Selecting Thermodynamic Functions — Select appropriate thermodynamic functions for a problem (ΔU vs ΔH vs ΔG) - Physical Chemistry 1

SECTION 1

Historical Context & Motivation

Thermodynamics did not emerge as a unified theory overnight; it grew from a series of practical and philosophical questions about heat, work, and the direction of natural change. The very notion that one might need different mathematical functions to describe energy changes under different constraints was itself a hard-won insight. Early engineers studying steam engines cared primarily about the total heat absorbed or released, while physicists probing fundamental laws wanted a quantity that captured the internal energy of a system independently of how energy was transferred. The realization that spontaneity required yet a third function — one that married energy with entropy — came only after decades of debate and experiment.

1840s • Joule and the Mechanical Equivalent of Heat

James Prescott Joule's paddle-wheel experiments demonstrated that mechanical work and heat are interconvertible, establishing the concept of internal energy (U) and laying the groundwork for the First Law of Thermodynamics: ΔU = q + w.

1850s • Clausius and the Second Law

Rudolf Clausius formalized the concept of entropy (S) and articulated the Second Law, showing that not all energy changes are equally useful — the direction of spontaneous change matters.

1870s • Gibbs and Thermodynamic Potentials

Josiah Willard Gibbs introduced the Gibbs free energy (G = H − TS) and systematized the use of Legendre transforms to generate new state functions suited to specific experimental constraints. His framework remains the backbone of chemical thermodynamics.

1882 • Helmholtz Free Energy

Hermann von Helmholtz articulated the Helmholtz free energy (A = U − TS), providing a spontaneity criterion for constant-temperature, constant-volume processes — essential for gas-phase reactions in rigid vessels.

1923 • Lewis and Randall's Textbook

Gilbert N. Lewis and Merle Randall published Thermodynamics and the Free Energy of Chemical Substances, popularizing the practical use of ΔG for predicting chemical equilibria and making thermodynamic function selection a routine part of problem-solving.

The central question this lesson addresses is deceptively simple: given a particular thermodynamic scenario, which state function — ΔU, ΔH, ΔG, or ΔA — should you choose as the starting point for your analysis? Getting this choice right is the single most important step in any thermodynamic calculation, because each function is naturally paired with a specific set of constraints, and using the wrong function leads to unnecessary complexity or outright errors.

SECTION 2

Core Principles & Definitions

Before selecting among thermodynamic functions, you must understand what each one physically represents and under what constraints it simplifies. All four functions — U, H, A, and G — are state functions, meaning their changes depend only on the initial and final states of the system, not on the path. The differences among them arise from the natural variables each function employs — the independent variables in terms of which the function's total differential is cleanly expressed without path-dependent quantities.

Internal Energy (U)

The total microscopic kinetic and potential energy of a system. Its natural variables are S and V (entropy and volume). Under constant volume with no non-PV work, ΔU = q_V.

Enthalpy (H = U + PV)

A Legendre transform of U that replaces V with P as a natural variable. Natural variables: S and P. Under constant pressure with no non-PV work, ΔH = q_P — the heat flow measured by calorimetry.

Helmholtz Free Energy (A = U − TS)

Replaces S with T as a natural variable via Legendre transform. Natural variables: T and V. At constant T and V, ΔA ≤ 0 for a spontaneous process; ΔA equals the maximum work extractable from the system.

Gibbs Free Energy (G = H − TS)

Natural variables: T and P. Since most chemistry occurs in open vessels (constant T, constant P), ΔG < 0 is the standard spontaneity criterion. ΔG also equals the maximum non-expansion work available from a process.

The Selection Principle

Choose the function whose natural variables match the constraints of the problem. If temperature and pressure are held constant, use G. If temperature and volume are fixed, use A. If the process is adiabatic at constant volume, use U. Matching constraints to natural variables eliminates cross-terms and simplifies the math.

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KEY TAKEAWAY

Think of thermodynamic functions like coordinate systems in mechanics. You can solve a pendulum problem in Cartesian coordinates, but polar coordinates make it trivial because they align with the system's constraints. In exactly the same way, choosing the thermodynamic potential whose natural variables match your experimental constraints turns a messy calculation into a clean one. ΔG is to constant-T-and-P problems what polar coordinates are to circular motion.

SECTION 3

Visual Explanation — The Decision Flowchart

The following decision flowchart provides a systematic approach to selecting the appropriate thermodynamic function. Begin at the top and answer the constraint questions sequentially: the path you follow determines which state function is optimal for your problem. This diagram encodes the logic that experienced physical chemists apply instinctively.

Decision flowchart for selecting thermodynamic functions. Begin at the top by identifying what the problem asks (energy/heat vs. spontaneity), then follow the branches corresponding to your system's constraints. The colored boxes at the termini indicate the appropriate state function.

The flowchart encodes two fundamental branching decisions. The first branch distinguishes between problems that ask for energy bookkeeping (how much heat flows, what is the energy change) and problems that ask about spontaneity or equilibrium. For energy bookkeeping, the constraint on mechanical variables (V or P) determines whether ΔU or ΔH is simpler. For spontaneity at constant temperature, the mechanical constraint selects ΔA or ΔG. In cases where temperature is not constant, the free energies lose their direct spontaneity interpretation and you must revert to the entropy criterion ΔS_total ≥ 0.

SECTION 4

Mathematical Framework

The mathematical justification for selecting specific thermodynamic functions rests on the concept of Legendre transforms and the behavior of exact differentials under constraints. Starting from the fundamental relation for internal energy, each successive function is obtained by trading one natural variable for its conjugate, producing a potential whose differential is naturally expressed in terms of experimentally controllable quantities.

FUNDAMENTAL RELATION — INTERNAL ENERGY

dU = TdS − PdV + Σᵢ μᵢ dnᵢ

Natural variables: S, V, {n_i}. At constant V and closed system (dn_i = 0), dU = TdS = δq_rev, so ΔU = q_V for any process (reversible or not).

ENTHALPY VIA LEGENDRE TRANSFORM

H ≡ U + PV → dH = TdS + VdP + Σᵢ μᵢ dnᵢ

Natural variables: S, P, {n_i}. At constant P (closed system), dH = TdS = δq_rev, yielding ΔH = q_P. This is why calorimeters at atmospheric pressure directly measure ΔH.

HELMHOLTZ FREE ENERGY

A ≡ U − TS → dA = −SdT − PdV + Σᵢ μᵢ dnᵢ

Natural variables: T, V, {n_i}. At constant T and V (closed), dA ≤ 0 for a spontaneous process. The inequality arises because dA = dU − TdS and the Second Law requires dS_total ≥ 0.

GIBBS FREE ENERGY

G ≡ H − TS → dG = −SdT + VdP + Σᵢ μᵢ dnᵢ

Natural variables: T, P, {n_i}. At constant T and P (closed), dG ≤ 0 for a spontaneous process. Since most laboratory reactions occur at constant T and P, ΔG is the workhorse function of chemical thermodynamics.

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WHY LEGENDRE TRANSFORMS MATTER

A Legendre transform does not create new physics — it repackages the same information into a form that is more convenient for a given set of constraints. In dU = TdS − PdV, the entropy S is a natural variable, but entropy is hard to control experimentally. By defining A = U − TS, we trade S for T (which we can control via a thermostat). The transform ensures that the new potential A has a clean differential in T and V with no leftover cross-terms — exactly what you need for a constant-T, constant-V problem.

SECTION 5

Constraint-to-Function Mapping

In practice, selecting the right function requires you to read the problem statement for constraint keywords and then map those constraints to the appropriate potential. The table below catalogues the most common physical scenarios you will encounter and the function best suited to each. Note that when non-PV work (electrical, surface, etc.) is present, the equality signs in the heat relations become inequalities, and the free-energy change equals the maximum non-PV work.

Scenario / Keywords	Constraints	Best Function	Key Relationship
Bomb calorimetry; rigid container; sealed vessel	Constant V, no w_non-PV	ΔU	ΔU = q_V = nC_VΔT
Coffee-cup calorimetry; open beaker; atmospheric pressure	Constant P, no w_non-PV	ΔH	ΔH = q_P = nC_PΔT
Phase transition at 1 atm; standard reaction enthalpy	Constant T, constant P	ΔH (for heat); ΔG (for spontaneity)	ΔG = ΔH − TΔS
Gas in a rigid isothermal vessel; explosion in closed container	Constant T, constant V	ΔA	ΔA = ΔU − TΔS ≤ 0 (spontaneous)
Electrochemical cell; maximum electrical work	Constant T, constant P, non-PV work	ΔG	ΔG = w_non-PV,max = −nFE
Chemical equilibrium constant	Constant T, constant P	ΔG°	ΔG° = −RT ln K
Adiabatic, constant-V process	q = 0, constant V	ΔU	ΔU = w (only PV work = 0 at const V, so ΔU = 0 for ideal gas)

Common scenarios and the thermodynamic function best suited to each.

The four principal thermodynamic potentials arranged in a Legendre-transform network. Dashed arrows indicate which natural variable is swapped and the corresponding term added or subtracted. U sits at the top with natural variables (S, V); H replaces V with P; A replaces S with T; and G replaces both, giving natural variables (T, P).

The Legendre-transform network above reveals a satisfying symmetry: U occupies the top node with the most fundamental (but least experimentally convenient) natural variables S and V. Moving left replaces V with P to obtain H; moving right replaces S with T to obtain A. Both paths converge at G, which has the most experimentally convenient pair, T and P. This is precisely why ΔG dominates chemical thermodynamics — temperature and pressure are the variables we most often control in the laboratory.

SECTION 6

Worked Example

Consider the following multi-part problem that requires careful function selection at each stage. A sample of 2.00 mol of an ideal gas initially at 300 K and 1.00 atm undergoes two sequential processes: (a) an isothermal, reversible expansion to twice the original volume in a piston-cylinder apparatus open to the atmosphere, and (b) the resulting gas is then transferred to a rigid, sealed bomb calorimeter and cooled to 250 K. For each step, determine the appropriate thermodynamic function and calculate the relevant energy change.

Selecting and Computing Thermodynamic Functions

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Step 1 — Analyze Constraints for Process (a)

The isothermal expansion occurs in a piston-cylinder apparatus, meaning pressure can change while the piston moves. The gas expands against a variable external pressure (reversible). Since the gas is in a piston open to the atmosphere, the pressure is not constant during expansion (the gas pressure changes as volume doubles), but the process is isothermal (constant T). The problem asks for an energy change. For an ideal gas undergoing an isothermal process, ΔU = 0 (internal energy depends only on T for an ideal gas). The heat absorbed equals the work done: q = −w. To find q or w, we use the First Law directly. However, if asked about the enthalpy change, ΔH = 0 as well for an ideal gas at constant T (since H = U + PV = U + nRT, and nRT is constant). The key point: for an ideal gas isothermal process, ΔU = ΔH = 0, so the interesting quantity is the work or heat.

ΔU = 0, ΔH = 0 for process (a). Use First Law: w = −nRT ln(V₂/V₁).
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Step 2 — Calculate Work for Process (a)

For a reversible isothermal expansion of an ideal gas: w_rev = −nRT ln(V₂/V₁). With n = 2.00 mol, R = 8.314 J mol⁻¹ K⁻¹, T = 300 K, and V₂/V₁ = 2:

w = −(2.00)(8.314)(300) ln 2 = −(4988.4)(0.6931) = −3457 J ≈ −3.46 kJ. Since ΔU = 0, q = +3.46 kJ.
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Step 3 — Analyze Constraints for Process (b)

The gas is now in a rigid, sealed bomb calorimeter — this means constant volume. The gas cools from 300 K to 250 K. There is no non-PV work. The appropriate function is ΔU because at constant volume with no non-PV work, ΔU = q_V. Using ΔH here would introduce an unnecessary PΔV correction term (although for an ideal gas in a rigid vessel, ΔV = 0 makes this zero anyway — but the conceptual point is that ΔU is the natural choice).

Select ΔU for process (b).
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Step 4 — Calculate ΔU for Process (b)

For an ideal gas, C_V = (3/2)R for a monatomic gas or (5/2)R for a diatomic gas. Assuming a monatomic ideal gas: C_V = (3/2)(8.314) = 12.47 J mol⁻¹ K⁻¹. Then ΔU = nC_VΔT = (2.00)(12.47)(250 − 300) = (2.00)(12.47)(−50).

ΔU = −1247 J ≈ −1.25 kJ. The negative sign confirms heat leaves the system (cooling).
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Step 5 — Reflect on Function Selection

In process (a), neither ΔU nor ΔH was the useful quantity because both were zero; the First Law expression for work was the operative equation. In process (b), ΔU was the natural choice because constant volume eliminates PdV work. Had we incorrectly tried to use ΔH for process (b), we would have obtained ΔH = nC_PΔT = (2.00)(20.79)(−50) = −2079 J — a different number that equals the heat only at constant pressure, not constant volume. For the bomb calorimeter, ΔH ≠ q_V, underscoring the importance of correct function selection.

Always match the function to the constraint: ΔU for constant V, ΔH for constant P.

SECTION 7

Strengths & Limitations of Each Function

No single thermodynamic function is universally superior; each has a domain where it excels and situations where it becomes awkward or misleading. Understanding these strengths and limitations helps you avoid the two most common errors in physical chemistry problem-solving: applying the wrong function to a given set of constraints, or interpreting a result (e.g., ΔG > 0) under conditions where that interpretation is invalid.

Function	Strengths	Limitations / Common Pitfalls
ΔU	Directly tied to First Law; no approximations needed; fundamental quantity from which all others derive; exact for any process at constant V.	Requires knowledge of both q and w in general; not directly measured at constant P; for condensed phases, ΔU ≈ ΔH, so choosing ΔU offers little advantage.
ΔH	Equals q at constant P (most common lab condition); extensive tabulated data (Hess's law); directly measured by coffee-cup and flow calorimeters.	Not equal to heat at constant V; for gas-phase reactions, ΔH ≠ ΔU (differ by ΔnRT); does not indicate spontaneity by itself.
ΔA	Spontaneity criterion at constant T, V; maximum work function; natural for statistical mechanics (partition function Z relates to A).	Rarely used in bench chemistry (constant V uncommon); less tabulated data; loses spontaneity meaning if T varies.
ΔG	Spontaneity at constant T, P; connects to K (ΔG° = −RT ln K); connects to E° (ΔG° = −nFE°); maximum non-PV work; most extensive tables.	Only valid as spontaneity criterion at constant T and P; ΔG° gives equilibrium position, not rate; students often confuse ΔG with ΔG°.

Comparative strengths and limitations of the four main thermodynamic potentials.

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KEY TAKEAWAY

Think of each thermodynamic function as a specialized tool in a toolkit. A wrench is perfect for bolts but useless for screws; similarly, ΔG is perfect for constant-T, constant-P spontaneity but meaningless as a spontaneity criterion if temperature is changing. The mark of a skilled thermodynamicist is not memorizing formulas but knowing instantly which tool to reach for when confronted with a given set of constraints.

SECTION 8

Connections to Advanced Theory

The framework of selecting thermodynamic potentials based on natural variables extends seamlessly into advanced physical chemistry. In statistical mechanics, the canonical partition function Q is naturally related to the Helmholtz free energy via A = −k_BT ln Q, because the canonical ensemble fixes T, V, and N — exactly A's natural variables. Similarly, the isothermal-isobaric partition function Δ connects to G. Choosing the right ensemble in statistical mechanics is the microscopic analog of choosing the right thermodynamic potential in classical thermodynamics.

Classical Potential	Natural Variables	Statistical Ensemble	Partition Function Link
U(S, V, N)	S, V, N	Microcanonical	S = k_B ln Ω
A(T, V, N)	T, V, N	Canonical (NVT)	A = −k_BT ln Q
G(T, P, N)	T, P, N	Isothermal-isobaric (NPT)	G = −k_BT ln Δ
PV (grand potential)	T, V, μ	Grand canonical (μVT)	PV = k_BT ln Ξ

Correspondence between classical thermodynamic potentials and statistical mechanical ensembles.

Beyond statistical mechanics, the function-selection framework extends to non-equilibrium thermodynamics and chemical kinetics. In transition-state theory, the Gibbs free energy of activation ΔG‡ is the relevant barrier height for reactions at constant T and P, while the Helmholtz free energy of activation ΔA‡ applies in constant-volume contexts. In molecular simulations (Monte Carlo and molecular dynamics), the choice of ensemble — and hence the thermodynamic potential being minimized — directly affects which experimental conditions are being modeled. Mastering function selection in this introductory course thus prepares you for every subsequent branch of theoretical and computational chemistry.

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LOOKING AHEAD

In your study of chemical thermodynamics and statistical mechanics, you will encounter the Maxwell relations — equalities between second partial derivatives that arise because U, H, A, and G are all exact differentials. These relations are most naturally derived from the potential whose natural variables match the partial derivatives you need. Function selection is therefore not just a problem-solving convenience but a prerequisite for deriving fundamental thermodynamic identities.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student claims that ΔG < 0 guarantees a reaction will proceed spontaneously regardless of the conditions. Identify the flaw in this reasoning and state the precise conditions under which ΔG < 0 serves as a valid spontaneity criterion. Additionally, explain what criterion you would use instead if the reaction occurs in a sealed, rigid container at constant temperature.

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The student's claim is incorrect because ΔG < 0 is a valid spontaneity criterion only at constant temperature and constant pressure with no non-PV work (or, more precisely, when the only work is PV work). The Gibbs free energy's natural variables are T and P; when these are held constant, dG ≤ 0 follows from the Second Law. If either T or P is not constant, the inequality dG ≤ 0 does not hold in general. For a sealed, rigid container at constant temperature, the appropriate criterion is ΔA ≤ 0 (Helmholtz free energy), because A has natural variables T and V, matching the constant-T, constant-V constraints of the rigid vessel.

Question 2: BASIC CALCULATION

A bomb calorimeter (constant volume) measures a temperature rise of 3.20 K when 1.50 g of benzoic acid (C₆H₅COOH, M = 122.12 g/mol) is combusted. The heat capacity of the calorimeter system is C_cal = 10.17 kJ/K. (a) Which thermodynamic function does this calorimeter directly measure? (b) Calculate the molar energy of combustion.

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(a) A bomb calorimeter operates at constant volume, so it directly measures ΔU (not ΔH). The relationship is ΔU = q_V = −C_calΔT. (b) q_V = −(10.17 kJ/K)(3.20 K) = −32.54 kJ for 1.50 g. Moles of benzoic acid = 1.50/122.12 = 0.01228 mol. Molar ΔU_comb = −32.54/0.01228 = −2650 kJ/mol. To convert to ΔH, one would add Δn_gasRT, but the direct measurement is ΔU.

Question 3: INTERMEDIATE

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K and 1 bar, ΔH° = −92.2 kJ/mol and ΔS° = −198.7 J/(mol·K). (a) Calculate ΔG° and determine spontaneity. (b) Calculate ΔU° given that Δn_gas = −2. (c) Would ΔU° or ΔH° be more appropriate if this reaction were run in a constant-volume reactor? Explain.

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(a) ΔG° = ΔH° − TΔS° = −92200 − (298)(−198.7) = −92200 + 59213 = −32987 J = −33.0 kJ/mol. Since ΔG° < 0, the reaction is spontaneous under standard conditions at constant T and P. (b) ΔU° = ΔH° − Δn_gasRT = −92200 − (−2)(8.314)(298) = −92200 + 4955 = −87.2 kJ/mol. (c) In a constant-volume reactor, ΔU is the appropriate function because ΔU = q_V. Using ΔH would overestimate the heat released by about 5 kJ/mol because it includes the PΔV work term that does not occur when volume is fixed.

Question 4: APPLIED

A hydrogen fuel cell operates at 298 K and 1 atm, electrochemically oxidizing H₂(g): 2H₂(g) + O₂(g) → 2H₂O(l). Given ΔH° = −571.6 kJ and ΔG° = −474.4 kJ: (a) Which function determines the maximum electrical work the cell can deliver? (b) Calculate the maximum cell potential E°. (c) Why is ΔG° less negative than ΔH°, and what happens to the difference?

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(a) The maximum non-PV (electrical) work at constant T and P is given by ΔG°, not ΔH°. The relationship is w_elec,max = ΔG° = −nFE°. (b) With n = 4 mol e⁻ and F = 96485 C/mol: E° = −ΔG°/(nF) = 474400/(4 × 96485) = 1.229 V. (c) ΔG° = ΔH° − TΔS°. Since ΔG° is less negative than ΔH°, TΔS° must be negative (ΔS° < 0), consistent with 3 mol of gas → 0 mol of gas (liquid product). The difference |ΔH° − ΔG°| = 97.2 kJ is the heat that must be released to the surroundings (TΔS_surr) to satisfy the Second Law; it cannot be converted to useful electrical work.

Question 5: CRITICAL THINKING

Consider an ideal gas undergoing a reversible adiabatic expansion (q = 0) against a piston. (a) Explain why neither ΔG nor ΔA serves as a useful spontaneity criterion for this process. (b) Which thermodynamic function(s) can still be meaningfully calculated, and what determines whether the process proceeds? (c) Show that for this process ΔH ≠ ΔU and derive the ratio ΔH/ΔU in terms of the heat capacity ratio γ = C_P/C_V.

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(a) ΔG requires constant T and P; ΔA requires constant T and V. In a reversible adiabatic expansion, temperature changes (T drops), volume changes (V increases), and pressure changes (P drops). Neither set of constraints is satisfied, so neither free energy serves as a spontaneity criterion. (b) Both ΔU and ΔH can be calculated. Since q = 0 and the process is reversible, ΔS_sys = 0 (isentropic). The process is driven by the mechanical constraint (external pressure equals gas pressure throughout). ΔU = w = nC_VΔT and ΔH = nC_PΔT for an ideal gas. (c) ΔH/ΔU = (nC_PΔT)/(nC_VΔT) = C_P/C_V = γ. For a monatomic ideal gas, γ = 5/3 ≈ 1.67, so ΔH is 67% larger in magnitude than ΔU. This difference reflects the PV work associated with the volume change; ΔH includes it while ΔU does not.

SUMMARY

Lesson Summary

Summary

Selecting the appropriate thermodynamic function is the critical first step in any physical chemistry problem. The four principal potentials — internal energy U, enthalpy H, Helmholtz free energy A, and Gibbs free energy G — are connected by Legendre transforms that swap natural variables for their conjugates. Each potential simplifies maximally when its natural variables are the quantities held constant in the problem: U for (S, V), H for (S, P), A for (T, V), and G for (T, P).

For energy bookkeeping, use ΔU at constant volume (bomb calorimetry) and ΔH at constant pressure (coffee-cup calorimetry, standard enthalpies). For spontaneity and equilibrium at constant T, use ΔA (constant V) or ΔG (constant P). The Gibbs free energy dominates chemical applications because laboratory conditions are overwhelmingly constant T and P. When temperature is not constant, revert to the universal criterion ΔS_total ≥ 0. This function-selection skill is foundational: it reappears in statistical mechanics (ensemble choice), electrochemistry (ΔG = −nFE), and computational simulations (NPT vs. NVT ensembles).