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AP Physics 2 Unit 9 Review: Thermodynamics

Review AP Physics 2 Unit 9 to build a complete picture of how energy moves, transforms, and spreads in thermal systems. From atomic collisions driving gas pressure to entropy limiting what heat engines can do, this unit connects microscopic particle behavior to macroscopic thermodynamic laws.

Use the topic guides, key terms, and practice questions available for every topic in this unit to work through each concept before exam day.

What is AP Physics 2 unit 9?

Thermodynamics in AP Physics 2 spans from the microscopic world of colliding gas atoms to the macroscopic constraints that govern every heat engine and energy transfer process. The unit builds in a logical sequence: kinetic theory explains what pressure and temperature mean at the atomic level, the ideal gas law ties those quantities together mathematically, and the first and second laws of thermodynamics set the rules for how energy can change and where it ultimately goes.

Unit 9 is about how energy behaves in thermal systems. Kinetic theory links atomic motion to pressure and temperature. The ideal gas law (PV = nRT) connects state variables. The first law (ΔU = Q + W) enforces energy conservation. Specific heat and thermal conductivity quantify how materials respond to and conduct heat. The second law introduces entropy, explaining why energy spontaneously spreads and why no process is perfectly efficient.

Kinetic theory and the ideal gas law

Gas pressure comes from atoms colliding with container walls. Temperature measures average kinetic energy: K_avg = (3/2)k_B T. The ideal gas law PV = nRT = Nk_B T ties pressure, volume, temperature, and amount together, and PV, PT, and VT graphs each reveal a different gas relationship.

Energy transfer and the first law

Thermal energy moves by conduction, convection, and radiation, always spontaneously from hot to cold. The first law, ΔU = Q + W, tracks internal energy changes across four key process types: isothermal, isobaric, isochoric, and adiabatic. PV diagrams let you read work as the area under the curve.

Entropy and the second law

Entropy measures the tendency of energy to spread out. The second law states that the total entropy of an isolated system never decreases. Spontaneous processes increase entropy, and maximum entropy corresponds to thermodynamic equilibrium. A closed system's entropy can decrease only if energy leaves it, but the surroundings gain at least as much entropy.

Energy conservation has limits

The first law says energy is always conserved, but the second law says not all of that energy is available to do useful work. Every real process increases the total entropy of the universe, which is why heat engines cannot be perfectly efficient and why energy spontaneously disperses rather than concentrating. These two laws together define what is physically possible in any thermal system.

AP Physics 2 unit 9 topics

9.1

Kinetic Theory of Temperature and Pressure

Gas pressure comes from atomic collisions with container walls. Temperature equals average kinetic energy: K_avg = (3/2)k_B T. Root-mean-square speed v_rms = sqrt(3k_B T / m). The Maxwell-Boltzmann distribution shows how speeds vary with temperature.

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9.2

The Ideal Gas Law

PV = nRT = Nk_B T connects pressure, volume, temperature, and amount for an ideal gas. Boyle's, Charles's, and Gay-Lussac's laws are special cases. PT graphs extrapolate to absolute zero. PV graphs show isothermal hyperbolas.

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9.3

Thermal Energy Transfer and Equilibrium

Energy transfers spontaneously from hot to cold by conduction, convection, or radiation. Thermal equilibrium is reached when both systems share the same temperature and net energy transfer stops.

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9.4

The First Law of Thermodynamics

ΔU = Q + W. Internal energy of an ideal monatomic gas is U = (3/2)nRT. Work done on a gas is W = -PΔV. PV diagrams track isothermal, isobaric, isochoric, and adiabatic processes; area under the curve equals work.

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9.5

Specific Heat and Thermal Conductivity

Q = mcΔT calculates energy needed to change an object's temperature. Fourier's law Q/Δt = kA(ΔT)/L gives the rate of conductive heat flow. Both specific heat and thermal conductivity are intrinsic material properties.

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9.6

Entropy and the Second Law of Thermodynamics

Entropy measures energy dispersal. The second law: total entropy of an isolated system never decreases. Entropy is a state function. Isolated systems move spontaneously toward thermodynamic equilibrium, the state of maximum entropy.

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practice snapshot

Hardest AP Physics 2 unit 9 topics

This snapshot uses Fiveable practice activity to show where students tend to miss questions and which review moves are worth prioritizing first.

61%average MCQ accuracy

Across 2.8k multiple-choice practice attempts for this unit.

2.8kMCQ attempts

Practice activity included in this snapshot.

29%average FRQ score

Across 17 scored free-response attempts for this unit.

Hardest topics in unit 9

MCQ miss rate
9.3

Review Thermal Energy Transfer and Equilibrium with attention to how the concept appears in AP-style source and evidence questions.

41%324 tries
9.1

Review Kinetic Theory of Temperature and Pressure with attention to how the concept appears in AP-style source and evidence questions.

38%938 tries
9.5

Review Specific Heat and Thermal Conductivity with attention to how the concept appears in AP-style source and evidence questions.

35%343 tries

Unit 9 review notes

9.1

Kinetic Theory of Temperature and Pressure

Gas pressure arises from atoms colliding with container walls and transferring momentum. Each collision exerts a perpendicular force on the surface, and pressure is the total perpendicular force divided by area: P = F_perp / A. Pressure exists throughout the gas, not only at the walls. Temperature is a measure of the average translational kinetic energy of the atoms, given by K_avg = (3/2)k_B T. The root-mean-square speed v_rms = sqrt(3k_B T / m) tells you the speed corresponding to that average kinetic energy. The Maxwell-Boltzmann distribution shows how speeds are spread across atoms at a given temperature; as temperature rises, the peak shifts right and the distribution broadens.

  • Pressure formula: P = F_perp / A; the sum of perpendicular force components from atomic collisions divided by surface area.
  • Average kinetic energy: K_avg = (3/2)k_B T; directly proportional to absolute temperature in kelvins.
  • Root-mean-square speed: v_rms = sqrt(3k_B T / m); increases with temperature and decreases with atomic mass.
  • Maxwell-Boltzmann distribution: A curve showing the spread of atomic speeds at a given temperature; higher T shifts the peak to higher speeds and flattens the curve.
  • Momentum transfer: Atom-wall collisions are analyzed with conservation of momentum; the perpendicular velocity component reverses in an elastic collision.
If you double the absolute temperature of an ideal gas, by what factor does the average kinetic energy of its atoms change, and what happens to v_rms?
QuantitySymbolDepends on
PressurePForce per area from collisions
Average kinetic energyK_avgTemperature T only
Root-mean-square speedv_rmsTemperature T and atomic mass m
9.2

The Ideal Gas Law

An ideal gas is modeled with four assumptions: random instantaneous velocities, negligible atomic volume, elastic collisions, and no forces between atoms except during collisions. These assumptions allow the equation PV = nRT = Nk_B T to relate pressure, volume, moles (or number of atoms), and absolute temperature. Graphs of gas behavior reveal specific relationships: a PV graph at constant temperature is a hyperbola (Boyle's law), a VT graph at constant pressure is linear through the origin (Charles's law), and a PT graph at constant volume is linear and can be extrapolated to absolute zero where pressure would reach zero.

  • PV = nRT: The ideal gas law; n is moles, R is the universal gas constant 8.314 J/(mol·K), T must be in kelvins.
  • Boyle's law: At constant T, P and V are inversely proportional: P1V1 = P2V2.
  • Absolute zero: Extrapolated from a PT graph; the temperature at which an ideal gas would have zero pressure, equal to 0 K or -273.15 degrees C.
  • Ideal gas assumptions: Random velocities, negligible atomic volume, elastic collisions, no intermolecular forces except during collisions.
  • Graph reading: PV diagrams show isothermal hyperbolas; PT graphs show linear relationships useful for extrapolating to absolute zero.
A gas is compressed at constant temperature from 2 L to 1 L. If the initial pressure was 100 kPa, what is the final pressure? Which gas law applies?
Gas LawConstant quantityRelationship
Boyle's lawTemperatureP inversely proportional to V
Charles's lawPressureV directly proportional to T
Gay-Lussac's lawVolumeP directly proportional to T
9.3

Thermal Energy Transfer and Equilibrium

Two systems in thermal contact can exchange energy by thermal processes. Energy flows spontaneously from the higher-temperature system to the lower-temperature system through three mechanisms: conduction (direct atomic collisions through a material), convection (bulk movement of a fluid carrying thermal energy), and radiation (electromagnetic energy transfer requiring no medium). At the atomic level, higher-energy atoms are more likely to transfer energy to lower-energy atoms during collisions. This continues until both systems reach the same temperature, which is thermal equilibrium. At equilibrium, no net energy is transferred by thermal processes.

  • Conduction: Energy transfer through direct atomic collisions within or between materials; requires physical contact.
  • Convection: Energy transfer by bulk fluid motion; warmer, less dense fluid rises and carries thermal energy.
  • Radiation: Energy transfer via electromagnetic waves; does not require a medium and occurs even in a vacuum.
  • Thermal equilibrium: The state reached when two systems in thermal contact have the same temperature and no net energy transfer occurs.
  • Spontaneous direction: Thermal energy always flows spontaneously from higher temperature to lower temperature, never the reverse on its own.
Two metal blocks at different temperatures are placed in contact. Describe what happens at the atomic level and what the final state looks like.
MechanismMedium requiredAtomic-level description
ConductionYes (solid or fluid)Collisions pass energy atom to atom through material
ConvectionYes (fluid)Bulk fluid motion carries higher-energy atoms to new regions
RadiationNoElectromagnetic waves carry energy through space
9.4

The First Law of Thermodynamics

The internal energy U of an ideal monatomic gas is the sum of the kinetic energies of all its atoms: U = (3/2)nRT = (3/2)Nk_B T. Because ideal gas atoms have no potential energy from interactions, U depends only on temperature. The first law, ΔU = Q + W, states that the change in internal energy equals the heat added to the system plus the work done on the system. Work done on a gas by a constant external pressure is W = -PΔV. PV diagrams represent thermodynamic processes visually; the area under the curve equals the magnitude of work done. Four standard process types each have a defining constraint that simplifies the first law.

  • Internal energy U: For an ideal monatomic gas: U = (3/2)nRT; depends only on temperature, not on pressure or volume independently.
  • First law ΔU = Q + W: Change in internal energy equals heat added to the system plus work done on the system.
  • Work W = -PΔV: Work done on the gas; positive when gas is compressed (ΔV negative), negative when gas expands.
  • PV diagram area: The area under a process curve on a PV diagram equals the magnitude of work done on or by the gas.
  • Adiabatic process: No heat exchange with surroundings (Q = 0); all internal energy change comes from work done on or by the gas.
For an isothermal expansion of an ideal gas, what is ΔU? Using the first law, what must be true about Q and W during this process?
ProcessConstraintFirst law simplification
IsothermalT constantΔU = 0, so Q = -W
IsobaricP constantW = -PΔV; ΔU = Q + W
IsochoricV constantW = 0, so ΔU = Q
AdiabaticQ = 0ΔU = W
9.5

Specific Heat and Thermal Conductivity

Specific heat c is an intrinsic material property that tells you how much energy is needed to raise one kilogram of a material by one kelvin. The equation Q = mcΔT lets you calculate the energy transferred when an object's temperature changes. In calorimetry problems, energy lost by a hot object equals energy gained by a cold object until they reach the same temperature. Thermal conductivity k describes how quickly energy flows through a material by conduction. Fourier's law gives the rate: Q/Δt = kA(ΔT)/L, where A is cross-sectional area, ΔT is the temperature difference across the material, and L is its thickness. Materials with high k (metals) conduct heat rapidly; materials with low k (wood, air) are insulators.

  • Q = mcΔT: Energy transferred equals mass times specific heat times temperature change; AP Physics 2 treats specific heat as temperature-independent.
  • Specific heat c: An intrinsic property of a material in J/(kg·K); depends on atomic arrangement and interactions.
  • Fourier's law: Q/Δt = kA(ΔT)/L; rate of conductive heat flow is proportional to conductivity, area, and temperature difference, and inversely proportional to thickness.
  • Thermal conductivity k: An intrinsic material property in W/(m·K); high for metals, low for insulators.
  • Calorimetry: Energy conservation applied to heat exchange: energy lost by hot object equals energy gained by cold object at thermal equilibrium.
A 0.5 kg aluminum block loses 2000 J of energy. If aluminum has a specific heat of 900 J/(kg·K), by how much does its temperature drop?
EquationWhat it calculatesKey variables
Q = mcΔTTotal energy to change temperatureMass, specific heat, temperature change
Q/Δt = kA(ΔT)/LRate of conductive energy flowConductivity, area, temperature difference, thickness
9.6

Entropy and the Second Law of Thermodynamics

Entropy is a state function that describes the tendency of energy to spread out and the unavailability of some energy to do useful work. The second law of thermodynamics states that the total entropy of an isolated system can never decrease; it increases for irreversible processes and stays constant only for reversible ones. Localized energy spontaneously disperses, and isolated systems move toward thermodynamic equilibrium, which is the state of maximum entropy. A closed system's entropy can decrease if energy leaves it, but the entropy gained by the surroundings is always at least as large, so the total entropy of the universe does not decrease. Entropy is a state function, meaning it depends only on the current state of the system, not on the path taken to reach it.

  • Second law: The total entropy of an isolated system never decreases; it increases for irreversible processes and is constant for reversible ones.
  • Entropy as energy dispersal: Entropy measures how spread out energy is; localized energy spontaneously disperses to increase entropy.
  • State function: Entropy depends only on the current state of the system, not on the process or path used to reach that state.
  • Reversible vs. irreversible: A reversible process leaves total entropy unchanged; all real processes are irreversible and increase total entropy.
  • Closed vs. isolated systems: An isolated system's entropy never decreases. A closed system's entropy can decrease if energy leaves, but the surroundings gain at least as much entropy.
A closed system's entropy decreases by 50 J/K. What must be true about the entropy change of its surroundings, and what does this imply about the total entropy of the universe?
System typeEnergy exchangeEntropy rule
IsolatedNo energy in or outEntropy never decreases
ClosedEnergy can enter or leaveEntropy can decrease if energy leaves, but universe total does not decrease
Reversible processAnyTotal entropy constant
Irreversible processAnyTotal entropy increases

Practice AP Physics 2 unit 9 questions

Try stimulus-based AP practice questions and written prompts after you review the notes.

Example stimulus-based MCQs

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graph

Stimulus-based practice question

A student heats a sealed rigid container of ideal monatomic gas. The temperature-versus-heat-added graph is shown. The student claims that because the container is rigid and no work is done, all of the heat added goes entirely into increasing the internal energy of the gas.

Question

Which of the following correctly evaluates the student's claim?

Correct, because in a rigid container the volume cannot change, so W = 0, and by the first law ΔU = Q, meaning all added heat increases internal energy.

Correct, because the linear graph confirms that temperature increases proportionally with heat, proving that no energy is lost to the surroundings.

Incorrect, because some of the added heat must be used to increase the pressure of the gas rather than the internal energy.

Incorrect, because the internal energy of an ideal gas depends on temperature alone, so adding heat at constant volume does not change the internal energy.

diagram

Stimulus-based practice question

A student investigates thermal radiation by placing two identical black metal plates facing each other across an air gap, as shown. Plate X is maintained at 600 K and Plate Y is maintained at 300 K by external reservoirs. The student claims that Plate Y does not emit any radiation because it is the cooler object and energy only flows from hot to cold.

Question

Which of the following best evaluates the student's claim?

Incorrect, because both plates emit thermal radiation, and Plate X emits more.

Correct, because energy cannot flow from a cold object to a hot one.

Incorrect, because Plate Y emits only toward the external reservoir.

Correct, because radiation requires a temperature difference and Y only absorbs.

Example FRQs

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FRQ

Thermodynamic process of ideal gas with movable piston

2. A sample of monatomic ideal gas is sealed in a vertical cylinder by a movable piston of mass M=3.00 kgM = 3.00\ \text{kg} and cross-sectional area A=2.00×103 m2A = 2.00\times10^{-3}\ \text{m}^2, as shown in Figure 1. The piston moves with negligible friction and the gas occupies an initial volume V0=2.40×103 m3V_0 = 2.40\times10^{-3}\ \text{m}^3. The space above the piston is open to the atmosphere at pressure Patm=1.01×105 PaP_{\text{atm}} = 1.01\times10^5\ \text{Pa}. Initially, the gas and piston are at rest and the gas is in thermal equilibrium with a large thermal reservoir at temperature T0=300 KT_0 = 300\ \text{K}.

Figure 1. Vertical cylinder of monatomic ideal gas sealed by a frictionless movable piston under atmospheric pressure, initially at thermal equilibrium with a 300 K reservoir.

Figure 1

Figure dot. Force diagram (piston represented as a dot).

Figure dot
A.

On the dot shown in Figure dot, representing the piston, draw and label the forces that are exerted on the piston. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

B.

Derive an expression for the internal energy U0U_0 of the gas in the initial state in terms of M, A, V0, PatmM,\ A,\ V_0,\ P_{\text{atm}}, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Figure 2. P–V diagram axes for the quasistatic compression process.

Figure 2
C.

On the axes provided in Figure 2, sketch the expected relationship between the pressure PP and volume VV of the gas for the thermodynamic process the gas undergoes while the mass is added. Draw an arrow on your sketch to represent the direction of the process. A mass m=2.00 kgm = 2.00\ \text{kg} is slowly placed on top of the piston. The gas remains in thermal contact with the reservoir at temperature T0T_0 during the entire process, and the piston comes to rest in a new equilibrium state.

D.

Indicate whether the entropy change ΔSgas\Delta S_{\text{gas}} of the gas during the first 20 s20\ \text{s} is greater than, less than, or equal to zero. After the mass mm is in place, the original reservoir at T0T_0 is removed. The cylinder wall is metal of thickness L=2.0×103 mL = 2.0\times10^{-3}\ \text{m} and contact area with the reservoir Ac=6.0×102 m2A_c = 6.0\times10^{-2}\ \text{m}^2. The gas (through the cylinder wall) is placed in thermal contact with a second large reservoir at temperature T2=360 KT_2 = 360\ \text{K}. The thermal conductivity of the metal wall is k=16 W/(m\cdotpK)k = 16\ \text{W/(m·K)}. During the first Δt=20 s\Delta t = 20\ \text{s} after contact, the temperature difference across the wall may be treated as constant and equal to T2T0T_2 - T_0.

ΔSgas>0\Delta S_{\text{gas}} > 0
ΔSgas<0\Delta S_{\text{gas}} < 0
ΔSgas=0\Delta S_{\text{gas}} = 0

Briefly justify your answer by referencing (i) the direction of energy transfer due to the temperature difference and (ii) a microscopic interpretation of temperature and pressure for an ideal gas.

FRQ

Gas moles determination through pressure, volume, temperature measurement

3. In an investigation of the thermodynamic behavior of a gas, a student uses a fixed amount of dry air confined in a cylinder with a movable piston. The student will design an experiment to determine a thermodynamic property of the gas and then analyze a second set of data to determine that property from a graph.

Figure 1. Cylinder-piston apparatus for gas measurement

Figure 1

Figure 2. Cartesian grid for moles determination

Figure 2
A.

Describe a procedure for collecting data that would allow the student to determine the number of moles n of gas in the cylinder. In your description, include the measurements to be made. Include any steps necessary to reduce experimental uncertainty.

B.

Describe how the collected data could be analyzed to determine n. Include references to appropriate equations and to relationships between measured and known quantities.

P (kPa)

V (x10^-4 m^3)

120

6.24

140

5.35

160

4.68

180

4.16

200

3.74

C.

In a second experiment, the student places the cylinder in the 300 K water bath and waits until the gas reaches thermal equilibrium at T = 300 K for each trial. The student then changes the external load on the piston and records the equilibrium pressure P and volume V of the gas. Table 1 shows the collected data.

i.

Indicate two quantities, either measured quantities from Table 1 or additional calculated quantities, that could be graphed to produce a straight line that could be used to determine n.

Vertical axis: Horizontal axis:

ii.

On the grid provided, create a graph of the quantities indicated in part C(i) that can be used to determine n.

Use Table 2 to record the data points or calculated quantities that you will plot.

Clearly label the axes, including units as appropriate.

Plot the points you recorded in Table 2.

iii.

Draw a best-fit line for the data graphed in part C(ii).

D.

Using the best-fit line that you drew in part C(iii), calculate an experimental value for the number of moles n of the gas. Use R = 8.31 J/(mol·K) and T = 300 K.

FRQ

Thermal equilibration between isolated gas compartments

1. A rigid, sealed cylindrical container with smooth, thermally insulating walls is divided into two equal compartments by a thin conducting metal partition of thickness L=2.0×103 mL = 2.0\times10^{-3}\ \text{m} and area A=4.0×103 m2A = 4.0\times10^{-3}\ \text{m}^2, as shown in Figure 1. Each compartment contains an ideal monatomic gas. Initially, Side 1 contains n1=0.50 moln_1 = 0.50\ \text{mol} of gas at temperature T1,i=600 KT_{1,i} = 600\ \text{K}, and Side 2 contains n2=0.50 moln_2 = 0.50\ \text{mol} of gas at temperature T2,i=300 KT_{2,i} = 300\ \text{K}. Each compartment has fixed volume V=8.0×103 m3V = 8.0\times10^{-3}\ \text{m}^3. The gases can exchange energy only by conduction through the partition. The partition has thermal conductivity k=50 W/(m\cdotpK)k = 50\ \text{W/(m·K)}. The ideal gas constant is R=8.31 J/(mol\cdotpK)R = 8.31\ \text{J/(mol·K)}.

Figure 1. Rigid insulated container with two fixed-volume ideal-gas compartments separated by a conducting metal partition (all given numerical parameters shown).

Figure 1

Figure 2. Student annotation diagram for initial energy-transfer direction and larger initial average molecular speed.

Figure 2

Figure 4. Reference equations for ideal gases and kinetic theory.

Figure 4
A.
i.

Complete the following tasks in Figure 2.

Indicate the direction of the net energy transfer through the partition.

Indicate which side (Side 1 or Side 2) has the larger average molecular speed at the initial instant.

ii.

At the initial instant, the gases exert pressures P1,iP_{1,i} and P2,iP_{2,i} on their respective compartment walls.

Derive an expression for the ratio P1,iP2,i\dfrac{P_{1,i}}{P_{2,i}} using only given quantities. Begin your derivation by writing a fundamental physics principle or an equation from the reference information in Figure 4.

Figure 3. P–V diagram illustrating a constant-volume process from initial state i to final state f at V = 8.0×10⁻³ m³.

Figure 3
B.

Indicate whether the total entropy change ΔStotal\Delta S_{\text{total}} for the two-gas system during this process is positive, negative, or zero. The partition is kept in place and energy is transferred between the gases until they reach thermal equilibrium. The total volume of each compartment remains V=8.0×103 m3V = 8.0\times10^{-3}\ \text{m}^3. The container is isolated from the environment, so the total internal energy of the two-gas system is constant. Assume both gases remain ideal and monatomic.

Positive
Negative
Zero

Justify your answer using Figure 3 as a reference for constant-volume behavior during the thermal equilibration process.

Key terms

TermDefinition
root-mean-square speedThe speed corresponding to the average kinetic energy of atoms in an ideal gas: v_rms = sqrt(3k_B T / m). Increases with temperature and decreases with atomic mass.
Maxwell-Boltzmann distributionA graph showing the spread of atomic speeds in an ideal gas at a given temperature. Higher temperatures shift the peak to higher speeds and broaden the curve.
Boyle's LawAt constant temperature, pressure and volume are inversely proportional: P1V1 = P2V2.
absolute zeroThe temperature at which an ideal gas would have zero pressure, extrapolated from a PT graph; equal to 0 K or -273.15 degrees C.
P-V diagramA graph of pressure versus volume used to represent thermodynamic processes. The area under the curve equals the magnitude of work done on or by the gas.
ΔU = Q + WThe first law of thermodynamics: change in internal energy equals heat added to the system plus work done on the system.
work done on a systemFor a gas under constant or average external pressure: W = -PΔV. Positive when the gas is compressed, negative when it expands.
constant volume processA thermodynamic process where volume does not change, so W = 0 and ΔU = Q entirely.
reversible processA theoretical process in which total entropy remains constant; all real processes are irreversible and increase total entropy.
state functionA property that depends only on the current state of the system, not on the path taken to reach it. Entropy and internal energy are both state functions.
Q=mcΔTEnergy transferred to or from an object equals its mass times its specific heat times its temperature change. Specific heat is treated as temperature-independent in AP Physics 2.
Fourier's Law of Heat ConductionQ/Δt = kA(ΔT)/L; the rate of conductive heat flow is proportional to thermal conductivity, cross-sectional area, and temperature difference, and inversely proportional to thickness.
degrees of freedomThe number of independent ways a particle can store energy. A monatomic ideal gas has 3 translational degrees of freedom, giving U = (3/2)nRT.
thermodynamic cycleA sequence of thermodynamic processes that returns a gas to its initial state, represented as a closed loop on a PV diagram. Net work equals the area enclosed by the loop.

Common unit 9 mistakes

Using Celsius instead of Kelvin in gas law equations

PV = nRT, K_avg = (3/2)k_B T, and v_rms = sqrt(3k_B T / m) all require absolute temperature in kelvins. Always convert: T(K) = T(°C) + 273.15 before substituting.

Confusing the sign of work in the first law

In AP Physics 2, W = -PΔV is work done on the gas. When a gas expands, ΔV is positive, so W is negative (the gas does work on surroundings, losing internal energy). When compressed, W is positive. Keep track of the sign convention consistently.

Assuming isothermal means no energy transfer

Isothermal means constant temperature, so ΔU = 0 for an ideal gas. But heat Q and work W are both nonzero; they are equal and opposite. Students often incorrectly set Q = 0 for an isothermal process, which applies to adiabatic processes instead.

Applying the second law only to isolated systems

The entropy of a closed system can decrease if energy leaves it. The second law says the total entropy of an isolated system never decreases, meaning you must account for the surroundings when analyzing a closed system's entropy change.

Treating pressure as existing only at the container walls

Pressure exists throughout the gas, not just where atoms hit the walls. This is a common conceptual error when explaining gas pressure using kinetic theory.

How this unit shows up on the AP exam

Qualitative atomic-level explanations

AP Physics 2 free-response questions frequently ask you to explain macroscopic phenomena using atomic or molecular reasoning. For Unit 9, this means describing why pressure changes when temperature rises, why energy flows from hot to cold, or why entropy increases in a given process. Practice writing explanations that explicitly reference atomic collisions, kinetic energy, and energy dispersal rather than just citing equations.

PV diagram analysis and the first law

Expect questions that present a PV diagram and ask you to identify the process type, determine the sign of Q, W, and ΔU, or compare work done in different processes by comparing areas under curves. You may also need to apply ΔU = Q + W quantitatively across a multi-step process or cycle.

Multi-part energy transfer problems

Questions often combine Q = mcΔT, Fourier's law, and thermal equilibrium in a single scenario, such as two objects exchanging heat through a conducting material until equilibrium is reached. You may need to calculate final temperatures, energy transferred, or conduction rates, and then explain the result using the second law or entropy reasoning.

Final unit 9 review checklist

  • Connect atomic motion to pressure and temperatureExplain how atomic collisions produce gas pressure using P = F_perp / A, and relate temperature to average kinetic energy using K_avg = (3/2)k_B T and v_rms = sqrt(3k_B T / m).
  • Apply the ideal gas law and read gas graphsUse PV = nRT to solve for unknown state variables. Interpret PV, PT, and VT graphs correctly, including extrapolating a PT graph to find absolute zero.
  • Identify and explain the three heat transfer mechanismsDistinguish conduction, convection, and radiation. Explain at the atomic level why energy flows from hot to cold and what thermal equilibrium means.
  • Apply the first law across all four process typesUse ΔU = Q + W and W = -PΔV for isothermal, isobaric, isochoric, and adiabatic processes. Read work from the area under a PV diagram curve.
  • Calculate heat transfer with Q = mcΔT and Fourier's lawSolve calorimetry problems using energy conservation. Apply Q/Δt = kA(ΔT)/L to find conduction rates given material properties and geometry.
  • Reason about entropy and the second lawExplain why isolated systems move toward equilibrium using entropy. Distinguish isolated from closed systems and apply the rule that total entropy of the universe never decreases.

How to study unit 9

Start with kinetic theory and the ideal gas law (Topics 9.1-9.2)Read the topic guides for 9.1 and 9.2. Practice deriving pressure from atomic collisions and using K_avg = (3/2)k_B T to find v_rms. Then work through ideal gas law problems involving all three graph types (PV, PT, VT) and practice extrapolating to absolute zero.
Work through thermal energy transfer (Topic 9.3)Review the three heat transfer mechanisms and practice explaining each at the atomic level. Write out a short explanation of why thermal equilibrium is the final state of two systems in thermal contact, using the language of energy transfer between atoms.
Understand the first law and PV diagrams (Topic 9.4)Study the four process types using the comparison table. Practice applying ΔU = Q + W and W = -PΔV to each. Draw PV diagrams for each process type and identify where work is positive, negative, or zero based on the area under the curve.
Practice specific heat and conduction calculations (Topic 9.5)Work through calorimetry problems using Q = mcΔT with energy conservation. Then apply Fourier's law Q/Δt = kA(ΔT)/L to problems that vary thickness, area, or temperature difference. Check that you can identify which variable changes the conduction rate most.
Review entropy and the second law (Topic 9.6)Focus on qualitative reasoning: explain why entropy increases in spontaneous processes, distinguish isolated from closed systems, and confirm you can apply the rule that total entropy of the universe never decreases. Use the available practice questions to test your reasoning on entropy scenarios.

More ways to review

Topic study guides

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FRQ practice

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Cram archive videos

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Frequently Asked Questions

What topics are covered in AP Physics 2 Unit 9?

AP Physics 2 Unit 9 covers 6 topics in thermodynamics: 9.1 Kinetic Theory of Temperature and Pressure, 9.2 The Ideal Gas Law, 9.3 Thermal Energy Transfer and Equilibrium, 9.4 The First Law of Thermodynamics, 9.5 Specific Heat and Thermal Conductivity, and 9.6 Entropy and the Second Law of Thermodynamics. Together these topics connect microscopic particle behavior to large-scale energy rules. You'll work through how temperature relates to molecular motion, how ideal gases behave under changing conditions, how energy moves between systems, and why entropy sets limits on what any heat engine can do. See AP Physics 2 Unit 9 for practice on all six topics.

How much of the AP Physics 2 exam is Unit 9?

AP Physics 2 Unit 9 makes up 15-18% of the AP exam, making thermodynamics one of the heavier-weighted units on the test. That means roughly 1 in 6 points traces back to topics like entropy, the ideal gas law, the first and second laws of thermodynamics, and thermal energy transfer. It's worth putting real time into this unit.

What's on the AP Physics 2 Unit 9 progress check (MCQ and FRQ)?

The AP Physics 2 Unit 9 progress check in AP Classroom includes both MCQ and FRQ parts drawn from all six thermodynamics topics. The MCQ section tests conceptual understanding of kinetic theory, the ideal gas law, thermal equilibrium, and entropy. The FRQ part asks you to apply the first and second laws of thermodynamics, often requiring you to explain energy conservation or justify why entropy increases in a given process. For matched practice before you take the progress check, head to AP Physics 2 Unit 9 to review each topic and work through similar question types.

How do I practice AP Physics 2 Unit 9 FRQs?

AP Physics 2 Unit 9 FRQs most often pull from entropy and the second law of thermodynamics, the first law of thermodynamics, and the ideal gas law. These questions typically ask you to analyze a thermodynamic process, calculate work or internal energy changes, or argue why a proposed process violates the second law. To practice, write out full explanations rather than just numbers, because AP Physics 2 FRQs reward clear reasoning. Start by reviewing each topic at AP Physics 2 Unit 9, then try past College Board FRQs on heat engines and gas processes. Check your answers against the scoring guidelines to see exactly where reasoning points are awarded.

Where can I find AP Physics 2 Unit 9 practice questions?

The best starting point for AP Physics 2 Unit 9 practice questions, including multiple-choice and practice test sets, is AP Physics 2 Unit 9. That page organizes practice by topic, so you can target weak spots like kinetic theory, the ideal gas law, or entropy separately before doing a full mixed set. For MCQ practice, focus on questions that ask you to compare thermodynamic processes or predict how changing pressure, volume, or temperature affects an ideal gas. For a practice test feel, work through a timed set covering all six topics in one sitting.

How should I study AP Physics 2 Unit 9?

Start AP Physics 2 Unit 9 by building a solid picture of entropy and the second law of thermodynamics, since that concept ties the whole unit together. Then work forward through the six topics in order: kinetic theory of temperature and pressure, the ideal gas law, thermal energy transfer and equilibrium, the first law of thermodynamics, and specific heat and thermal conductivity. Here's a concrete plan: - Sketch PV diagrams for isothermal, adiabatic, and isochoric processes until the shapes feel automatic. - Practice applying the ideal gas law with changing variables, not just plugging in numbers. - For entropy questions, always ask: does disorder increase? That single check handles most second-law FRQs. - After each topic, do a short MCQ set to catch gaps before moving on. All six topics with practice are at AP Physics 2 Unit 9. Since this unit is 15-18% of the exam, a few focused study sessions here pay off more than spreading that time across lighter units.

Ready to review Unit 9?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.