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AP^® Physics 2

All seven College Board units, the three science practices, and the equation sheet: built for the 42 MC and the 4 FRQs.

Start Unit 1 free. Unit 9: Thermodynamics is open to everyone, no account needed. Other topics are locked.

Unit 9: Thermodynamics

THE BIG PICTURE. Unit 9 (15–18%, the joint-largest unit) opens AP Physics 2 by reframing energy at the microscopic level. Where Physics 1 said "energy is conserved across a process," Physics 2 says "energy can be transferred as HEAT (random thermal motion) or as WORK (organized force-times-displacement)" and the first law of thermodynamics governs the bookkeeping. The unit's second great idea, entropy, explains WHY heat flows from hot to cold and never spontaneously reverses.

KINETIC THEORY OF GASES

Gas pressure arises from countless molecular collisions with container walls. For an ideal monatomic gas: $\bar K = \tfrac{1}{2} m \bar{v^2} = \tfrac{3}{2} k_B T,$ where $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann's constant and $T$ is the absolute (Kelvin) temperature.

Implications:

Temperature is a measure of the AVERAGE kinetic energy of molecules, NOT total energy.
The root-mean-square speed: $v_{\text{rms}} = \sqrt{3 k_B T / m}$ . Lighter molecules move faster at the same $T$ .
Maxwell-Boltzmann distribution: most molecules near $v_{\text{rms}}$ , with a long high-speed tail.

Maxwell-Boltzmann speed distributions for argon at 300 K and 900 K, with the rms speed of each marked by a dashed line; the hotter curve peaks at a higher speed and is lower and broader. — **Speed distributions at two temperatures** Raising the temperature shifts the distribution to higher speeds and flattens it; the area under each curve stays the same. The rms speed grows with the square root of T.

WORKED EXAMPLE: A SEALED TANK OF HELIUM (MATHEMATICAL ROUTINES)

A rigid 0.010 m $^3$ tank holds 0.50 mol of helium (monatomic, atomic mass 4.0 u) at 300 K. (a) Find the pressure. (b) Find the average kinetic energy per atom and the rms speed. (c) The tank is heated to 600 K. By what factor does each quantity change? Justify.

(a) Ideal gas law. $P = \frac{nRT}{V} = \frac{(0.50)(8.31)(300)}{0.010} \approx 1.25 \times 10^5$ Pa, about 1.2 atm.
(b) Kinetic theory. $K_{\text{avg}} = \tfrac{3}{2}k_B T = \tfrac{3}{2}(1.38 \times 10^{-23})(300) = 6.2 \times 10^{-21}$ J. With $m = 4.0 \times 1.66 \times 10^{-27} = 6.64 \times 10^{-27}$ kg, $v_{\text{rms}} = \sqrt{3k_BT/m} \approx 1.37 \times 10^3$ m/s.
(c) Functional dependence. The volume and the number of moles are fixed, so $P \propto T$ : the pressure doubles. $K_{\text{avg}} \propto T$ , so it also doubles. But $v_{\text{rms}} \propto \sqrt{T}$ , so the rms speed rises only by $\sqrt{2} \approx 1.41$ , to about $1.93 \times 10^3$ m/s. The figure above shows the same effect for argon: the whole distribution shifts right and flattens.

Equations used ( $PV = nRT$ , $K_{\text{avg}} = \tfrac{3}{2}k_BT$ , $v_{\text{rms}}$ ) are all on the AP equation sheet.

THE IDEAL GAS LAW

The fundamental equation of state for an ideal gas: $P V = n R T = N k_B T,$ where $n$ = moles, $R = 8.314$ J/(mol K), $N$ = number of molecules. Two equivalent forms (per mole vs per molecule).

Useful special cases:

Isothermal (constant $T$ ): $P V =$ constant (Boyle's law). $P_1 V_1 = P_2 V_2$ .
Isobaric (constant $P$ ): $V/T =$ constant (Charles's law). $V_1/T_1 = V_2/T_2$ .
Isochoric / Isovolumetric (constant $V$ ): $P/T =$ constant (Gay-Lussac). $P_1/T_1 = P_2/T_2$ .
Adiabatic (no heat exchange, $Q = 0$ ): $P V^\gamma =$ constant, where $\gamma = C_P/C_V$ . Adiabatic compression heats the gas; expansion cools it.

Scope note: the CED names the four special processes (isovolumetric, isothermal, isobaric, adiabatic) and asks you to read them from $PV$ diagrams, but the $PV^\gamma$ relation, $C_P$ , and $C_V$ are not on the equation sheet. Treat them as background: for an adiabatic process, use $Q = 0$ and $\Delta U = W$ .

Always use absolute (Kelvin) temperature: $T_K = T_C + 273.15$ . Common error: using Celsius.

Graph of gas pressure versus Celsius temperature with six data points from 0 to 100 degrees and a best-fit line extrapolated down to zero pressure near minus 274 degrees Celsius. — **Extrapolating to absolute zero** At fixed volume, pressure is linear in temperature. Extending the best-fit line to zero pressure locates absolute zero, about 273 degrees below the freezing point of water.

WORKED EXAMPLE: FINDING ABSOLUTE ZERO FROM DATA (EXPERIMENTAL DESIGN AND ANALYSIS)

A student seals air in a rigid metal sphere attached to a pressure gauge and places it in water baths at different temperatures, waiting for the reading to settle each time.

Bath temperature (°C)	0	20	40	60	80	100
Pressure (kPa)	100.2	107.1	114.8	121.8	129.5	136.4

Design. Independent variable: bath temperature (thermometer). Dependent variable: gas pressure (gauge). Controlled: the volume (rigid sphere) and the amount of gas (sealed). Wait for thermal equilibrium before each reading.
Linearize. $PV = nRT$ with $V$ and $n$ fixed predicts $P$ is proportional to the ABSOLUTE temperature, so $P$ vs $T$ in °C should be a straight line that reaches $P = 0$ at absolute zero.
Analyze. The best-fit line has slope $\approx 0.365$ kPa/°C and vertical intercept $\approx 100.1$ kPa. Setting $P = 0$ : $T = -100.1/0.365 \approx -274$ °C, within about 0.5% of the accepted $-273.15$ °C.
Interpret the slope. Slope $= nR/V$ , so with a measured volume it gives the number of moles of trapped air.

A common error is to force the line through the origin of a °C graph. The point of the experiment is that the intercept is NOT at 0 °C.

CASE STUDY: KELVIN'S ABSOLUTE SCALE (1848)

In 1848 William Thomson (later Lord Kelvin) published "On an Absolute Thermometric Scale," proposing a temperature scale that does not depend on the properties of any one substance. In a footnote he estimated the coldest possible temperature as about $-273$ °C, using the fact that a gas's pressure or volume changes by about 1/273 of its 0 °C value per degree. The kelvin, the SI unit of temperature, is named after him.

Concept: temperature measures average atomic kinetic energy, which has a true zero (topic 9.1 and 9.2).
Graph: the extrapolated $P$ vs $T$ line above, which crosses zero pressure near $-273$ °C.
Lesson: always convert to kelvins before using $PV = nRT$ or $K_{\text{avg}} = \tfrac{3}{2}k_BT$ .

THERMAL ENERGY AND HEAT TRANSFER

Internal energy $U$ of an ideal monatomic gas: $U = \tfrac{3}{2} N k_B T = \tfrac{3}{2} n R T$ . (Diatomic adds rotational degrees: $U = \tfrac{5}{2} n R T$ at room temperature. AP Physics 2 focuses on monatomic.)

Q

Conduction: molecular collisions transfer energy through a solid.
Convection: bulk fluid motion (warm air rising).
Radiation: electromagnetic waves (Sun warming Earth).

For solids/liquids, the heat-temperature relationship is $Q = m c \Delta T,$ where $c$ is specific heat capacity (J/(kg K)). Water: $c = 4186$ J/(kg K). High $c$ → temperature changes slowly for given heat input (why water moderates climate).

Phase changes absorb/release latent heat $L$ at constant $T$ : $Q = m L.$ $L_{\text{fusion}}$ for water (ice ↔ water): $3.34 \times 10^5$ J/kg. $L_{\text{vaporization}}$ for water (water ↔ steam): $2.26 \times 10^6$ J/kg.

Thermal conductivity (CED 9.5). The equation sheet gives the rate at which energy flows by conduction through a slab. $\frac{Q}{\Delta t} = \frac{kA\,\Delta T}{L},$ where $k$ is the material's thermal conductivity (an intrinsic property), $A$ the cross-sectional area, $L$ the thickness, and $\Delta T$ the temperature difference across it. Double the area or the temperature difference and the rate doubles; double the thickness and the rate halves.

WORKED EXAMPLE: HEAT LEAKING THROUGH A WINDOW

A single glass pane ( $k = 0.80$ W/(m K)) is 1.5 m $^2$ in area and 4.0 mm thick. Inside it is 20 °C, outside 0 °C. (a) Find the rate of energy loss. (b) Predict the rate for a pane twice as thick.

(a) $\frac{Q}{\Delta t} = \frac{(0.80)(1.5)(20)}{0.0040} = 6.0 \times 10^3$ W. (A temperature DIFFERENCE is the same in °C and K, so no conversion is needed here.)
(b) Rate $\propto 1/L$ , so doubling the thickness halves the rate to $3.0 \times 10^3$ W. Real double-glazed windows do far better because the trapped air layer has a much smaller $k$ than glass.

FIRST LAW OF THERMODYNAMICS

Energy conservation for a thermodynamic system, in the form on the AP equation sheet: $\Delta U = Q + W,$ where $\Delta U$ is the change in internal energy, $Q$ is heat ADDED to the system, and $W$ is work DONE ON the system by its surroundings. (Some textbooks write $\Delta U = Q - W$ with $W$ = work done BY the gas; the AP sheet uses the opposite sign, so a positive $W$ means the gas is COMPRESSED. Match the sheet.)

Sign conventions (memorize):

$Q > 0$ : heat enters system. $Q < 0$ : heat leaves.
$W > 0$ : gas COMPRESSED (surroundings do work on it). $W < 0$ : gas EXPANDS (does work on surroundings).
$\Delta U > 0$ : internal energy (and $T$ , for ideal gas) increases.

Pressure-volume diagram with four paths leaving state A at 2 liters and 300 kilopascals: isobaric expansion, isovolumetric pressure drop, isothermal expansion along an isotherm, and a steeper adiabatic expansion, with dashed isotherms in the background. — **Four special processes on a PV diagram** Each path is labeled with the signs of W (work done on the gas), Q, and ΔU. The adiabat falls below the isotherm because the gas cools as it does work with no heat input.

Work done ON the gas: $W = -P \Delta V$ for constant pressure (isobaric); for variable pressure, $W = -\int P dV$ . The area under the $P$ - $V$ curve is the work done BY the gas, which is $-W$ on the sheet's convention.

PV DIAGRAMS AND CYCLES

A $P$ - $V$ diagram shows a gas's state. The area enclosed by a CYCLIC process equals the net work done per cycle.

Here $W$ is the work done ON the gas (AP-sheet convention).

Process	Constraint	$Q$	$W$ (on gas)	$\Delta U$
Isothermal	$T$ const	$Q = -W$	$-\int P dV$	0
Isobaric	$P$ const	$n C_P \Delta T$	$-P\Delta V$	$n C_V \Delta T$
Isochoric	$V$ const	$n C_V \Delta T$	0	$Q$
Adiabatic	$Q = 0$	0	$\Delta U$	$n C_V \Delta T$

For a complete cycle returning to its starting state, $\Delta U_{\text{cycle}} = 0$ , so $Q_{\text{net}} = -W_{\text{net}}$ . The area enclosed by the cycle is the net work done BY the gas: positive for a clockwise loop (a heat engine, $Q_{\text{net}} > 0$ ), negative for counterclockwise (a refrigerator).

Rectangular clockwise cycle on a PV diagram through A, B, C, D with the enclosed area shaded and the work done on the gas labeled for each leg. — **Work around a cycle** Only the constant-pressure legs involve work. For a clockwise cycle the net work done on the gas is negative and equals minus the enclosed area; over the full cycle ΔU = 0.

WORKED EXAMPLE: FILLING IN A CYCLE TABLE (TRANSLATION BETWEEN REPRESENTATIONS)

A monatomic ideal gas is taken around the rectangular cycle A → B → C → D → A shown above: A (1 L, 300 kPa), B (4 L, 300 kPa), C (4 L, 100 kPa), D (1 L, 100 kPa). Find $W$ , $\Delta U$ , and $Q$ for each leg using the AP convention ( $W$ = work done ON the gas).

Step 1, work. Only the constant-pressure legs have $\Delta V \neq 0$ . A → B: $W = -P\Delta V = -(300 \text{ kPa})(3 \text{ L}) = -900$ J. C → D: $W = -(100)(-3) = +300$ J. The vertical legs have $W = 0$ . (Useful shortcut: 1 kPa × 1 L = 1 J.)
Step 2, internal energy. For a monatomic ideal gas $U = \tfrac{3}{2}nRT = \tfrac{3}{2}PV$ , so $\Delta U = \tfrac{3}{2}\Delta(PV)$ . The $PV$ products are A 300 J, B 1200 J, C 400 J, D 100 J.
Step 3, heat. From the first law, $Q = \Delta U - W$ .

Leg	Process	$W$ (J)	$\Delta U$ (J)	$Q$ (J)
A → B	isobaric expansion	$-900$	$+1350$	$+2250$
B → C	isovolumetric cooling	0	$-1200$	$-1200$
C → D	isobaric compression	$+300$	$-450$	$-750$
D → A	isovolumetric heating	0	$+300$	$+300$
Cycle		$-600$	0	$+600$

Step 4, check against the graph. $\Delta U_{\text{cycle}} = 0$ as it must be for a return to the start. The net work done ON the gas, $-600$ J, is minus the enclosed area (200 kPa × 3 L = 600 J), consistent with a clockwise loop in which the gas does net work.

PRACTICE: SIGNS OF Q, W, AND ΔU

Process for an ideal gas	$W$ (on gas)	$\Delta U$	$Q$
Isobaric expansion	negative	positive ( $T$ rises)	positive
Isothermal compression	positive	zero	negative (heat leaves)
Isovolumetric heating	zero	positive	positive
Adiabatic expansion	negative	negative ( $T$ falls)	zero
Clockwise full cycle	negative (net)	zero	positive (net)
Gas pushed into a smaller volume very quickly	positive	positive	about zero (no time for heat flow)

CASE STUDY: JOULE'S PADDLE WHEEL (1845)

In 1845 James Prescott Joule reported an experiment in which falling weights turned a paddle wheel inside an insulated container of water. The churning raised the water's temperature, and he found the temperature rise was proportional to the work done by the falling weights. His value for the "mechanical equivalent of heat" was within about 1% of the modern 4.186 J per calorie.

Concept: the first law. Work and heating are two different ways of transferring the same thing: energy.
Representation: for the water, $Q \approx 0$ (insulated) and $W > 0$ , so $\Delta U = W$ and the temperature rises.
Lesson: temperature can rise with no heating at all. "Heat" is a transfer, not something an object contains.

ENTROPY AND THE SECOND LAW

Entropy ( $S$ ) is a measure of the number of microscopic states (microstates) consistent with a macroscopic state. High entropy = many microstates = "disordered."

Second law (two equivalent statements):

Heat spontaneously flows from hot to cold, never the reverse without external work.
The total entropy of an isolated system never decreases ( $\Delta S_{\text{total}} \geq 0$ ).

For a reversible heat transfer at constant $T$ : $\Delta S = Q/T$ .

Examples of entropy increase:

Ice melting: rigid lattice → mobile molecules (more microstates).
Gas expanding into vacuum: more possible positions per molecule.
Mixing two gases: combined system has more microstates than separated.

Heat engines and efficiency. A heat engine takes in $Q_H$ from a hot reservoir, does work $W$ , dumps $Q_C$ to a cold reservoir. Efficiency $\eta = W/Q_H = 1 - Q_C/Q_H$ . Carnot (ideal) limit: $\eta_{\max} = 1 - T_C/T_H$ (both in Kelvin). No real engine can exceed this.

Scope note: the CED limits the second law to a QUALITATIVE treatment (topic 9.6). $\Delta S = Q/T$ and the efficiency formulas above are not on the equation sheet; know the ideas (energy spreads out, entropy of an isolated system never decreases, entropy is a state function) rather than the calculations.

PRACTICE: WHAT HAPPENS TO ENTROPY?

Situation	System type	Entropy of the system	Why
Hot coffee and cold milk mix in a sealed, insulated flask	isolated	increases	energy spreads until one temperature is reached
A freezer turns water into ice	closed, not isolated	decreases	energy leaves the system; the kitchen's entropy rises more
Gas expands to fill an evacuated box	isolated	increases	more possible arrangements
A gas returns to its exact starting state after a cycle	any	net change zero	entropy is a state function
Two blocks at the same temperature touch	isolated	no change	already in thermal equilibrium

EXAM CONNECTIONS. Unit 9 dominates the Translation Between Representations FRQ (Q2) with $PV$ -diagram problems: identify the process type along each leg, compute $Q$ , $W$ , $\Delta U$ per leg, then sum. MC favors quick proportional reasoning on ideal gas law (" $P$ doubled, $V$ tripled, what is $T$ ?") and on heat-transfer problems with mixing.

Key Terms

Temperature ( $T$ )

A measure of the average kinetic energy per molecule in a system. ALWAYS use Kelvin ( $T_K = T_C + 273.15$ ) in thermodynamic formulas.

Ideal Gas Law

$PV = nRT = Nk_B T$ . The equation of state for an ideal gas. $R = 8.314$ J/(mol K); $k_B = 1.38 \times 10^{-23}$ J/K.

Internal Energy ( $U$ )

Total kinetic energy of all molecules in a system. Monatomic ideal gas: $U = \tfrac{3}{2} n R T$ . Depends only on $T$ for an ideal gas (NOT on $P$ or $V$ separately).

Heat ( $Q$ )

Energy transferred between systems due to a temperature difference. Positive if entering system; SI unit joule.

Work ( $W$ ) in Thermodynamics

Energy transferred via volume change. AP-sheet convention: $W$ = work done ON the gas, $W = -P\Delta V$ (isobaric) or $-\int P dV$ . So $W > 0$ under COMPRESSION; the area under the $P$ - $V$ curve is the work done BY the gas ( $= -W$ ).

First Law of Thermodynamics

$\Delta U = Q + W$ (AP sheet). Energy conservation: change in internal energy = heat added PLUS work done ON the gas. (Equivalent to $\Delta U = Q - W_{\text{by gas}}$ .)

Specific Heat Capacity ( $c$ )

$Q = mc\Delta T$ . Energy needed to raise 1 kg by 1 K. Water $c = 4186$ J/(kg K) is unusually high.

Latent Heat ( $L$ )

$Q = mL$ . Energy for phase change at constant $T$ . $L_{\text{fusion, water}} = 3.34 \times 10^5$ J/kg; $L_{\text{vap, water}} = 2.26 \times 10^6$ J/kg.

Isothermal Process

$T$ constant. Ideal gas: $PV =$ const, $\Delta U = 0$ , so $Q = -W$ (heat absorbed equals work done BY the gas). On $P$ - $V$ diagram: hyperbola.

Adiabatic Process

No heat exchange ( $Q = 0$ ). $\Delta U = W$ (work done on the gas). Compression heats the gas; expansion cools it. $PV^\gamma =$ const.

Isobaric Process

$P$ constant. $W = -P\Delta V$ (work on gas). On $P$ - $V$ diagram: horizontal line.

Isochoric Process

$V$ constant. No work done. $\Delta U = Q$ . On $P$ - $V$ diagram: vertical line.

Entropy ( $S$ )

A measure of microstates: higher $S$ = more disorder. For reversible heat transfer at constant $T$ : $\Delta S = Q/T$ .

Second Law of Thermodynamics

Total entropy of an isolated system never decreases. Equivalently: heat flows spontaneously from hot to cold, never the reverse without work.

Heat Engine Efficiency

$\eta = W/Q_H = 1 - Q_C/Q_H$ . Carnot limit: $\eta_{\max} = 1 - T_C/T_H$ (Kelvin). No real engine can exceed this.

Exam Tips

Always use Kelvin in ideal gas law and entropy calculations. Forgetting to convert from Celsius is the most common Unit 9 error.
For $P$ - $V$ diagram problems, identify each leg (isothermal, isobaric, isochoric, adiabatic) and apply the appropriate formula for $Q$ , $W$ , $\Delta U$ . Sum across legs to get totals.
For a cyclic process, $\Delta U = 0$ regardless of path. The area enclosed by the loop is the net work done BY the gas (equal to $Q_{\text{net}}$ ): positive for a clockwise loop (engine), negative for counterclockwise (refrigerator). On the sheet's $W$ = work-on-gas convention, $Q_{\text{net}} = -W_{\text{net}}$ .
The area under the curve on a $P$ - $V$ diagram is the work done BY the gas for any process (the sheet's $W$ , work done ON the gas, is the negative of it). For constant $P$ , work by gas $= P\Delta V$ .
For ideal gas, $\Delta U$ depends only on $\Delta T$ , NOT on the path. So $\Delta U = nC_V \Delta T$ for any process (not just isochoric).
For mixing-water problems, $Q_{\text{lost}} = Q_{\text{gained}}$ (in an insulated container). Set up the equation symbolically before plugging in.
Carnot efficiency $\eta = 1 - T_C/T_H$ is the absolute upper limit. Always use Kelvin. Any real-engine efficiency higher than Carnot is a sign of an error.

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AP® Physics 2

Unit 9: Thermodynamics

KINETIC THEORY OF GASES

Implications:

WORKED EXAMPLE: A SEALED TANK OF HELIUM (MATHEMATICAL ROUTINES)

THE IDEAL GAS LAW

Useful special cases:

WORKED EXAMPLE: FINDING ABSOLUTE ZERO FROM DATA (EXPERIMENTAL DESIGN AND ANALYSIS)

CASE STUDY: KELVIN'S ABSOLUTE SCALE (1848)

THERMAL ENERGY AND HEAT TRANSFER

Heat (QQQ) is energy transferred due to a temperature difference. SI unit: joule. Three mechanisms

WORKED EXAMPLE: HEAT LEAKING THROUGH A WINDOW

FIRST LAW OF THERMODYNAMICS

Sign conventions (memorize):

PV DIAGRAMS AND CYCLES

WORKED EXAMPLE: FILLING IN A CYCLE TABLE (TRANSLATION BETWEEN REPRESENTATIONS)

PRACTICE: SIGNS OF Q, W, AND ΔU

CASE STUDY: JOULE'S PADDLE WHEEL (1845)

ENTROPY AND THE SECOND LAW