Thermodynamics — AP Physics 2 Phys 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Temperature and absolute scale, specific heat and latent heat, ideal gas law, first and second laws of thermodynamics, and PV diagram work calculations aligned to the College Board AP Physics 2 CED.

You should already know: AP Physics 1 or equivalent.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Thermodynamics?

Thermodynamics is the branch of physics that studies the relationships between heat, work, temperature, and energy, focusing on how energy transfers between a defined system and its surroundings, and how energy is converted between different forms. It builds on your AP Physics 1 knowledge of work and conservation of energy, applying these rules to macroscopic systems like gases, liquids, and solid materials. Per the College Board CED, thermodynamics makes up 15-20% of your AP Physics 2 exam score, appearing in both multiple-choice and multi-part free-response questions. It is sometimes referred to as thermal physics, though thermodynamics specifically emphasizes energy conversion processes like those in heat engines, refrigerators, and gas systems.

2. Temperature and absolute scale

Temperature is a measure of the average translational kinetic energy of the particles in a system, and it dictates the direction of spontaneous heat transfer: heat always flows from a higher-temperature system to a lower-temperature system unless external work is applied. Three temperature scales are commonly referenced, but only the absolute Kelvin scale is valid for thermodynamic calculations:

• Celsius (°C): Defined by the freezing (0°C) and boiling (100°C) points of water at 1 atm of pressure. Negative values are common for everyday conditions.
• Fahrenheit (°F): Used primarily in the US, not required for AP Physics 2 calculations.
• Kelvin (K): The absolute temperature scale, where 0 K (absolute zero) is the theoretical temperature at which all molecular motion stops. No negative values exist on this scale.

The conversion between Celsius and Kelvin is: $$T(K)=T{(}^{\circ }C)+273.15$$ You may use 273 instead of 273.15 for all AP exam calculations with no point deduction.

Worked Example

A lab report lists the temperature of a carbon dioxide sample as -33°C. What is its temperature in Kelvin, appropriate for use in thermodynamic calculations? Solution: $T(K)=-33+273=240K$ Exam tip: Examiners frequently test this conversion as a hidden trap: if you use Celsius values in the ideal gas law or thermodynamic energy calculations, you will get a drastically incorrect answer, so always convert temperature first.

3. Specific heat capacity and latent heat

Heat ($Q$) is energy transferred between systems due to a temperature difference, measured in Joules (J). Two key relationships describe how heat transfer affects a system:

1. Specific heat capacity ($c$): The amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1°C, since temperature changes are identical on both scales). The formula for heat transfer that causes a temperature change is: $$Q=mc\Delta T$$ Where $m$ is mass in kg, and $\Delta T$ is the change in temperature in K or °C. Units of $c$ are $J/(kg\cdot K)$.
2. Specific latent heat ($L$): When a substance undergoes a phase change (melting, freezing, vaporization, condensation), heat transfer breaks or forms intermolecular bonds instead of changing temperature. The formula for heat transfer during a phase change is: $$Q=mL$$ $L_f$ (latent heat of fusion) is the heat required to melt 1 kg of solid to liquid at constant temperature, while $L_v$ (latent heat of vaporization) is the heat required to vaporize 1 kg of liquid to gas at constant temperature.

Worked Example

How much total heat is required to convert 0.2 kg of ice at -20°C to liquid water at 30°C? Use $c_{ice}=2100J/(kg\cdot K)$, $L_{f,water}=3.34\times 10^{5}J/kg$, $c_{water}=4186J/(kg\cdot K)$. Solution: Split the process into 3 discrete steps:

1. Heat ice from -20°C to 0°C: $Q_1=0.2\ast 2100\ast 20=8400J$
2. Melt ice at constant 0°C: $Q_2=0.2\ast 3.34\times 10^5=66800J$
3. Heat liquid water from 0°C to 30°C: $Q_3=0.2\ast 4186\ast 30=25116J$ Total heat: $Q_{total}=8400+66800+25116=100316J\approx 1.0\times 10^{5}J$ Exam note: AP free-response questions often ask you to interpret or draw heating curves: the flat horizontal regions of these curves correspond to phase changes where temperature is constant.

4. Ideal gas law

The ideal gas law describes the behavior of ideal gases, which follow four core assumptions: (1) gas molecules are point particles with negligible volume, (2) no intermolecular attractive or repulsive forces exist between molecules, (3) all collisions between molecules and container walls are perfectly elastic, (4) molecules move in random straight lines between collisions. These assumptions hold for most real gases at low pressure and moderate temperature. The law combines Boyle’s Law ($P\propto 1/V$ at constant $n,T$), Charles’ Law ($V\propto T$ at constant $n,P$), and Avogadro’s Law ($V\propto n$ at constant $P,T$) into two standard forms:

1. Molar form (for calculations using moles of gas): $$PV=nRT$$ Where $P$ = pressure in Pascals (Pa), $V$ = volume in $m^3$, $n$ = number of moles, $R=8.31J/(mol\cdot K)$ (universal gas constant), $T$ = absolute temperature in K.
2. Molecular form (for calculations using number of individual molecules): $$PV=N{k}_{B}T$$ Where $N$ = number of molecules, $k_B=1.38\times 10^{-23}J/K$ (Boltzmann constant). A key relation from kinetic molecular theory links temperature directly to particle motion: the average kinetic energy per gas molecule is $⟨KE⟩=\frac{3}{2}{k}_{B}T$.

Worked Example

A 0.01 $m^3$ sealed tank holds 0.5 mol of helium gas at a temperature of 127°C. What is the pressure inside the tank? Solution: First convert temperature to Kelvin: $T=127+273=400K$. Rearrange the ideal gas law to solve for $P$: $$P=\frac{nRT}{V}=\frac{0.5\ast 8.31\ast 400}{0.01}=166200Pa\approx 1.7\times 10^{5}Pa$$ Exam trap: Always convert volume from liters to cubic meters if using SI units: $1L=0.001m^3$. Using liters with the standard value of $R$ will give you an incorrect pressure by a factor of 1000.

5. First and second laws of thermodynamics

First Law of Thermodynamics

This is the law of conservation of energy applied to thermal systems, stated as: $$\Delta U=Q-W$$ Where $\Delta U$ is the change in internal energy of the system, $Q$ is heat added to the system, and $W$ is work done by the system on the surroundings. The sign conventions are critical for AP Physics 2:

• $Q>0$ if heat flows into the system, $Q<0$ if heat flows out of the system
• $W>0$ if the system does work on its surroundings (e.g., gas expands pushing a piston out), $W<0$ if work is done on the system (e.g., piston is pushed in compressing the gas)

Second Law of Thermodynamics

This law defines the direction of spontaneous processes, stated in two testable forms for AP Physics 2:

1. Clausius statement: Heat cannot spontaneously flow from a colder body to a hotter body without input of external work (the operating principle of refrigerators).
2. Kelvin-Planck statement: No heat engine operating in a cycle can convert 100% of input heat to work; some heat will always be lost to the low-temperature reservoir. The second law can also be stated using entropy ($S$), a measure of the disorder of a system: the total entropy of an isolated system always increases over time, $\Delta S_{total}\ge 0$, with equality only for perfectly reversible processes.

Worked Example

A gas absorbs 300 J of heat from its surroundings, and 100 J of work is done on the gas to compress it. What is the change in internal energy of the gas? Solution: $Q=+300J$ (heat added to system), $W=-100J$ (work done on the system means work done by the system is negative). Apply first law: $$\Delta U=300J-(-100J)=400J$$ The internal energy of the gas increases by 400 J.

6. PV diagrams and work

A pressure-volume (PV) diagram plots the pressure of a gas against its volume as it undergoes thermodynamic processes. The area under the curve between two volume values is equal to the work done by the gas during that process: $$W={\int }_{V_1}^{V_2}PdV$$ For an isobaric (constant pressure) process, this simplifies to $W=P\Delta V$, where $\Delta V=V_2-V_1$. Four common processes you will be asked to identify on PV diagrams:

1. Isobaric: Constant pressure, horizontal line on PV diagram, $W=P\Delta V$
2. Isochoric (isovolumetric): Constant volume, vertical line on PV diagram, $\Delta V=0$ so $W=0$, all heat transfer changes internal energy
3. Isothermal: Constant temperature, hyperbolic curve ($PV=constant$), $\Delta U=0$ for ideal gases so $Q=W$
4. Adiabatic: No heat transfer, $Q=0$, steeper curve than isothermal, $\Delta U=-W$

Worked Example

A gas undergoes the cycle shown below: isobaric expansion from $V=0.02m^3$ to $V=0.05m^3$ at $P=2\times 10^{5}Pa$, then isochoric cooling to $P=1\times 10^{5}Pa$, then isobaric compression back to $V=0.02m^3$, then isochoric heating back to the initial state. What is the net work done by the gas per cycle? Solution: Net work is the area enclosed by the cycle. The cycle forms a rectangle with height $\Delta P=1\times 10^{5}Pa$ and width $\Delta V=0.03m^3$. $$W_{net}=\Delta P\times \Delta V=1e5\ast 0.03=3000J$$ Exam tip: If volume increases during a process, work done by the gas is positive; if volume decreases, work done by the gas is negative.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using Celsius instead of Kelvin in ideal gas law or thermodynamic calculations. Why you might do it: You forget the conversion step, or assume temperature changes are the same on both scales. Correct move: Always convert to Kelvin first, even if the question only gives temperature differences; ratios of temperature (e.g., doubling temperature to double pressure) require absolute values.
• Wrong move: Mixing up the sign convention for work in the first law. Why you might do it: Other curricula use the opposite convention where $W$ is work done on the system. Correct move: Memorize the AP Physics 2 rule: $\Delta U=Q-W$, where $W$ is work done by the system: expanding gas = positive $W$, compressed gas = negative $W$.
• Wrong move: Using $Q=mc\Delta T$ for phase change calculations. Why you might do it: You associate all heat transfer with temperature change. Correct move: Split heating/cooling problems into discrete steps: temperature change steps use $Q=mc\Delta T$, phase change steps use $Q=mL$, then add all heat values together.
• Wrong move: Calculating work from PV diagrams as area above the curve instead of below it. Why you might do it: You mix up work done on vs by the system. Correct move: Area under the P-V curve = work done by the gas; for closed cycles, the area inside the loop is net work per cycle.
• Wrong move: Assuming ideal gas behavior applies to all gas systems. Why you might do it: Most exam problems use ideal gases. Correct move: Real gases deviate from ideal behavior at high pressure and low temperature, where molecular volume and intermolecular forces become significant.

8. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A 2 kg block of iron ($c=450J/(kg\cdot K)$) at 200°C is placed in an insulated container with 5 kg of water ($c=4186J/(kg\cdot K)$) at 20°C. What is the approximate final equilibrium temperature of the system? A) 25°C B) 45°C C) 75°C D) 110°C Solution: Heat lost by iron = heat gained by water, no heat lost to surroundings: $$m_{Fe}{c}_{Fe}(200-T_f)=m_{water}{c}_{water}(T_f-20)$$ $$2\ast 450\ast (200-T_f)=5\ast 4186\ast (T_f-20)$$ $$900\ast (200-T_f)=20930\ast (T_f-20)$$ $$180000-900T_f=20930T_f-418600$$ $$598600=21830T_f⟹T_f\approx 27.4°C$$ Closest to 25°C, so answer A.

Question 2 (Free Response Part A)

A 0.2 mol sample of ideal nitrogen gas is held at a constant volume of 0.005 $m^3$ at 300 K. Calculate the pressure of the gas. Solution: Use $PV=nRT$, rearrange for $P$: $$P=\frac{nRT}{V}=\frac{0.2\ast 8.31\ast 300}{0.005}=99720Pa\approx 1.0\times 10^{5}Pa$$

Question 3 (Free Response Part B)

The gas from Question 2 is heated at constant volume until its pressure doubles. (i) What is the new temperature of the gas? (ii) How much work is done by the gas during this process? Justify your answer. Solution: (i) Constant volume, so $P/T=constant$. If $P$ doubles, $T$ doubles: $T_{new}=2\ast 300=600K$, or 327°C. (ii) Work done by the gas is 0 J. The process is isochoric (constant volume), so $\Delta V=0$, and $W=P\Delta V=0$.

9. Quick Reference Cheatsheet

Key Rules

1. Heat flows spontaneously from higher to lower temperature only.
2. Total entropy of an isolated system always increases; no heat engine is 100% efficient.
3. Common PV processes: isobaric (constant P), isochoric (constant V, W=0), isothermal (constant T, ΔU=0), adiabatic (Q=0).

10. What's Next

Thermodynamics is a foundational cross-cutting topic for the rest of the AP Physics 2 curriculum. It directly connects to later units including fluid dynamics (where pressure and temperature relationships govern buoyancy and flow behavior), electricity and magnetism (where thermal energy loss in resistive circuits is a key application of heat transfer), and modern physics (where thermodynamics principles apply to nuclear reactions and stellar energy production). You will also see thermodynamics concepts integrated into multi-topic free-response questions on the AP exam, so mastering the sign conventions and formula relationships here will save you significant points on later topics.

If you struggle with any of the concepts, sign conventions, or practice problems in this guide, you can ask Ollie for personalized explanations, additional practice questions, or step-by-step walkthroughs of any problem type at any time by visiting the homepage. You can also access our full library of AP Physics 2 study guides, full-length practice tests, and exam strategy tips to help you earn a 5 on your test date.