First Law Problem Patterns (IPhO)
IPhO thermodynamics lesson on first-law bookkeeping: system choice, sign conventions, and common process templates.
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Most thermo marks are earned (or lost) before you do any algebra: system choice, sign convention, and a clean first-law line. This lesson turns common IPhO setups into reusable bookkeeping patterns.
Mind-stretcher feedback
- For a gas-only system, frictional energy delivered to the piston leaves as work and any return heating enters as heat; for gas plus piston, the frictional heating is an internal conversion. The terms change with the boundary, while total energy accounting does not.
- An electrical heater inside the gas may be counted as electrical work across the system boundary rather than heat. State the convention and boundary; moving the same transfer between Q and W must not change Δ U.
- Q is positive when heat flows into the system.
- W is positive when the system does work on the surroundings (e.g. expansion work).
- First law: Δ U = Q - W.
1. Definitions (Must Know)
- System / surroundings: decide what you track energy for. A “good” system makes unknown transfers disappear (for example, choose “gas + piston” if piston friction is central).
- State function: depends only on the state, not the path. Key ones: U, H, T, P, V.
- Path function: depends on the process. Key ones: Q and W.
- First law (closed system): Δ U = Q - W.
- P dV work (quasi-static): W = ∫ P dV (work done by the system). For non-quasi-static problems, use W = ∫ Pₑₓₜ dV.
- Enthalpy: H = U + PV. For a process at constant external pressure with only P dV work, Q = Δ H.
- Ideal gas internal energy: Δ U = nC_VΔ T (depends only on Δ T).
2. Key Ideas (What Earns Marks)
- Start with words, then one equation: “System = gas”, “quasi-static”, “ideal gas”, “insulated”, then write Δ U = Q - W.
- Use state functions to skip paths: if you know initial and final states, compute Δ U (and sometimes Δ H) without caring how the process happened.
- Compute W from the process geometry: on a P-V diagram, W is the signed area under the curve.
- Prefer ratios and differences: in multi-step problems, solve each step with the same sign convention, then sum Q, W, and Δ U.
- Always do one physical check: expansion should give W positive (with this convention), compression gives W negative.
3. Detailed Explanations
A. The 20-second setup (what to write every time)
- System: gas only, or gas plus container, or gas plus piston plus weights.
- Known constraints: constant V, constant P, constant T, insulated (Q = 0), in contact with a reservoir (T fixed), etc.
- First law: Δ U = Q - W.
- A state equation (if allowed): ideal gas PV = nRT and Δ U = nC_VΔ T.
B. Work: which pressure goes into ∫ P dV?
- Reversible / quasi-static: the gas stays close to equilibrium, so you can use the gas pressure P in W = ∫ P dV.
- Irreversible but with known external pressure: use W = ∫ Pₑₓₜ dV. This is where many IPhO “sudden expansion” and “piston with weights” traps live.
Quick pattern: constant external pressure
If Pₑₓₜ is constant, then W = Pₑₓₜ(V_f - Vᵢ). You do not need the gas path to get W.
C. High-yield process templates (ideal gas)
| Process | What stays fixed | Fast facts |
|---|---|---|
| Isochoric | V | W = 0, so Q = Δ U = nC_VΔ T |
| Isothermal | T | Δ U = 0, so Q = W and W = nRT ln(V_f/Vᵢ) (reversible) |
| Isobaric | P | W = PΔ V = nRΔ T, Q = Δ H = nC_PΔ T |
| Adiabatic (reversible) | Q = 0 | Δ U = -W = nC_VΔ T, plus PV^γ = const |
Derivation: isothermal ideal-gas work (reversible)
For an isothermal ideal gas, P = nRT/V.
W = ∫_Vᵢ^(V_f)P dV = nRT∫_Vᵢ^(V_f)dV/V = nRT ln(V_f/Vᵢ).
D. Cycles: the fastest marks in thermo
- For a complete cycle, the system returns to its initial state, so Δ U_cycle = 0.
- Therefore, Qₙₑₜ = Wₙₑₜ (with this sign convention).
- On a P-V diagram, Wₙₑₜ is the signed enclosed area (clockwise cycle gives positive net work by the gas).
E. “System choice” patterns that remove pain
- Gas only is best when the question asks for Q, W, Δ U for the gas and gives a clean process constraint.
- Gas + piston is best when friction or piston kinetic energy is mentioned; then “mechanical dissipation” stays inside the system energy bookkeeping instead of appearing as a mysterious heat leak.
- Gas + surroundings is sometimes best in “sudden mixing” or “free expansion” questions, where internal transfers cancel and only external work matters.
4. Common Mistakes
- Writing the first law before defining the sign convention (and then silently switching conventions mid-solution).
- Using W = ∫ P dV with the gas pressure when the process is not quasi-static and only Pₑₓₜ is justified.
- Treating Q = 0 as “temperature stays constant” (adiabatic does not mean isothermal).
- Forgetting that U for an ideal gas depends only on T, so Δ U ignores the path.
- Mixing up C_P and C_V, or using Q = nC_VΔ T for an isobaric process where Q = nC_PΔ T is the clean route.
- Dropping a sign on compression versus expansion when evaluating V_f-Vᵢ.
5. Exam Tips
- Spend a line on the setup: “system, constraints, sign convention, one diagram”. It is usually worth more than one page of algebra.
- If you see “cycle”, write Δ U = 0 immediately and aim for area on the P-V plot.
- If you see “constant pressure”, consider Δ H and Q = Δ H (PV-only work).
- If numbers are ugly, keep everything symbolic until the end: W = nRT ln(V_f/Vᵢ) is safer than early decimals.
- Do a sanity check with words: “expansion does work on the outside”, “heating increases T so U should increase for an ideal gas”.
6. Worked Examples
Example 1: Isobaric heating (ideal gas)
One mole of a monatomic ideal gas is heated at constant pressure from Tᵢ = 300 K to T_f = 450 K. Find W, Δ U, and Q.
- Use C_V = (3/2)R and C_P = (5/2)R.
- Work: W = nRΔ T = R(150 K).
- Internal energy: Δ U = nC_VΔ T = (3/2)R(150 K).
- Heat: Q = Δ U + W = nC_PΔ T = (5/2)R(150 K).
Example 2: Isochoric then isothermal (bookkeeping by steps)
One mole of ideal gas goes from state A to B at constant volume, increasing temperature from T to 2T. Then it goes from B to C isothermally at temperature 2T, doubling its volume. Find total W and total Q.
- Step A to B (isochoric): W_AB = 0, so Q_AB = Δ U_AB = nC_V(T).
- Step B to C (isothermal at 2T): Δ U_BC = 0, so Q_BC = W_BC = nR(2T) ln 2 (reversible).
- Totals: W = W_AB + W_BC = 2nRT ln 2, and Q = Q_AB + Q_BC = nC_VT + 2nRT ln 2.
Example 3: A rectangular P-V cycle (use area)
A cycle consists of two isochoric legs and two isobaric legs, forming a rectangle on the P-V diagram between pressures P₁ and P₂ and volumes V₁ and V₂.
- Net work by the gas over one cycle is the enclosed area: Wₙₑₜ = (P₂-P₁)(V₂-V₁) (clockwise orientation).
- Because it is a cycle, Δ Uₙₑₜ = 0, so Qₙₑₜ = Wₙₑₜ.
7. Mind Stretchers
- Free expansion into vacuum (insulated): argue what W is, then use Δ U = Q-W to infer what happens to U and (for an ideal gas) to T.
- Piston with friction: if the gas expands and the piston heats up due to friction, where does that energy appear in the first-law terms for different system choices?
- Two-step “same endpoints” trap: compare Q for two different paths between the same states; explain why Δ U is the same but Q is not.
- Non-P dV work: add an electrical heater inside the gas. How does that change what you call Q versus W?
8. Practice
- Drill these four templates until automatic: isochoric, isothermal (ideal gas), isobaric, adiabatic (Q = 0).
- For each past-paper thermo problem, force yourself to write: system, constraints, sign convention, and a P-V sketch before equations.
Syllabus and review details
No official syllabus alignment is listed for this lesson.