Enthalpy, Helmholtz energy, and Gibbs energy

One gas, a bath and a load

A sealed sample of 0.200 mol of an assigned ideal gas is held at 300 K by a large thermal bath. A horizontal piston is held at a volume of about 2.49434 L; the gas pressure is 200 kPa absolute. A constant remote resistance corresponds to 100 kPa. Remove the restraint and let the gas and piston settle. With the bath still at 300 K, the final equilibrium volume is about 4.98868 L. If the ideal gas starts and ends at the same temperature, its internal energy is unchanged. Has nothing useful been lost?

The answer depends on the surroundings as well as the gas. In the actual release the load receives about 249.434 J. A carefully controlled reversible comparison between the same gas endpoints could deliver about 345.789 J of total work. The difference, about 96.355 J, is an opportunity to do useful work beyond moving that same load. It is unavailable after the uncontrolled release has settled.

Our selected system is the gas. The piston is horizontal and frictionless, starts and ends at rest, and has no other retained apparatus energy; there is no external damper. Internal viscous and acoustic dissipation in the gas, together with heat exchange with the bath, permits settling. The gas need not have one uniform temperature or pressure during the motion. Only the initial and final equilibrium states are specified. The later curves compare constrained equilibrium trial states; they are not a movie of this release.

For a first reading, follow the definitions, enthalpy and free-energy work bounds, the constraint table, and the gas comparison. Return to the Legendre geometry and sections marked “Further” when comfortable with partial derivatives. Published preparation is “Physical systems, states, boundaries, and environments”, “Work, kinetic energy, potential energy, and power”, and “Pressure, buoyancy, and fluid statics”. We introduce heat, entropy and the required first-law sign here. The neighboring thermodynamics subjects provide fuller mechanisms and return paths, rather than assumed live prerequisites.

Name the quantities and the work sign

Internal energy U is a state property. For the closed systems here, changes of macroscopic motion and height vanish between the endpoints. Write ΔU = Q − W_by: heat Q is positive into the system, and work W_by is positive outward. Heat and work describe transfers during a process; they are not quantities stored in the sample. The product pV has units of energy, since 1 Pa m³ = 1 J.

Entropy S is another state property. For a reversible heat transfer to a closed system, dS = δQ_rev/T, with absolute temperature T in kelvin. For a closed system interacting thermally with one ideal reservoir at a fixed positive temperature T₀, the complete endpoint entropy balance is ΔS = Q/T₀ + S_gen, where S_gen ≥ 0 includes the irreversibility of the system and its thermal contact. The actual heat of an irreversible process cannot simply replace δQ_rev in the definition.

Define enthalpy H = U + pV, Helmholtz energy F = U − TS, and Gibbs energy G = H − TS = U + pV − TS. All have units of joules. Some sources use A for Helmholtz energy, so always check the definition. The word “free” does not mean energy can be created or all internal energy can be converted to work. These combinations are useful because particular surroundings hold particular variables fixed.

The gas uses pV = nRT, with R = 8.314462618… J/(mol K), and the separate caloric assignment c_v = 2.50R. We choose U = 0 at 300 K and S = 0 at the final reference state. An entropy difference and an available-work difference do not depend on this arbitrary reference choice when it is used consistently. This local reference is not a claim of zero absolute entropy or a third-law model.

Natural variables tell us what is held fixed

For neighboring equilibrium states of a simple compressible bulk phase with fixed composition and no additional reversible work coordinate, dU = T dS − p dV. Here T is the coefficient of dS and −p is the coefficient of dV. In the representation U(S,V), entropy and volume are called natural variables. This is a relation among equilibrium state properties, not permission to identify actual irreversible heat with T dS or actual gas work with p dV.

Apply the product rule to the definitions: dH = dU + p dV + V dp, dF = dU − T dS − S dT, and dG = dU + p dV + V dp − T dS − S dT. Substituting the equilibrium expression for dU cancels the unwanted terms. The resulting differentials explain the natural variables in the table. Fixed amounts and any extra work coordinates are suppressed in this compact table only.

“Natural” does not mean that a state cannot be described using another independent pair. It means that the differential directly exposes the conjugate properties in that representation. For example, the volume derivative of F at fixed T gives −p. The pressure derivative of G at fixed T gives V. A held-variable subscript is part of the statement, not optional decoration.

The four functions encode the same equilibrium information on a branch where the necessary variable changes can be inverted. They are not four independent energy stores. A pressure equation alone is still insufficient: a thermal equation of state and compatible caloric information are needed to build an actual potential function.

Enthalpy makes pressure-controlled heating convenient

Let a closed sample expand quasistatically at a uniform constant pressure, with pressure–volume work as its only work channel. The first law gives ΔU = Q − pΔV. Adding pΔV shows that ΔH = Q. The useful statement is a change under stated conditions; H is not a quantity of “heat contained” in the material.

For our ideal gas, c_p = c_v + R = 3.50R and ΔH = n c_p ΔT. Heating from 300 K to 450 K at 100 kPa gives ΔU ≈ 623.585 J, volume work ≈ 249.434 J, and Q = ΔH ≈ 873.019 J. Heat both raises internal energy and makes room against the load. A rigid-container comparison would instead have Q = ΔU if all other work were excluded.

Electrical output, stirring, surface formation or another work channel changes the equality. With a constant-pressure load and additional outward work W_other,by, the corresponding complete endpoint balance is ΔH = Q − W_other,by when both endpoint pressures equal the load pressure and the load-work model is valid. Thus Q need not equal ΔH even at constant pressure.

Enthalpy also groups internal energy and pressure flow work in an open-flow energy balance: specific enthalpy h = u + pv. Pressure flow work is then counted in h once. This packaging does not make a flowing device a closed system or remove its kinetic, potential, heat and shaft-work terms. The first-law subject develops that full accounting; it is a different use from the fixed-entropy equilibrium criterion for H below.

Helmholtz energy and work from one bath

For a closed system exchanging heat only with a bath at T₀, combine ΔU = Q − W_by with ΔS = Q/T₀ + S_gen. If the two equilibrium endpoints have T = T₀, then ΔF = ΔU − T₀ΔS = −W_by − T₀S_gen. Therefore total outward work obeys W_by ≤ −ΔF. Equality is the reversible limit of this complete bath-and-system accounting, not a promise that a practical apparatus can attain it.

This total-work bound can include pressure–volume work if volume changes. In a rigid container, volume work is zero, so the same bound concerns the remaining work channels. If volume is fixed and no work is exchanged, ΔF = −T₀S_gen ≤ 0. This is the familiar Helmholtz decrease condition for unforced relaxation at fixed T and V.

For the assigned gas between the opening endpoints, ΔU = 0 and ΔS = nR ln 2 ≈ 1.15263 J/K. Thus ΔF = −nRT₀ ln 2 ≈ −345.789 J. A reversible isothermal comparison could deliver that much total work, while absorbing the same amount of heat from the bath. The gas changes its state and entropy; this is not a cyclic conversion of heat from one bath entirely into work.

An isothermal compression between the reversed endpoints has ΔF > 0 and needs at least that much work input under the same ideal bath conditions. Nor does F have to decrease when a work source drives the sample: negative outward work represents input. The constraints and the transfer signs determine the inequality that applies.

Gibbs energy separates useful work from moving the load

Now include a constant mechanical reservoir pressure p₀. Suppose the closed system starts and ends in equilibrium at T₀ and p₀, and its pressure–volume work delivered to that reservoir is p₀ΔV. Write W_by = p₀ΔV + W_other,by. Combining the energy and entropy balances gives ΔG = −W_other,by − T₀S_gen. Hence W_other,by ≤ −ΔG. Gibbs energy bounds nonexpansion work under these conditions; it does not bound total work by the same number.

For unforced relaxation with no nonexpansion work exchange, ΔG ≤ 0 at the fixed reservoir T₀ and p₀. A suitable electrical work-storage device can receive at most −ΔG in the reversible limit. Work input can instead drive a change with ΔG > 0. The sign of ΔG supplies a conditional direction or work bound, not a speed of reaction, a guarantee of access, or a complete description of an open machine.

Consider stipulated endpoint differences for a closed device at 300 K and 100 kPa: ΔG = −240 J, ΔH = −600 J and ΔV = 0.000400 m³. If it delivers 180 J of nonexpansion work, the load also receives 40 J, so total work is 220 J. The heat is Q = ΔH + W_other,by = −420 J. Its own entropy decreases by 1.20 J/K, while the bath gains 1.40 J/K; total production is 0.20 J/K. The 60 J shortfall from maximum useful work equals T₀S_gen.

These device values are an assigned thermodynamic endpoint case, not measurements of a battery or a material. Internal reactions or structural changes may occur within a matter-closed sample; the allowed amounts must obey its conservation constraints. A complete treatment of chemical potentials and phase or reaction equilibrium belongs to the next related subject.

What is being minimized?

A minimum principle compares admissible trial states under declared constraints. Some internal parameter must still be able to vary: a partition position, redistribution between subsystems, phase amount, structure or reaction extent consistent with conserved quantities. At equilibrium the correct potential has no downhill allowed variation. It is not minimized with respect to a control variable that the surroundings are holding fixed.

For a fully equilibrated simple one-phase sample with fixed amount, specifying T and p already fixes its state. Writing dG = 0 at dT = dp = 0 is then not an independent prediction of a relaxation mechanism. To study one, expose the remaining internal variable or the constrained subsystems and compare their admissible states. Temperature and pressure derivatives of the equilibrium G surface are response relations, not directions over which G must be a minimum.

The U and H entries in the table are constrained equilibrium variational statements at fixed entropy. An actual isolated irreversible process does not generally lower U: it conserves U while its entropy can rise. A pressure reservoir alone does not hold the system entropy fixed either. The commonly accessible thermal-bath experiments instead lead to F and G decrease conditions, with no uncounted work source.

When several states are allowed, stable equilibrium is a minimum over the relevant allowed variations. A stationary point is only a candidate. A local minimum can be metastable behind a barrier, and a flat direction can give nonunique equilibrium. The Maxwell-relations and stability subject develops response functions and curvature tests; chemical and phase equilibrium develops the admissible composition and phase variations.

Include the bath and load explicitly

The opening gas is initially at twice the remote load pressure, so the whole release is not a constant-gas-pressure process. It would be wrong to invoke a fixed-p Gibbs decrease rule merely because the external resistance is constant. Introduce instead Ψ = U − T₀S + p₀V, using reservoir values T₀ and p₀ rather than silently identifying them with the gas values throughout the event.

For equilibrium endpoints of our closed system, with the one-bath entropy balance and total outward work p₀ΔV + W_other,by, subtraction gives ΔΨ = −W_other,by − T₀S_gen. No uniform intermediate gas p or T is needed for this endpoint identity. When both endpoints actually match T₀ and p₀, Ψ equals G at those endpoints. When the volume is fixed and the endpoints match T₀, its change equals ΔF instead.

The work equality for the actual piston also needs a mechanical reason. The work exerted by the gas at its moving boundary is ∫p_b dV, where p_b is area-averaged normal traction. A horizontal frictionless piston obeys ∫(p_b − p₀)dV = ΔK_piston = 0 between rest endpoints. With no other apparatus energy term, the complete gas work equals p₀ΔV. This equality of integrals does not require instantaneous p_b = p₀.

Piston friction, an external damper, a spring retaining energy, or a final moving apparatus can change that balance. Thermal contact alone does not supply mechanical settling in an ideal undamped model; our stated internal gas dissipation does. If extra transfers, several bath temperatures or matter flows are introduced, rebuild the energy and entropy balances rather than reusing this particular Ψ identity.

Return to the gas: a lost opportunity, not lost energy

The assigned energy scale is nRT₀ ≈ 498.868 J, and V* = nRT₀/p₀ ≈ 4.98868 L. The release starts at V*/2 and ends at V*. The actual load receives p₀(V_f − V_i) = nRT₀/2 ≈ 249.434 J. Since ΔU = 0, the gas receives Q ≈ 249.434 J from the bath over the complete event.

The gas entropy increases by about 1.15263 J/K, while the bath entropy changes by −Q/T₀ ≈ −0.831446 J/K. Their total production is about 0.321183 J/K. Consequently ΔΨ ≈ −96.3549 J. In the actual release no separate useful-work device receives energy, so this whole decrease is T₀S_gen.

A reversible comparison between the same gas endpoints can balance the gas with the 100 kPa load plus an ideal adjustable work-storage device. Its total gas work is nRT₀ ln 2 ≈ 345.789 J. The same load displacement still requires 249.434 J; the remaining 96.3549 J can go to the device. Reversibility requires controlled mechanical loading and thermal driving differences approaching zero. This alternative protocol is not a hidden trajectory of the actual release.

Both protocols conserve energy and have the same gas state-function differences. They differ in heat, work and entropy production. Here ΔG happens to equal ΔF because pV = nRT₀ has the same value at both ideal-gas endpoints. It still does not equal ΔΨ: the initial gas pressure is not p₀. Numerical coincidence between some state-function changes cannot replace checking the reservoir conditions.

Further: a tangent explains the Legendre transform

Consider equilibrium trial states of the same gas at bath temperature T₀ while its volume is held at different values. Write x = V/V* > 0. With our reference choice, F/(nRT₀) = −ln x. The slope is −1/x; at x = 1 it is −1, corresponding to the actual dimensional relation (∂F/∂V)_T = −p₀. The tangent there is 1 − x.

The vertical intercept of a tangent is F + pV at its point of tangency, because its slope is −p. Thus the intercept, expressed as a function of the selected pressure, is G(T,p). A Legendre transform retains slope and intercept information together. Simply replacing V by V(T,p) inside F would leave out the pV term and would not produce G.

The same geometry gives an equilibrium calculation. Add p₀V to every trial F value: Ψ/(nRT₀) = x − ln x at T = T₀. Its derivative is 1 − 1/x, so the stationary volume is x = 1. Its second derivative is 1/x² > 0 for x > 0, and the function rises without bound toward either x → 0 or x → ∞. The equilibrium is the unique global minimum of this assigned model.

Subtracting the equilibrium value gives [Ψ − Ψ_eq]/(nRT₀) = x − 1 − ln x ≥ 0. Geometrically this is also the gap between −ln x and its supporting tangent 1 − x. The initial x = 1/2 has the positive gap responsible for the 96.3549 J useful-work opportunity. The curve compares states that could be held at the bath temperature; it supplies neither the pressure history nor the temperature history of the finite release.

For a smooth stable branch with an invertible conjugate relation, one may write G(T,p,n) = min_V[F(T,V,n) + pV]. A singular or nonconvex branch needs more care: a slope may fail to identify one state, or global equilibrium may select phase coexistence and a convex envelope. Do not infer a globally invertible transform or a stable phase from algebra alone. Those extensions connect to the neighboring stability and phase-equilibrium lessons.

Read properties from a potential

From dF = −S dT − p dV at fixed amounts, S = −(∂F/∂T)_V and p = −(∂F/∂V)_T. From dG = −S dT + V dp, S = −(∂G/∂T)_p and V = (∂G/∂p)_T. These are measurable thermodynamic responses encoded in derivatives. They do not require calculating a reversible path for every new state difference.

Our compatible ideal-gas model can be written U = n c_v(T − T₀) and S = n c_v ln(T/T₀) + nR ln[V/(n v*)], where v* = RT₀/p₀ is a fixed reference molar volume. Substituting into F = U − TS and taking the two derivatives recovers the assigned S and p = nRT/V. The logarithms have dimensionless arguments, and the chosen U and S references remain consistent.

The differential of a smooth state function has equal mixed derivatives. For example, differentiating F gives (∂S/∂V)_T = (∂p/∂T)_V; both equal nR/V in this gas. This single relation shows why the choice of potential is productive. The full set of Maxwell relations and stability restrictions is the next subject, rather than another collection to memorize here.

Further: temperature and pressure dependence

At fixed T and amounts, dG = V dp. For this ideal gas, integration gives G(T,p₂) − G(T,p₁) = nRT ln(p₂/p₁). Raising pressure from 100 kPa to 200 kPa at 300 K increases G by about 345.789 J. The pressure ratio is dimensionless, and both pressures are absolute. This is a state-function comparison across different pressures, not an invocation of a constant-pressure spontaneous-change rule.

Temperature changes require the product rule, not replacing Δ(TS) by TΔS with an arbitrary temperature. At fixed p and amounts, differentiate G/T: (∂(G/T)/∂T)_p = [T(−S) − G]/T² = −H/T². This is the Gibbs–Helmholtz identity. Similarly, at fixed V and amounts, (∂(F/T)/∂T)_V = −U/T². Other independent work coordinates are held fixed in these derivatives.

These relations allow temperature integration when the corresponding H(T,p) or U(T,V) is supplied. One cannot hold enthalpy or heat capacity constant without a model or property data. If composition changes with temperature, specify whether amounts, reaction extent or equilibrium constraints are held; the derivative then follows that specification.

For the assigned 100 kPa heating from 300 K to 450 K, ΔH ≈ 873.019 J while the model gives G changing from about 498.868 J to 309.951 J. The change follows the varying T and S in H − TS. A falling G in this externally heated experiment is not a proof of spontaneous relaxation at a single fixed bath temperature. Absolute values depend on the declared reference; consistent differences and derivative identities are the useful checks.

Further: composition and open systems

When species amounts can vary, extend the equilibrium differential to dU = T dS − p dV + Σ μ_i dn_i. Here n_i is an amount in moles and μ_i is its conjugate chemical potential in J/mol. The same Σ μ_i dn_i term remains in dH, dF and dG. In particular, μ_i = (∂G/∂n_i)_(T,p,n_j≠i). This defines the coefficient; calculating how it depends on phase or composition is the work of the chemical-potential subject.

A matter-closed reacting sample can change individual species amounts while conserving its atoms. A multiphase closed sample can redistribute a conserved component between phases. Minimum principles compare only these allowed changes. “All species amounts fixed” and “all conserved elemental amounts fixed” are different constraints. If all relevant amounts and internal coordinates are fixed already, there may be no remaining change to minimize over.

For a subsystem allowed to exchange particles with ideal reservoirs at fixed T, V and chemical potentials μ_i,res, a relevant potential is Ω = F − Σ μ_i,res n_i. The reservoir contributions must be included when comparing trial subsystem states. At equilibrium, the subsystem chemical potentials match those reservoir values, and the grand-potential representation has dΩ = −S dT − p dV − Σ n_i dμ_i, with no extra work coordinates. This does not make an arbitrary open subsystem’s own G decrease monotonically.

The term Σ μ_i dn_i in an equilibrium differential is not by itself the energy flux carried by an actual stream. An open-flow first law also includes enthalpy, entropy transport, motion and other transfers at the boundaries. A driven open steady state can continuously receive useful work or matter and need not be an equilibrium potential minimum. Exchange, phase equilibrium and reaction mechanisms require their own complete system choices.

Further: extensivity and what G = Σμn assumes

For a homogeneous, additive bulk system with pressure–volume work only, U is first-degree homogeneous in S, V and the species amounts. Scaling all those extensive quantities by the same factor scales U by that factor. Euler’s relation then gives U = TS − pV + Σ μ_i n_i, and therefore G = Σ μ_i n_i. For a pure phase, G/n = μ.

The distinction between the total amount n in moles and particle number N matters: their conjugate chemical potentials differ by the corresponding conversion factor. In our extensive ideal-gas extension, the reference volume is n v*, so scaling n and V together leaves the entropy per mole unchanged. Using a fixed total reference volume while varying n would need an additional consistent reference term.

Surface area, elastic strain, polarization, magnetization, strong finite-size effects or long-range nonadditivity can change the independent variables and the simple Euler form. For example, a surface term γ dA_s introduces a surface-work coordinate; one must state what is included in the chosen bulk-plus-interface system before using an extensive relation. Adding a new coordinate is a physical change in the model, not a harmless choice of notation.

Further: an imposed force changes the potential

Write an additional reversible work-on term as Y dX. Then dU = T dS − p dV + Σ μ_i dn_i + Y dX, and the same Y dX appears in dF and dG when X remains an independent coordinate. For a stretched sample, X can be an extension and Y a tensile force. A dissipative stirrer angle or an accumulated frictional distance is not automatically an equilibrium state coordinate with a reversible conjugate.

If the environment fixes the force at f₀ instead of fixing X, include its work source by comparing F − f₀X at fixed T and the other declared controls. Its derivative with respect to X is Y − f₀. Mechanical equilibrium requires Y = f₀. A blanket “F must fall” statement would omit the energy supplied by the force source.

As a separate assigned linear isothermal model, let F_el(X) = kX²/2 with k = 100 N/m over 0 ≤ X ≤ 0.040 m, while T, volume, composition and all other coordinates are fixed. An imposed f₀ = 2 N gives X_eq = f₀/k = 0.020 m. From X = 0 to this extension, F_el increases by 0.020 J while F_el − f₀X decreases by 0.020 J. The second derivative k is positive, so the transformed potential has its unique minimum there in the stated domain.

The force plot compares constrained equilibrium extensions. It is not a claim about oscillation frequency, relaxation time or an actual trajectory. An electrostatic or magnetic example also requires an explicit system, field-source contribution and work convention before selecting a transform; sign conventions copied between different field-controlled ensembles can be misleading.

Equilibrium, reversibility and rate are different questions

A state-function differential relates nearby equilibrium states. A reversible process is a limiting protocol that can be reversed without leaving an uncompensated change in the system and surroundings. An irreversible relaxation can connect the same equilibrium endpoints while generating entropy. The gas comparison shows why identical endpoints do not imply identical work.

For the one-bath and constant-load model, the useful-work shortfall is T₀S_gen. It vanishes in the reversible comparison. In the no-other-work fixed-T,p relaxation, a strict G decrease reflects entropy production; at equilibrium there is no downhill allowed variation. Measurements of ΔG alone do not establish that every apparatus transfer has been accounted for, nor that a real process is reversible.

A minimum principle does not supply an equation of motion or a reaction rate. A metastable configuration can persist because a transition needs nucleation or activation across a barrier. External driving, finite gradients and transport may prevent global equilibrium. Kinetics and irreversible heat and mass transfer connect thermodynamic direction to rates through additional mechanisms and constitutive information.

Do not extend a one-phase smooth formula through a phase transition without checking its domain. The potential remains useful, but derivatives can become discontinuous and global minimization can choose another phase. The neighboring lessons develop the response and phase information needed for those decisions.

Choose the system, controls and useful application

For calorimetry at controlled pressure with pressure–volume work only, use ΔH to connect to heat. For a rigid thermally controlled device, F supplies the relevant no-work relaxation criterion or the maximum remaining work. For a constant-T,p device with a pressure reservoir, G bounds nonexpansion work. In each case, specify the boundary and transfers before choosing the formula.

For the gas initially out of mechanical balance with its environment, use U − T₀S + p₀V and the explicit load model. For particle exchange or an imposed force, add the corresponding reservoir contribution and constraints. Check absolute T and p, amount units, reference consistency, and whether a curve represents actual equilibrium states, a constrained trial family or an unsupplied transient.

Return to “Maxwell relations and thermodynamic stability” for mixed derivatives, response functions and stability tests, and to “Chemical potential and phase equilibrium” for composition, phase and reaction mechanisms. The engines subject supplies cyclic performance limits; the first-law and entropy subjects supply the complete balances used here. The connecting idea is to select the potential whose variables and reservoir terms match the physical question.

Worked examples

Try it yourself

1. Does H mean the amount of heat stored in a sample?

   Show the answer and reasoning

   No. H = U + pV is a state property.

   Only under specified closed-system constant-pressure, pressure–volume-only conditions does its change equal heat transferred. Heat itself is a process transfer.

2. A closed rigid sample can relax internally while contacting one bath at fixed T. No work source acts. Which free energy supplies the decrease condition?

   Show the answer and reasoning

   Helmholtz energy F at fixed T and V.

   The one-bath balance becomes ΔF = −T S_gen ≤ 0 when no work is exchanged. Allowed internal variations and conserved amounts still have to be specified.

3. At fixed bath T and mechanical-reservoir p, does −ΔG bound total gas work or nonexpansion work?

   Show the answer and reasoning

   Nonexpansion work under the complete closed-system load model.

   The load term pΔV has already been included in G. Total work also contains that pressure–volume contribution; counting it again as useful output would double-count the grouping.

4. For work-on term Y dX in dU, which transform is appropriate when the source fixes Y = Y₀ instead of X?

   Show the answer and reasoning

   Subtract Y₀X from the appropriate temperature-controlled potential.

   At fixed T and other controls, ∂(F − Y₀X)/∂X = Y − Y₀. The source contribution is essential; F alone need not decrease.

5. At fixed T and species amounts, what property is (∂G/∂p)_T?

   Show the answer and reasoning

   Volume V.

   The coefficient of dp in dG = −S dT + V dp + Σμ_i dn_i is V. Holding the other natural variables fixed is part of the derivative statement.

6. Why does the volume derivative of F give −p rather than +p?

   Show the answer and reasoning

   Because dF = −S dT − p dV at fixed amounts.

   The sign follows the pressure–volume work term in dU and the declared work convention. At fixed T only the −p dV term remains.

7. At 300 K, raising the assigned gas pressure from 100 kPa to 200 kPa changes G by how much?

   Show the answer and reasoning

   ΔG = nRT ln 2 ≈ +345.789 J.

   Integrate V dp = nRT dp/p at fixed T and amount. The endpoints have different pressures, so this calculation does not satisfy the fixed-p spontaneity test.

8. During the actual finite release, may p_b be replaced by p₀ at every instant?

   Show the answer and reasoning

   No. Only the integrated work equality follows from the stated complete piston model.

   A frictionless horizontal piston can accelerate. Rest endpoints give ∫(p_b − p₀)dV = 0; they do not assert instantaneous equality or a uniform gas trajectory.

9. Why is the lost useful-work opportunity in the opening release about 96.3549 J?

   Show the answer and reasoning

   It is the reversible total work minus the same load work, and equals T₀S_gen.

   345.789 − 249.434 ≈ 96.3549 J. Actual energy is conserved; entropy production removes this opportunity for work beyond moving the load.

10. Must U decrease during irreversible relaxation of an isolated system with fixed outer volume?

    Show the answer and reasoning

    No. U is conserved, while entropy may increase.

    The minimum-U principle compares trial states at fixed S,V and conserved amounts. Those are different constraints from actual isolated relaxation at fixed U,V.

11. Does a zero first derivative prove a unique stable global equilibrium?

    Show the answer and reasoning

    No.

    The point could be unstable, a local metastable minimum, or part of a flat set. Check admissible variations, curvature and competing allowed states; barriers concern kinetics.

12. In the assigned pressure-controlled device, how can its entropy fall by 1.20 J/K without violating the second law?

    Show the answer and reasoning

    The bath gains 1.40 J/K, so total entropy increases by 0.20 J/K.

    The second law concerns the complete system-plus-surroundings entropy balance. A non-isolated subsystem may export entropy with heat.

13. Would inserting V(T,p) into F(T,V) alone produce G(T,p)?

    Show the answer and reasoning

    No. The conjugate product pV is also required.

    G = F + pV cancels the dV term and makes dp appear with coefficient V. The tangent construction preserves its intercept as well as its slope.

14. May the closed fixed-T,p condition ΔG ≤ 0 be applied to an arbitrary matter-exchanging subsystem?

    Show the answer and reasoning

    No. Include particle-reservoir, entropy-transport and work terms.

    For ideal particle reservoirs at fixed T,V,μ_i,res a relevant equilibrium transform is Ω = F − Σμ_i,res n_i. An actual flowing or driven subsystem still needs its full balances.

15. In a closed reacting sample, are all individual species amounts necessarily fixed?

    Show the answer and reasoning

    No. Conserved atoms constrain the allowed changes.

    Species may be converted into other species without matter crossing the outer boundary. A Gibbs minimum must be tested only along allowed composition changes, developed in the chemical-potential lesson.

16. Can ΔG always be calculated as ΔH − TΔS when initial and final temperatures differ?

    Show the answer and reasoning

    No. In general ΔG = ΔH − Δ(TS).

    The product difference is T_fS_f − T_iS_i. A single TΔS expression requires the same fixed temperature at both endpoints; temperature integration can instead use the Gibbs–Helmholtz identity.

17. Does an adiabatic boundary automatically justify the fixed-S enthalpy minimum as an actual time law?

    Show the answer and reasoning

    No. An irreversible adiabatic process can produce entropy.

    Fixed S is an extra constraint. Reversible adiabatic changes of the closed model are isentropic; insulation alone does not remove entropy production.

18. Is G = Σμ_i n_i an unrestricted identity for every finite sample with arbitrary surfaces and fields?

    Show the answer and reasoning

    No. Its simple form uses homogeneous additive bulk extensivity and the stated work coordinates.

    Interfaces, extra generalized work and nonadditivity can change the Euler relation. State the full system and independent coordinates before using it.

What to carry forward

• H = U + pV, F = U − TS and G = U + pV − TS are state functions whose natural variables match different controls; they are not independent stores of energy or heat.
• For the complete one-bath closed-system model, total outward work is at most −ΔF between endpoints at the bath temperature. At fixed T,p with a valid constant-load model, nonexpansion work is at most −ΔG.
• Use Ψ = U − T₀S + p₀V when bath and load values must be kept distinct from the system’s intermediate or initial values. The useful-work shortfall is T₀S_gen under that stated model.
• Potential minima compare allowed internal variations at fixed controls and conserved amounts. Actual isolated relaxation conserves U; stationarity alone is not a global-stability or rate test.
• Legendre transforms exchange conjugate independent variables while retaining the matching product and derivative information; singular or nonconvex branches need domain and phase checks.
• Composition changes, particle exchange, surfaces and other work coordinates require their corresponding terms and sometimes another reservoir transform. Choose the physical model before choosing the potential.