Important Thermodynamic Processes to Know for AP Physics 2 (2025) (AP)

What You Need to Know

Thermodynamic processes are the “standard moves” a gas can make (expand, compress, heat up, cool down) under specific constraints like constant T, P, or V. On AP Physics 2, you’re expected to:

• Read/interpret P\text{-}V diagrams and connect them to work.
• Use the First Law of Thermodynamics to relate heat, work, and internal energy.
• Recognize the “signature” equations/results for the big 4 processes: isothermal, isobaric, isochoric, adiabatic.
• Handle cycles (net work, net heat, efficiency) using those processes.

Core rules (the backbone)

• Ideal gas law: PV = nRT
• Work done by the gas (quasi-static): W = \int_{V_i}^{V_f} P\,dV
• First Law (AP convention: W is work done by the gas): \Delta U = Q - W
  • Q>0: heat into the gas
  • W>0: gas expands (does work on surroundings)

Critical: Many textbooks flip the sign convention (use \Delta U = Q + W with W done on the gas). On AP problems, be consistent and watch their wording.

Ideal-gas internal energy fact (huge shortcut)

For an ideal gas, internal energy depends only on temperature:
\Delta U = nC_V\Delta T
So if you can decide whether \Delta T=0, you instantly know \Delta U=0.

Step-by-Step Breakdown

Use this every time you see “process” or a P\text{-}V graph.

1. Identify the process constraint

   • Isothermal: T constant
   • Isobaric: P constant
   • Isochoric (isovolumetric): V constant
   • Adiabatic: Q=0 (often insulated/very fast)
   • Cyclic: returns to initial state

2. Write the right “process relationship”

   • Isothermal (ideal gas): PV = \text{const}
   • Adiabatic (reversible ideal gas): PV^{\gamma} = \text{const} where \gamma = \dfrac{C_P}{C_V}
   • Isobaric: P = \text{const}
   • Isochoric: V = \text{const}

3. Decide \Delta U using temperature

   • Use PV=nRT to see if T changes.
   • Then apply \Delta U = nC_V\Delta T.

4. Compute work W from the path

   • Graph given? Work is area under the curve on a P\text{-}V diagram.
   • If constant pressure: W = P\Delta V
   • If constant volume: W=0
   • If isothermal ideal gas: W = nRT\ln\!\left(\dfrac{V_f}{V_i}\right)
   • If adiabatic reversible ideal gas: use one of the adiabatic work forms (see formulas section).

5. Finish with the First Law
   Q = \Delta U + W

   • Often easiest: find \Delta U and W, then get Q.

Mini worked walk-through (annotated)

“An ideal gas expands isothermally from V_i to V_f at temperature T.”

• Isothermal \Rightarrow \Delta T=0 \Rightarrow \Delta U=0
• Work: W = nRT\ln\!\left(\dfrac{V_f}{V_i}\right)
• First Law: Q = \Delta U + W = W
  So for an isothermal ideal-gas expansion, all heat in becomes work out.

Key Formulas, Rules & Facts

Process “cheat table” (ideal gas)

Heat capacities + useful identities

• Mayer’s relation (ideal gas): C_P - C_V = R
• Ratio: \gamma = \dfrac{C_P}{C_V} (always >1)
• Internal energy change (ideal gas): \Delta U = nC_V\Delta T

On AP, you’re often given or can assume the gas is ideal. If it’s explicitly monatomic, then C_V=\tfrac{3}{2}R and C_P=\tfrac{5}{2}R and \gamma=\tfrac{5}{3}.

P\text{-}V diagram facts you must use correctly

• Work done by the gas equals area under the curve:
  W = \int P\,dV
• Expansion to the right: \Delta V>0 \Rightarrow W>0
• Compression to the left: \Delta V

Adiabatic relationships (reversible) — know the trio

For an ideal gas undergoing a reversible adiabatic process:

• PV^{\gamma}=\text{const}
• TV^{\gamma-1}=\text{const}
• T^{\gamma}P^{1-\gamma}=\text{const} (equivalently TP^{\frac{1-\gamma}{\gamma}}=\text{const})

Warning: PV^{\gamma}=\text{const} is not for just “any insulated change.” It assumes a reversible (quasi-static) adiabatic path.

Examples & Applications

Example 1: Isochoric heating (vertical line on P\text{-}V)

Setup: A sealed rigid tank (constant V) containing n moles of ideal gas is heated so temperature rises by \Delta T.

• Isochoric \Rightarrow W=0
• \Delta U = nC_V\Delta T
• First Law: Q = \Delta U + W = nC_V\Delta T
  Key insight: At constant volume, heat only increases internal energy (no boundary work).

Example 2: Isobaric expansion (horizontal line)

Setup: Gas expands at constant pressure P from V_i to V_f.

• Work: W = P\left(V_f - V_i\right)
• Temperature change from ideal gas law: \Delta T = \dfrac{P\left(V_f-V_i\right)}{nR}
• Internal energy: \Delta U=nC_V\Delta T
• Heat in: Q = \Delta U + W = nC_P\Delta T
  Key insight: At constant pressure, heat goes into both raising U and doing expansion work.

Example 3: Isothermal expansion (curved hyperbola)

Setup: Ideal gas expands isothermally at temperature T, doubling its volume.

• \Delta U=0
• Work: W = nRT\ln\!\left(\dfrac{2V_i}{V_i}\right)=nRT\ln 2
• Heat: Q=W=nRT\ln 2
  Exam variation: They might give a P\text{-}V graph and ask for a comparison: isothermal curve is less steep than adiabatic during expansion.

Example 4: Adiabatic compression (steep curve)

Setup: Ideal gas is compressed adiabatically (reversible) so volume decreases.

• Adiabatic \Rightarrow Q=0
• Compression \Rightarrow W

Common Mistakes & Traps

1. Mixing up sign conventions for the First Law

   • Wrong move: Using \Delta U = Q + W while also taking W as work done by the gas.
   • Fix: If you use \Delta U = Q - W, then W>0 for expansion.

2. Using W=P\Delta V when pressure isn’t constant

   • Wrong move: Treating any expansion as constant pressure.
   • Fix: Only use W=P\Delta V for **isobaric** processes. Otherwise use W=\int P\,dV or geometry/known formula.

3. Assuming “adiabatic” means “temperature constant”

   • Wrong move: Thinking Q=0 \Rightarrow \Delta T=0.
   • Why wrong: Adiabatic means no heat transfer; temperature often changes because work changes internal energy.
   • Fix: Use \Delta U=nC_V\Delta T and \Delta U = -W (since Q=0).

4. Assuming “isothermal” means “no heat transfer”

   • Wrong move: Setting Q=0 because \Delta T=0.
   • Fix: For isothermal ideal gas, \Delta U=0, so Q=W (usually nonzero).

5. Using PV^{\gamma}=\text{const} for any adiabatic process

   • Wrong move: Applying it to rapid, irreversible processes or free expansion.
   • Fix: That relation is for reversible adiabatic paths. If not reversible, lean on Q=0 and the First Law, not the reversible curve equation.

6. Forgetting that on a full cycle \Delta U_{\text{net}}=0

   • Wrong move: Adding internal energy changes across a cycle and getting nonzero.
   • Fix: State variables (like U and T) return to start, so Q_{\text{net}}=W_{\text{net}}.

7. Reading P\text{-}V graph area incorrectly

   • Wrong move: Using “area under the curve” but mixing up whether you use the region to the axes or the loop area.
   • Fix:
     • Single path: W is area under that path to the V-axis.
     • Closed loop: W_{\text{net}} is area enclosed by the loop (sign from direction).

8. Confusing state variables vs path variables

   • Wrong move: Treating Q or W like they depend only on endpoints.
   • Fix: \Delta U depends only on endpoints; Q and W depend on the path/process.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can write and use: \Delta U = Q - W and PV=nRT.
• You remember: W = \int P\,dV and “area under curve = work.”
• For isothermal ideal gas: \Delta U=0 and Q=W=nRT\ln\!\left(\dfrac{V_f}{V_i}\right).
• For isochoric: W=0 and Q=\Delta U=nC_V\Delta T.
• For isobaric: W=P\Delta V and Q=nC_P\Delta T.
• For adiabatic: Q=0 and (reversible) PV^{\gamma}=\text{const} plus \Delta U=-W.
• On a cycle: \Delta U_{\text{net}}=0 so Q_{\text{net}}=W_{\text{net}}; direction sets the sign.
• You’re alert for traps: not every expansion is isobaric; not every adiabatic obeys PV^{\gamma}.

You’ve got this—identify the process first, then let \Delta U and the P\text{-}V area do most of the work for you.