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: $\Delta V<0 \Rightarrow W<0$
• On a cycle, the enclosed area is $|W_{\text{net}}|$
  • Clockwise loop: $W_{\text{net}}>0$ (engine)
  • Counterclockwise loop: $W_{\text{net}}<0$ (refrigerator/heat pump)

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 $\Rightarrow W<0$
• First Law: $\Delta U= -W >0$ so temperature rises.
• Use $TV^{\gamma-1}=\text{const}$ to relate temperature change:
  $T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}$
  Key insight: In adiabatic compression, work done on the gas becomes internal energy (heats it up without heat transfer).

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.