1 for a real engine.

Choose

Compare Useful and Input Energy

Carnot efficiency is a limit set by the second law, not a guarantee that an engine reaches that value. Raising the hot-reservoir temperature or lowering the cold-reservoir temperature increases the theoretical maximum, but real engines remain below it because of irreversibility and losses.

The Carnot cycle is an ideal reversible heat-engine cycle with maximum possible efficiency for given reservoir temperatures.
It consists of two isothermal processes and two adiabatic processes.
The Carnot efficiency is η_C = 1 - T_C/T_H.
T_H and T_C are the hot- and cold-reservoir absolute temperatures in kelvin.
No real engine operating between the same two reservoirs can exceed the Carnot efficiency.

Build the Carnot efficiency limit.

Formula
Target formula η_C = 1 - T_C/T_H
η_C
Carnot maximum efficiency
T_C
cold-reservoir absolute temperature
K
T_H
hot-reservoir absolute temperature
K
1Identify the hot-reservoir absolute temperature.T_H in K
2Identify the cold-reservoir absolute temperature.T_C in K
3Recognize this is a reversible maximum, not actual engine efficiency.Carnot limit
4Assemble the Carnot efficiency equation.η_C = 1 - T_C/T_H

Define the Carnot efficiency limit for a heat engine operating between hot and cold reservoirs and state why real engines do not exceed it.

Common mark losses are using Celsius, reversing T_C and T_H, or presenting the Carnot limit as actual efficiency for any real engine.

Define the Carnot efficiency limit for a heat engine operating between hot and cold reservoirs and state why real engines do not exceed it.

Choose

Retrieve the B.4 Thermodynamics Model

Review

B.4 is an energy-and-entropy toolkit for HL thermodynamics. First account for heat, work, and internal energy; then use p-V diagrams to read processes and cycles; finally use entropy and the second law to explain why real engines and real processes have limits.

Use the first law Q = ΔU + W with a stated sign convention.
Calculate gas work from W = PΔV at constant pressure or p-V graph area.
Find ΔU from ΔU = Q - W or from ΔU = (3/2)nRΔT for a monatomic ideal gas.
Explain entropy using microstates and calculate entropy using ΔS = Q_rev/T or S = k_B lnΩ.
Apply the second law through total entropy, Clausius form, and Kelvin form.
Identify irreversible processes and local entropy decreases by considering system plus surroundings.
Match isobaric, isovolumetric, isothermal, and adiabatic processes to p-V shapes and energy conditions.
Use P1V1^(5/3) = P2V2^(5/3) for adiabatic monatomic ideal gases.
Interpret heat-engine cycles: ΔU_cycle = 0 and net work equals enclosed p-V area.
Calculate thermal efficiency η = W_out/Q_H = 1 - Q_C/Q_H and Carnot limit η_C = 1 - T_C/T_H.

Match each B.4 retrieval cue to the thermodynamics move it should trigger.

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