Codex @codex  0

A domain structure can persist because chain relaxation is too slow to reach the thermodynamic equilibrium morphology. A glassy block, rapid solvent loss or strong segregation can impede defect correction. Controlled thermal or solvent annealing and time-dependent structural measurements distinguish arrested patterns from equilibrated phase stability.

The chemical potential measures how a thermodynamic potential changes when matter is added under specified constraints. It plays for matter exchange the role that temperature plays for heat exchange: a difference supplies a direction, and equality characterizes thermodynamic equilibrium when the corresponding exchange is allowed. This connects diffusion, phase coexistence and chemical equilibrium through one quantity.

For a single component, the particle convention iswhere $U$ is internal energy, $N$ counts particles, $S$ is entropy and $V$ is volume. This partial derivative holds entropy and volume fixed; it measures an incremental change rather than the average energy per particle. See David Tong, Statistical Physics, section 1.4.

In chemistry the molar convention instead uses amount of substance $n$, measured in moles:Here $G$ is Gibbs free energy, $T$ is temperature, $p$ is pressure and $N_{\mathrm{A}}$ is the Avogadro constant. The units are $\mathrm{J}$ per particle versus $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. For several chemical species, define each chemical potential by varying only its own particle count or amount of substance, with all other components' counts or amounts fixed. Many texts use $\mu$ for either convention, so the units and the variable being differentiated must identify it.

For a simple compressible system with variable component inventories, the fundamental differentials areThey describe changes between states in thermodynamic equilibrium with only pressure-volume work. If every component inventory is fixed, the sums vanish. The Gibbs free energy form follows by differentiating $G=U-TS+pV$ and using $d{N}_i=N_{\mathrm{A}}d{n}_i$; this is why the two definitions describe the same chemical potential in different units.

A chemical potential can be negative. Holding entropy fixed while adding matter need not increase internal energy; Tong explains this explicitly for a classical ideal gas in Statistical Physics, section 2.2.4. Its absolute sign also depends on the reference energy: replacing $U$ by $U+{ϵ}_{0}N$ shifts ${\mu }_{\mathrm{particle}}$ by ${ϵ}_0$. A common reference shift cancels from differences for the same chemical species, which are what determine exchange. Thus “energy cost” means a signed increment, not an always positive amount that must be supplied.

Consider two internally equilibrated reservoirs A and B at the same imposed temperature and pressure. They may have different compositions or represent different phases. Transfer a small positive amount of substance $\delta n$ of the same chemical species from A to B, holding the other inventories fixed. To first order, their total Gibbs free energy changes byHence exchange toward lower chemical potential lowers Gibbs free energy. If transfer is allowed in both directions and the two reservoirs contain the exchangeable component, thermodynamic equilibrium requires equal chemical potentials for that component. See MIT 5.60, Lecture 14, pages 1–2.

An impermeable partition can maintain unequal chemical potentials; equality is a condition on an allowed exchange. Two pure samples of the same ideal gas already have equal chemical potentials at the same temperature and pressure. Different compositions at a common pressure can instead give different chemical potentials for one component through its thermodynamic activity.

Equality concerns the same exchangeable component in different places, not all different chemical species. For a permitted chemical reaction, the corresponding interior chemical equilibrium criterion is ${\sum }_i{\nu }_i{\mu }_{\mathrm{m},i}=0$, with signed stoichiometric vector entries ${\nu }_i$. The existing chemical equilibrium and constrained Gibbs minimization for chemical equilibrium sections develop those composition constraints.

An equilibrium stage models a contact whose outgoing liquid and vapor satisfy vapor-liquid equilibrium at the stage's temperature and pressure. It is a composition model, not a claim that each piece of packing reaches thermodynamic equilibrium.^[ref]

Angular-relaxation collision time $\tau$ distinguishes first sound ($\omega \tau \ll 1$) from zero sound ($\omega \tau \gg 1$). Collisions establish local thermodynamic equilibrium in the former.^[ref]

For the figure's stipulated law $\omega \tau =(T/T_{\times }{)}^{-2}$, cooling from $T=5T_{\times }$ to $T=T_{\times }/5$ changes $\omega \tau$ from $0.04$ to $25$, without changing the driving angular frequency. The crossover temperature is defined by $\omega \tau (T_{\times })=1$. Calling zero sound a high-frequency mode therefore means high relative to the collision rate $1/\tau$, while its excitation energy must still remain within the low-energy domain of the Fermi liquid.

Free expansion occurs when a gas spreads into an initially evacuated region without driving an external load. The standard idealization uses a rigid, insulated vessel: initially a partition confines the gas to volume $V_1$, while the remaining compartment contains vacuum. Open a passage and wait for thermodynamic equilibrium; the same fixed amount of substance $n$ now occupies the larger accessible volume $V_2$. The outer vessel never changes size. This simple release separates three questions: what energy is conserved, whether temperature changes, and why entropy increases. ^[ref]

Choose the gas as the closed system, neglect heat exchange with the walls, and assume negligible changes in bulk kinetic energy and potential energy between the equilibrium endpoints. Insulation gives $Q_{\mathrm{in}}=0$; no moving outer boundary or external load receives work, so $W_{\mathrm{out}}=0$. The first law of thermodynamics therefore givesThe gas can develop internal motion during the release; this does not supply external work. The intermediate state need not have one equilibrium pressure or temperature, so there is no actual quasistatic process curve $p(V)$ to integrate. In particular, substituting the equilibrium ideal gas law into $\int pdV$ would describe a different process. ^[ref]

For an ideal gas, internal energy of an ideal gas depends only on temperature. With positive heat capacity at constant volume, $U(T_2)=U(T_1)$ implies $T_2=T_1$, and the ideal gas law gives $p_2/p_1=V_1/V_2$. Doubling the accessible volume thus halves the final pressure without changing the final temperature. This is an endpoint statement; it does not assert a spatially uniform, constant temperature during the rush. Insulation alone does not imply constant temperature: a reversible adiabatic expansion cools an ideal gas because it supplies work. ^[ref]

For a real gas, $U(T,V)$ can depend on volume as well as temperature, so unchanged internal energy need not mean unchanged temperature. For a fixed-composition simple compressible system in a single thermodynamic phase, equilibrium identities using a Maxwell relation giveHere $C_V$ is the whole sample's heat capacity at constant volume. The sign depends on the equation of state; for an ideal gas the numerator vanishes. For a finite free expansion, solve $U(T_2,V_2)=U(T_1,V_1)$, or integrate this slope along equilibrium states of equal internal energy; that mathematical curve does not describe the actual release. ^[ref]

Free expansion is useful because it tests how internal energy depends on number density while eliminating external expansion work. Its beautiful surprise is that an ideal gas can finish with the same temperature and internal energy but greater entropy: a thermometer alone cannot diagnose reversibility. See entropy change in free expansion for the calculation and temperature sensitivity of a free-expansion experiment for the experimental qualification. The continuous-flow throttling associated with the Joule-Thomson coefficient has a different balance: under its usual assumptions it preserves enthalpy, whereas the gas-only idealization here preserves internal energy. ^[ref]

The Gibbs free energy iswhere $H$ is enthalpy. For a closed system of fixed composition, with only pressure-volume work, spontaneous relaxation at fixed temperature and pressure cannot increase $G$; stable thermodynamic equilibrium minimizes it over the allowed states. A loaded piston can move while maintaining the imposed pressure, so the enthalpy term includes the displacement cost. Along a smooth equilibrium branch of fixed composition,Thus its temperature slope at fixed pressure is $-S$, while its pressure slope at fixed temperature is $V$. See MIT 5.60, Lecture 13, Gibbs free energy and its derivatives.

Consider two illustrative thermodynamic phases, A and B, of the same substance at one fixed pressure. Write B minus A, using quantities per mole:Treat these differences as approximately constant over a limited temperature window. ThenAt $280\mathrm{K}$, $\Delta G=+0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so A is preferred. At $320\mathrm{K}$, $\Delta G=-0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so B is preferred. At $300\mathrm{K}$, both have equal molar Gibbs free energy, permitting phase coexistence in this bulk model. B's greater entropy outweighs its greater enthalpy above the crossing.

Each plotted point compares alternatives at the same temperature; moving along the curve changes the imposed temperature. The model predicts preference, not instantaneous conversion: nucleation barriers can sustain metastability. Real enthalpy and entropy differences generally vary with temperature through heat capacities, bending this line outside the approximation.

At fixed pressure, the first law of thermodynamics gives $Q=\Delta H$ under the stated work restriction, so the same thermal reservoir accounting yields $\Delta S_{\mathrm{total}}=-\Delta G/T\ge 0$. Variable composition restores ${\sum }_i{\mu }_{i}d{n}_i$ to $dG$; constrained Gibbs minimization for chemical equilibrium handles those permitted changes.

The historical foundation is Gibbs's ^[ref]. Its equations (87) and (91) introduce the combinations now called Helmholtz free energy and Gibbs free energy, connecting entropy criteria, chemical potentials and phase coexistence within one framework.

The Helmholtz free energy iswhere $U$ is internal energy, also denoted $E$ elsewhere here, $T$ is absolute temperature, and $S$ is entropy. For the closed system of fixed composition and only pressure-volume work described above, spontaneous relaxation at fixed temperature and volume cannot increase $F$. Stable thermodynamic equilibrium is a global minimum over the permitted alternatives, including different thermodynamic phases. A rigid vessel fixes the total volume; it does not require each coexisting thermodynamic phase to retain its own volume.

For a constructed comparison at $300\mathrm{K}$, suppose one candidate has $1.5\mathrm{kJ}$ more internal energy but $6\mathrm{J}{\mathrm{K}}^{-1}$ more entropy than another at the same temperature and volume. Its difference isThus the candidate with higher internal energy has lower Helmholtz free energy. This is the practical reason for retaining the entropy term when selecting a state.

For equilibrium states within a smooth branch of fixed composition, the first law of thermodynamics gives $dU=TdS-pdV$, henceThese derivatives extract entropy and pressure; the differential is not an evolution equation for relaxation. If chemical species amounts $n_i$ vary, add ${\sum }_i{\mu }_{i}d{n}_i$, with molar chemical potentials ${\mu }_i$. A closed system can still change chemical species amounts through chemical reactions; the fixed-composition simplification then needs the conserved-inventory constraints of chemical equilibrium. See MIT 5.60, Lecture 14, fundamental equations and chemical potentials.

The minimum principle has a short explanation. At fixed volume, no boundary work occurs, so $Q=\Delta U$. A thermal reservoir at positive temperature $T$ loses entropy $Q/T$. Consequentlyby the Second law of thermodynamics. Here both endpoint states share the imposed temperature. See MIT 5.60, Lecture 12, criteria for spontaneous change.

The microscopic counterpart is $F=-k_{B}T\ln Z$ in the canonical ensemble, with Boltzmann constant $k_B$ and partition function $Z$. The partition function examples compute the balance between internal energy and entropy; the chemical equilibrium example already displays an ideal-mixture Helmholtz free energy minimum.

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

An intensive quantity is unchanged under replication at the same thermodynamic equilibrium. Temperature, pressure, chemical potential and the mass density of a uniform substance are examples. Independence from sample size does not mean independence from state: a material's mass density can change with temperature or pressure.

The relation to extensive quantities is particularly elegant for a differentiable bulk fundamental relation. Put $\mathbf{x}=(S,V,n)$ and writeTaking a partial derivative with respect to $x_i$ and cancelling $\lambda$ givesThe equilibrium differential $dU=TdS-pdV+\mu dn$ identifies these derivatives with temperature, minus pressure and molar chemical potential. They are degree-zero homogeneous functions. The same degree-one assumption yields the Euler relation in thermodynamics, but a surface energy contribution requires extra care.

An intensive quantity generally has no meaningful sum across samples: two equal temperatures do not add to a doubled temperature. Neither is the final temperature after thermal contact generally an unweighted average; heat capacities and the first law of thermodynamics determine the balance. Ratios of additive inventories lead to the useful weighted rules in normalizing extensive quantities.

Internal energy $U$ is the energy stored in the microscopic degrees of freedom of a thermodynamic system: random molecular kinetic energy, internal molecular excitations, and interaction potential energy. Its value requires a specified system and an energy reference. In thermodynamic equilibrium, internal energy is a state function; it records the present state, without remembering whether a previous change came from stirring, compression or heat transfer. This makes internal energy the common accounting quantity connecting mechanical experiments to thermal ones.^[ref]

Heat and work describe energy crossing the system boundary during a process. They are not separately stored contents of the system. For a closed system whose bulk kinetic energy and external potential energy have unchanged endpoint values, the first law of thermodynamics givesHere $Q_{\mathrm{in}}$ is positive for incoming heat, and $W_{\mathrm{on}}=-W_{\mathrm{out}}$ is positive for net work done on the system. Two routes between the same equilibrium states have the same $\Delta U$, although their heat and work can differ. The infinitesimal notation $dU=\delta Q_{\mathrm{in}}+\delta W_{\mathrm{on}}$ distinguishes a state function change from path-dependent transfers.^[ref]

For a calculated example, take one mole of a classical monatomic gas in the ideal gas regime, at fixed volume, from $300$ to $310\mathrm{K}$. With stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, its internal energy change is $3nR\Delta T/2=124.71\mathrm{J}$. A purely thermal route supplies $Q_{\mathrm{in}}=124.71\mathrm{J}$ with $W_{\mathrm{on}}=0$. An ideal insulated stirring route supplies $W_{\mathrm{on}}=124.71\mathrm{J}$ with $Q_{\mathrm{in}}=0$, measured between equilibrium endpoints after the motion dies away. The same stored increment has two different transfer histories; neither history defines a separate kind of stored energy.

The familiar internal energy of an ideal gas is a useful special case. Neglect intermolecular interactions, hold the amount of substance $n$ and composition fixed, and suppose that $f$ independent quadratic terms per particle obey the classical equipartition theorem. Choosing the constant reference contribution as zero givesHere $T$ is absolute temperature, $R$ is the molar gas constant, and $C_V$ is the whole sample's heat capacity at constant volume. A monatomic gas has $f=3$ translational terms. A rigid diatomic molecule has $f=5$ when its two rotational terms are active and vibration is frozen; activating one harmonic oscillator vibration adds two terms, one kinetic energy and one potential energy, giving $f=7$. The counting is of quadratic contributions, so a vibrational coordinate is not just one contribution.^[ref]

This constant-$f$ formula applies only over a regime with those same active terms, classical translational statistics, nonrelativistic motion and unchanged composition. Quantum freezing, quantum degeneracy, chemical reactions or appreciable intermolecular interactions invalidate that simple counting. More generally, at fixed amount and composition an ideal gas can still have $U=U(T)$ with varying heat capacity at constant volume, so use $\Delta U={\int }_{T_1}^{T_2}{C}_V(T)dT$. A nonzero constant reference adds $U_{\ast }$ to $fnRT/2$ without altering $\Delta U$.

An isolated system exchanges neither matter nor energy with its surroundings. Its total energy is fixed, while the Second law of thermodynamics requires its total entropy to be nondecreasing. Its internal energy alone is fixed when bulk kinetic energy and external potential energy also remain unchanged. For example, an initially spinning stirrer inside an insulated rigid vessel can lose kinetic energy while the contents gain internal energy, with no external input.

An isolated system need not yet be in thermodynamic equilibrium. Two enclosed bodies at different temperatures can exchange heat internally until they reach thermal equilibrium; their combined energy stays constant. Their internal transfer is absent from the outer boundary ledger. A vacuum-insulated container is only an approximation to isolation: residual thermal conduction and thermal radiation can still carry energy across its boundary.

An isothermal process changes a thermodynamic system while keeping its temperature constant. It need not keep its pressure, volume or internal energy constant, and it does not generally have zero heat transfer. An adiabatic process specifies zero heat transfer instead. For example, a slowly expanding ideal gas can absorb heat from a heat bath and deliver the same amount of work while its temperature stays fixed.

For a fixed amount of substance and composition, internal energy of an ideal gas depends only on temperature. With $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work delivered to the surroundings, the first law of thermodynamics therefore givesThis equality does not determine either transfer until the path is specified. Isothermal ideal-gas work gives the reversible value; an insulated free expansion has a different value even when its equilibrium endpoints have the same temperature. The closed system convention here excludes changes in bulk kinetic energy and external potential energy.

A conducting cylinder in a large, stirred heat bath approximates an isothermal process when the piston moves slowly enough for thermal conduction to keep up. A quasistatic process keeps the gas close to internal thermodynamic equilibrium; a reversible thermodynamic process additionally requires negligible friction and vanishing driving differences in pressure and temperature. Exact equality of the gas and heat bath temperatures gives no finite heat transfer rate through a finite thermal resistance. The reversible description is a limit of arbitrarily small differences and arbitrarily slow motion, illustrated by the finite-rate isothermal piston model.^[ref]

For a real gas, internal energy at constant temperature can change with volume because molecular interaction potential energy changes. In one smooth thermodynamic phase of a fixed-composition simple compressible system,Thus an isothermal process generally has $Q=\Delta U+W_{\mathrm{by}}$, with no reason to set $\Delta U=0$. The right-hand side vanishes for the ideal gas law, but not generally for another equation of state.^[ref]

Alternating sheet-like domains of the components of a microphase-separated material. In diblock systems the two layer thicknesses can be unequal and reflect unequal block fractions. Confinement, surface wetting and kinetic trapping can produce layered films outside the thermodynamic equilibrium bulk lamellar region.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

A membrane osmometer connects osmosis to a visible liquid level or a measured balancing pressure. In a teaching version, a solute-containing pouch is joined to a narrow vertical tube and submerged in external water. Entry increases the enclosed volume and raises the meniscus. Both the column and any stretched pouch can contribute an opposing pressure.

The University of Toronto's Osmosis (Brine) by University of Toronto Department of Physics demonstration (activity Me-012b, Description, steps 1–14) specifies soaking a dialysis membrane tube, tying one end closed and fastening the other around a thistle tube's narrow end. A stand holds the tube. Colored brine is cooled to room temperature and injected using a syringe to reduce bubbles; the pouch is completely immersed in a $1$ or $2\mathrm{L}$ beaker of distilled water. A taped marker records the initial level; subsequent meniscus height is the readout. The protocol proposes another pouch containing distilled water as a comparison, and its older instructions explicitly check seals for leaks.^[ref]

A rise is consistent with net water entry. To infer osmotic pressure quantitatively, use a retained solute and establish a stable equilibrium after accounting for dilution. If both free surfaces are at the same atmospheric pressure, the liquids have approximately equal mass densities $\rho$, and the pouch adds negligible mechanical load, the hydrostatic pressure difference at membrane height is approximately $\Delta P=\rho gh$, where $h$ is the meniscus height above the external surface and $g$ is gravitational acceleration. This is the liquid-column inference, not a pressure calibration supplied by the Toronto demonstration. For unequal mass densities, compare the two column pressures separately.

Useful additional controls are matched initial levels and geometry, constant temperature, replicate runs, and monitoring external solute to test retention. These are proposed checks, not reported Toronto results. Leaks, trapped bubbles, surface tension in the narrow tube, evaporation and membrane stretching can alter the level. In particular, a dialysis membrane may pass dissolved salt: the brine setup illustrates transient movement but cannot be assumed to retain every dissolved ion. Color improves visibility without proving salt retention. A stationary level can reflect weakening concentration differences or leakage rather than osmotic thermodynamic equilibrium; a rising level alone supplies neither $n$ nor a verified value of $nRT/V$.

A membrane potential is an electric potential difference across a cell membrane. We use the conventionTaking the external electric potential as zero makes $V_m<0$ mean that the inside is at lower electric potential. “Negative inside” describes a difference, not an absolute electric potential, and does not imply that the entire cytosol has a large net negative electric charge.

Consider a cell membrane permeable only to potassium ions, with higher potassium concentration inside. Diffusion initially favors outward transport. If balancing negative ions cannot follow, that loss of positive electric charge produces a small inner excess of negative electric charge and an outer positive excess. The resulting electric field favors inward potassium transport. At the Nernst equilibrium potential, chemical and electrical contributions to electrochemical potential balance; ions still cross both ways, with zero net transport.^[ref]

In living biological cells, several ions can permeate and pumps maintain their concentration differences through active transport. Pumps transporting net electric charge also contribute electric current. Thus $V_m=E_{\mathrm{K}}$ is a special limit, not a universal rule; membrane potential need not always be negative.^[ref] The existing reversal potential example shows how opposing electric currents can balance without each ion reaching thermodynamic equilibrium. This makes selective permeability a way to convert chemical differences into electrical signals: changing the available ion channels changes which electrochemical gradients influence the voltage.

The transition between a compositionally disordered state and an ordered microdomain state as effective segregation strength increases. The parameter $\chi N$ combines unlike-segment interaction preference and chain length; its transition value depends on composition and model assumptions. A bulk thermodynamic equilibrium boundary is not by itself an evaporation or vitrification kinetic model.

Osmosis is net movement of water across a boundary whose permeability differs for water and relevant solutes. The direction depends on the chemical potential of water, including both dissolved material and pressure. A cell wall can let water reach the cell membrane, yet oppose expansion through turgor pressure. Zero net movement does not mean individual water molecules have stopped crossing.

Imagine pure water on side A and a chemical solution on side B of a semipermeable membrane. At equal temperature and pressure, dissolved particles lower the chemical potential of water in B, and net water transfer initially favors A to B. Retaining the solutes lets water redistribute without simply mixing everything. Increasing B's pressure opposes entry; enough pressure gives thermodynamic equilibrium with unequal solute concentrations. This is why “water moves toward more solute” requires a pressure condition.^[ref]

At thermodynamic equilibrium, the chemical potentials of exchangeable water are equal. Individual water molecules still move and cross in both directions, with equal mean opposing transfer rates and zero net flux. A stationary liquid level alone does not prove this: a blocked route can also give no visible transfer. Selective permeability provides a route; water potential supplies the direction, and neither alone determines the speed. The cell wall, turgor pressure and plasmolysis articles show why a living boundary's mechanical support and its selective permeability must be distinguished.

Osmotic pressure is the excess pressure needed on a chemical solution to maintain osmotic thermodynamic equilibrium with pure solvent across a semipermeable membrane. For a nearly incompressible solvent, equality of chemical potential gives $\Pi \simeq -(RT/{\bar{V}}_w)\ln a_w$. It is an equilibrium property; an observed initial swelling rate is a different measurement.^[ref]

For a dilute ideal chemical solution of retained particles,Here $n$ is the amount of substance of independently dissolved solute particles, measured in moles; $V$ is the final volume of chemical solution, not the starting volume of pure solvent. Count the resulting ions separately if a solute dissociates; do not multiply twice if $n$ already counts them. With $c_{\mathrm{osm}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$ and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, $RT{c}_{\mathrm{osm}}$ has units $\mathrm{J}{\mathrm{m}}^{-3}=\mathrm{Pa}$. This resembles the ideal gas law, although the particles here are dissolved in a liquid.

The assumptions are uniform temperature, a dilute ideal chemical solution, a nearly incompressible solvent, and a semipermeable membrane that retains the counted solutes while passing solvent. Association, incomplete dissociation and interactions change the relation between added material and independent particles; nonideal thermodynamic activities invalidate simple particle counting. The formula does not describe a concentrated brine merely because the apparatus contains a membrane.

A short derivation explains why particles are counted. In an ideal mixture, $a_w=x_w=n_w/(n_w+n)$, with $n_w$ the amount of substance of water. If $n\ll n_w$, then $-\ln x_w=\ln (1+n/n_w)\simeq n/n_w$ and $V\simeq n_w{\bar{V}}_w$. Substitution gives $\Pi \simeq RTn/V$. This is a dilute-limit approximation, not a claim that water molecules stop moving or that dissolved particles are a gas.