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Gas outside the uniform-temperature bulb of a gas thermometer contributes to its conserved particle inventory. An idealized calculation divides the connecting tube into small regions with volumes $V_j$ and local temperatures $T_j$. Neglect hydrostatic pressure offsets and thermal transpiration, assume a common pressure $P$ and a local ideal gas law in each region, and let the bulb have volume $V_b$ and temperature $T_b$. Adding the particle inventories givesThus even an ideal gas in a perfectly rigid sealed apparatus need not obey $P\propto T_b$: some of its particles remain in regions with other temperatures. For a stipulated bulb of $100\mathrm{c}{\mathrm{m}}^3$, a connecting volume of $1\mathrm{c}{\mathrm{m}}^3$ maintained at $300\mathrm{K}$, and bulb reference and test temperatures of $300$ and $400\mathrm{K}$, conservation of $N$ givesThese are toy dimensions and a calculated bias, not measured data. Reducing $N$ reduces both pressures proportionally and leaves this ratio unchanged: extrapolating away gas nonideality does not eliminate the connecting-volume error. Actual apparatus also requires the tube's thermal expansion, a measured temperature profile, and the pressure corrections omitted in this model.

The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

The Gibbs-Duhem equation constrains changes of the intensive quantities describing one uniform thermodynamic phase. For a mixture with $k$ independently specified component amounts, its molar form isHere $S$ is entropy, $V$ is volume, $T$ is temperature, $p$ is pressure, $n_i$ is the amount of substance of component $i$, and ${\mu }_i$ is its molar chemical potential. This assumes a simple compressible system whose bulk internal energy is an extensive quantity; additional work coordinates require additional terms.^[ref]

The proof exposes the source of the constraint. For internal energy $E(S,V,n_1,\dots ,n_k)$, replication of the whole thermodynamic phase by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\dots ,\lambda n_k)=\lambda E(S,V,n_1,\dots ,n_k)$. The Euler theorem for homogeneous functions and the fundamental differential of internal energy therefore giveTaking the differential of the first equation produces all the terms in the second, plus $SdT-Vdp+{\sum }_i{n}_{i}d{\mu }_i$. Subtraction proves the Gibbs-Duhem equation. Thus the constraint comes from two descriptions of the same state function, rather than an assumed microscopic interaction model.

At fixed temperature and pressure, divide by $n={\sum }_i{n}_i>0$ and introduce the mole fractions $x_i=n_i/n$:The chemical potentials cannot be varied independently. In a two-component mixture, put $x_A=x$ and $x_B=1-x$. Within a smooth single-thermodynamic phase region with $0<x<1$,One chemical potential fixes the other's change, but determining its absolute value still requires a reference value. This does not say ${\mu }_A={\mu }_B$: equality at phase coexistence concerns the same chemical species in different thermodynamic phases. Nor does the constraint say $dG=0$ whenever temperature and pressure are fixed; changing composition still contributes ${\sum }_i{\mu }_{i}d{n}_i$ to the Gibbs free energy.

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.

For the equation of state $V=N{k}_{B}T/p-NA(T)$, integrate $(\partial G/\partial p{)}_{T,N}=V$ along an isotherm. This gives the displayed Gibbs free energy difference. Its negative temperature derivative at fixed endpoint pressures gives $\Delta S=-N{k}_B\ln (p_1/p_0)+N{A}^{\prime }(T)(p_1-p_0)$. The physically admissible branch must retain positive volume.

The Gibbs free energy of mixing compares the final mixture's Gibbs free energy with that of its initially separate components at the same temperature and pressure. For an ideal solution, subtract the pure-component reference from $G={\sum }_i{n}_i{\mu }_i$:For one mole of the chosen A/B ideal solution at $300\mathrm{K}$ and $x_A=0.4$,The individual chemical potential changes are $RT\ln 0.4\approx -2.28554\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for A and $RT\ln 0.6\approx -1.27417\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for B. Weighting these by $0.4\mathrm{mol}$ and $0.6\mathrm{mol}$ reproduces the total Gibbs free energy of mixing.

If ideality holds in a neighborhood of the specified temperature and pressure, differentiation at fixed component amounts givesThus the ideal mixing Gibbs free energy decrease comes entirely from the entropy increase; the mixing enthalpy and volume changes vanish. Ideality at a single isolated state alone does not determine temperature or pressure derivatives.^[ref]

The same formula rules out spontaneous separation into two liquid thermodynamic phases within this model. Per mole, the mixing Gibbs free energy $g_{\mathrm{mix}}=RT[x\ln x+(1-x)\ln (1-x)]$ hasAdding the pure-component reference contributions changes only a term linear in $x$. The total molar Gibbs free energy is therefore a strictly convex function: a separated pair of different compositions has a larger composition-weighted Gibbs free energy than the uniform sample with the same inventory. Such phase separation requires physics beyond this ideal solution model.

Adding sodium hydroxide to acidified glycine successively removes its two titratable protons. Two broad buffer solution regions appear around the two pKa values, separated by a steep pH rise near the isoelectric point. This is useful evidence for two separated acid-base chemistry steps even though the changing chemical species are invisible. In the approximation that added hydroxide only changes glycine protonation, $b=1-\bar{z}$ is the added base amount of substance per initial amount of substance of glycine. Then $b\approx 0.5$, $1$ and $1.5$ correspond to $\mathrm{p}{K}_{a1}$, $\mathrm{pI}$ and $\mathrm{p}{K}_{a2}$; endpoints are not exactly pure chemical species. ^[ref]

For the solid-curve model, start with $n_G=0.00200\mathrm{mol}$ of glycine in $V_0=0.0200\mathrm{L}$ and exactly $n_G$ of fully dissociated hydrochloric acid; this fixes the chloride inventory, but does not force every glycine molecule to remain $G^{+}$. Let the sodium hydroxide concentration be $C_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$ and $b=n_{\mathrm{NaOH}}/n_G$. Assume volume is additive and no other ions are present; stipulate $K_w=10^{-14}$ for self-ionization of water at a temperature of $25{}^{\circ }\mathrm{C}$. ThenBalancing all positive and negative electric charges requiresSolving this equation gives the solid curve. It remains finite at $b=0$ and beyond $b=2$, where ignoring free hydronium and hydroxide would fail. At the first pKa, the full model gives $b\approx 0.436$, rather than $0.5$: appreciable free hydronium makes that half-equivalent shortcut inaccurate. At the second pKa, $b\approx 1.502$. Known acid inventory, dilution, and thermodynamic activity assumptions are therefore part of any quantitative interpretation, not optional corrections inferred from the curve's shape.

The Amrita Virtual Lab procedure uses $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ glycine, hydrochloric acid and sodium hydroxide, two burettes, a pH meter, beaker and stirrer. A pipette transfers $20\mathrm{mL}$ of glycine into a $100\mathrm{mL}$ beaker. Calibrate the meter with standard buffer solutions at pH $4$, $7$ and $10$, then record the starting pH. Add hydrochloric acid in $0.3\mathrm{mL}$ increments, recording pH after each addition until $1.6$. Rinse the electrode in distilled water, take a separate fresh $20\mathrm{mL}$ portion, and add sodium hydroxide in $0.3\mathrm{mL}$ increments until pH $12.5$. Its observed readout is paired added volume and pH, plotted as an acid-base titration curve; it does not measure each glycine form directly. ^[ref]

These are two titrations from the starting glycine chemical solution, rather than the single acidified-sample protocol assumed by the solid curve in glycine titration. Retain each portion's initial volume and the signed added acid/base amount of substance; do not concatenate raw burette readings and assume the result begins at $b=0$. For this procedure's $0.00200\mathrm{mol}$ of glycine, each $0.3\mathrm{mL}$ addition supplies $0.000030\mathrm{mol}$ of acid or base, a $0.015$ increment relative to the glycine inventory. This calculated resolution is not a reported measurement.

Useful additional controls are a fresh repeat portion, independently checked acid/base concentrations, a water blank, and a calibration-check buffer solution after the run. Stir without striking the glass electrode, keep its sensing region immersed and wait for stable readings; record temperature and background salts. Reduce addition size near a steep pH change. These are suggested controls for testing the model, beyond the published steps. A water blank can reveal contamination but must not be subtracted directly from pH, which is a logarithm.

The standards at pH $4$, $7$ and $10$ do not bracket either endpoint of this protocol. Extreme readings therefore require independently checked instrument performance, and alkaline samples are sensitive to atmospheric carbon dioxide uptake. Electrode drift, incomplete mixing, volume error, dilution and changing ionic strength can shift apparent pKa values. A titration constrains the acid/base model; it cannot establish chirality or separate every microscopic protonation form. The two original diagrams show calculated predictions to compare with measured points, not reconstructed experimental data.

A hard-sphere gas is a classical model of identical impenetrable spheres, each of diameter $\sigma$ and radius $\sigma /2$. Its pair potential is $u(r)=+\infty$ for centre separation $r<\sigma$ and $u(r)=0$ for $r\ge \sigma$; the convention at exact contact has no effect on the integral. It isolates the effect of particle size without an attractive potential energy.

The upper-left panel is a cross-section of a three-dimensional forbidden region: the second centre cannot enter a sphere of radius $\sigma$ around the first centre. The particles themselves have radius $\sigma /2$. Consequently, with particle volume $v_{\mathrm{s}}$ and pair excluded volume $v_{\mathrm{ex}}$,The last equality follows directly from the Mayer f-function: $f=-1$ throughout the forbidden volume and zero outside it. Confusing particle radius with contact separation changes the answer by a factor of eight; omitting the unordered-pair factor changes it by a further factor of two.

Define the packing fraction $\eta =n{v}_{\mathrm{s}}$, where $n$ is number density. Then the virial expansion beginsFor example, at the toy value $\eta =0.01$, the leading correction predicts $Z\approx 1.04$; the omitted terms still matter if accuracy beyond this approximation is needed. The classical second virial coefficient is independent of temperature because the geometry of forbidden configurations does not change when the gas is warmed. The pressure itself still scales with temperature at fixed number density.

This model makes the positive correction particularly clear: eliminating overlapping configurations changes the dependence of the available configuration space on volume, even though the allowed configurations all have zero interaction energy. The figure's other panels add a toy attractive shell and show how its Boltzmann factor competes with that geometry. The hard-sphere gas is a useful reference model, while a low-order virial expansion remains a dilute-gas approximation rather than a dense sphere packing formula.

Heat capacity measures the heat supplied per small temperature increase along a specified process:The subscript $X$ records the constraint, such as fixed volume or fixed pressure. The units are $\mathrm{J}{\mathrm{K}}^{-1}$. The symbol $\delta Q$ emphasizes that heat is a transfer along a path, rather than a state function with a differential $dQ$. Consequently, an unspecified heat capacity does not determine the required heat.

For a closed system of fixed composition undergoing a quasistatic process, assume that only pressure-volume work occurs, with no friction and no change in bulk kinetic energy or potential energy. The first law of thermodynamics and $H=U+pV$, where $U$ is internal energy and $H$ is enthalpy, then giveThese partial derivatives define heat capacity at constant volume and heat capacity at constant pressure, with the amount and composition also held fixed. Integrating along the selected constraint gives $Q={\int }_{T_1}^{T_2}{C}_X(T)dT$ within a single thermodynamic phase; $Q=C_X\Delta T$ requires constant $C_X$. ^[ref]

For a computed example, take a monatomic gas in the classical ideal gas regime, with amount of substance $n=1\mathrm{mol}$ and stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The equipartition theorem gives $U=3nRT/2$, and the ideal gas law gives $pV=nRT$. ThusFor the same temperature increase $\Delta T=10\mathrm{K}$, the two processes requireHere work done by the gas is positive. At fixed pressure, the extra heat supplies expansion work; the internal energy increase is the same because internal energy of an ideal gas depends only on temperature. Their final volume values differ. ^[ref]

The practical value of heat capacity is that a measured heat input and temperature response can determine it. Once its constraint and temperature dependence are known, it predicts heating requirements and supplies entropy changes through entropy from heat capacity. ^[ref]

For a fixed-composition closed system, the heat capacity at fixed pressure is $C_p=(\partial H/\partial T{)}_p$, with enthalpy $H$. Under the quasistatic process assumptions in heat capacity, only pressure-volume work is permitted, so $dp=0$ gives $\delta Q=dH=C_{p}dT$. Thus supplied heat is $Q=\Delta H$, or $Q=C_p\Delta T$ within a single thermodynamic phase when $C_p$ is constant. For an ideal gas, the expansion work explains why this exceeds the required heat at fixed volume for the same temperature increase. Its per-unit-mass counterpart is specific heat capacity at constant pressure.

For a fixed-composition closed system, the heat capacity at fixed volume is $C_V=(\partial U/\partial T{)}_V$, with internal energy $U$. Under the quasistatic process assumptions in heat capacity, only pressure-volume work is permitted, so $dV=0$ gives $\delta Q=dU=C_{V}dT$. A rigid vessel therefore needs $Q=\Delta U$, or $Q=C_V\Delta T$ when $C_V$ is constant. Its per-unit-mass counterpart is specific heat capacity at constant volume.

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat $Q_h>0$ at fixed temperature $T_h$, and the cold thermal reservoir receives heat of magnitude $Q_c>0$ at fixed temperature $T_c<T_h$. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus $Q_h$ enters with a plus sign and $Q_c$, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy $U$, returns to its starting value:The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation, $Q_h>Q_c$ and $W>0$. Its thermal efficiency isUse net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies $P_{\mathrm{out}}={\dot{Q}}_h-{\dot{Q}}_c$ and $\eta =P_{\mathrm{out}}/{\dot{Q}}_h$, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at $T_h=600\mathrm{K}$ and $T_c=300\mathrm{K}$, let $Q_h=1000\mathrm{J}$ and $Q_c=600\mathrm{J}$ per thermodynamic cycle. Then $W=400\mathrm{J}$ and $\eta =2/5=40\%$. At two thermodynamic cycles per second, these become $2000\mathrm{W}$ incoming heat transfer, $1200\mathrm{W}$ outgoing heat transfer, and $800\mathrm{W}$ net power. These are constructed values, not measurements.

Different signs describe different tasks. With the same directions of heat transfer, $Q_h=Q_c$ gives $W=0$: a device transmitting heat without net work output. If $Q_c>Q_h$, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work $W_{\mathrm{in}}$ and the balance $Q_h=Q_c+W_{\mathrm{in}}$. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have $Q=W$ and $\Delta U=0$ while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance toEquality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the $50\%$ bound: its total entropy production is $-1000/600+600/300=1/3\mathrm{J}{\mathrm{K}}^{-1}$ per thermodynamic cycle.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.

Heat-engine calorimetry measures heat transfers so that observed work can be compared with the correct energy input. Measuring motion alone cannot supply the denominator of thermal efficiency.

The institutional Laboratory 16: Stirling Engine describes a modern teaching arrangement: an ethanol lamp drives a Stirling engine coupled to an electrical output device, which supplies a resistor immersed in water in a calorimeter. The prescribed readouts include fuel mass consumed, the pressure-volume loop, electrical readings, and water temperature. The sequence separates fuel input, gas-cycle work, electrical output, and Joule heating of the water. Warming of the water is the final useful output; it is not the engine's rejected $Q_c$. The handout distinguishes warm-up from steady operation and asks students to compare the final warming with direct heating by the lamp. It gives a procedure, not a table of measured efficiencies. Its pp. 1–3 therefore establish an apparatus and measurement design without supplying a new experimental result.

For a proposed calibration, use a known input of electrical work to determine the effective heat capacity $C_{\mathrm{eff}}$ of the water, vessel and immersed parts; stir to make the measured temperature representative. A matched run with no input estimates exchange with the room. For constant $C_{\mathrm{eff}}$ and no phase transition, the resulting model iswhere $Q_{\mathrm{leak},\mathrm{out}}$ is the signed net heat lost to the room during the measurement. Here $E_{\mathrm{in}}$ is supplied electrical work when the resistor lies inside the calibration boundary; for a separate external heat input it is the received heat. These are proposed controls, not controls reported in that handout. Slow cooling, changing material properties and imperfectly matched room exchange limit this inference; the approach to thermal equilibrium and heat capacity matter as much as the thermometer reading.

To measure the actual engine-input $Q_h$, one must instrument its hot-side transfer separately, or close a calibrated energy balance around a specified heater boundary. A heater's electrical work includes changes in its stored internal energy and heat transfer that bypasses the engine. At periodic operation a cold-side flowing-water measurement can instead estimate ${\dot{Q}}_c=\dot{m}{c}_p(T_{\mathrm{out}}-T_{\mathrm{in}})$, if mass flow rate $\dot{m}$ and specific heat capacity at constant pressure $c_p$ are known, with negligible storage and other heat exchange in that measurement boundary. It measures only the rejection captured by that stream. Include all remaining transfers before using the residual to infer ${\dot{Q}}_h=P_{\mathrm{out}}+{\dot{Q}}_c$. This measurement discipline makes a comparison with Carnot's theorem meaningful.

A published teaching experiment used ATCC CCL-2 HeLa to make the distinction between a cell line and the condition of individual biological cells visible. Cultures were expanded in treated flasks and supplied as adherent dishes. A wash with phosphate-buffered saline preceded trypsin treatment to detach them; added fetal bovine serum stopped the digestion. The dispersed preparation was mixed with an equal volume of trypan blue and loaded into a counting chamber under an optical microscope.^[ref]

For the ethanol comparison, treatment groups and an untreated control were brought to the same final volume; the control received phosphate-buffered saline. Dye-excluding counts were compared across replicates. The paper reports representative student measurements in Figure 3A and documents the untreated control in its discussion. Its alternative temperature comparison used three replicates per condition and a $30\min$ exposure.^[ref]

The paper's disposable chamber had a different count-to-number density calibration from the hemocytometer above. Dye exclusion does not establish future division; recovery and mixing also affect counts. The constructed hemocytometer example is independent of those observations.^[ref]

Pyotr Leonidovich Kapitza's 1938 test compared helium flow through optically flat glass disks. Suspension set the tube-to-bath level difference; a cathetometer read levels. The gap was estimated as $0.5\mu \mathrm{m}$. Levels barely moved over minutes above the transition, but equalized in seconds below. A laminar flow viscosity calculation gave only an upper bound.^[ref]

The essential control is to compare the same restriction and level readout at different temperatures. A changed pressure head is not itself evidence for a changed material response: the driving head must be included when interpreting the flow. For an idealized comparison at equal vapor pressure, negligible surface tension corrections and nearly constant mass density $\rho$, hydrostatic equilibrium givesHere $\Delta h$ is the difference in free-surface heights and $g$ is gravitational acceleration. This explains why following a liquid level provides both a driving pressure and a transported volume; it does not turn a finite-resolution observation into proof of zero viscosity.

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.

A hemocytometer has a calibrated grid and depth. ATCC describes $1{\mathrm{mm}}^2$ squares $0.1\mathrm{mm}$ deep, each sampling $10^{-4}\mathrm{mL}$.^[ref]

An original constructed hemocytometer count uses four such squares after mixing the biological cell preparation with the same volume of stain. The dye-excluding counts are $36,44,40,40$; the dye-stained counts are $4,6,5,5$. The dilution factor is $D=2$. With $v$ the sampled volume per square and $\bar{n}$ the arithmetic mean count per square,Hence the undiluted preparation has dye-excluding number density $2(40)/10^{-4}=800000{\mathrm{mL}}^{-1}$ and total number density $2(45)/10^{-4}=900000{\mathrm{mL}}^{-1}$. Its dye-excluding fraction is $160/180=8/9\approx 88.9\%$. If the well-mixed preparation contains $3\mathrm{mL}$, it contains an estimated $2400000$ dye-excluding biological cells. The chamber sample itself contains only $160$ counted dye-excluding biological cells.

The relation $c=D\bar{n}/v$ follows from counting material in a known volume, then undoing dilution. It assumes a representative mixed sample and correctly identified single biological cells. Clumps, inconsistent decisions at grid edges, wrong chamber depth and nonrepresentative sampling introduce measurement error. A fixed convention such as including top and left edges while excluding bottom and right edges assigns shared-edge centers to just one square. Adding ten times as much suspension to a correctly filled chamber does not make each square's calibrated volume ten times larger.

In homogeneous catalysis, the catalyst and reactants occupy the same thermodynamic phase.

For a liquid example, specify hydrogen peroxide as the reactant, dissolved iodide as the catalyst, and water as the solvent. If both reacting chemical species remain dissolved in that one liquid thermodynamic phase, this is homogeneous catalysis. The University of Wisconsin–Madison's The Rate of a Reaction identifies iodide-catalysed hydrogen peroxide decomposition in its final example; the aqueous setting here makes the phase roles explicit.^[ref] Its net chemical equation isThe molecular oxygen is a reaction product: its bubbles do not make the dissolved catalytic route heterogeneous catalysis. The classification compares the participating catalyst and reactants, rather than demanding that everything in the vessel share a phase.^[ref]

Being in the same thermodynamic phase describes where the chemical species are; it does not specify their reaction mechanism. A uniform-looking liquid can contain unresolved solid particles, so appearance alone is insufficient. Conversely, suspended solid particles do not establish that their surfaces supply the observed catalysis: dissolved material released from them could contribute. Phase identification and catalytic participation need separate evidence.

The bookkeeping of a catalyst in homogeneous catalysis is particularly transparent in the schematic catalytic cycle A plus C forming AC, followed by AC releasing B plus C. This is the generic model in catalyst, not an asserted iodide reaction mechanism. Suppose a fixed volume initially contains $1.00\mathrm{mmol}$ of A and $0.20\mathrm{mmol}$ of C. At a chosen snapshot, let $0.60\mathrm{mmol}$ of B and $0.12\mathrm{mmol}$ of AC be present, with no other chemical reactions. Each AC contains one A unit and one C unit, soThus free A is $0.28\mathrm{mmol}$ and free C is $0.08\mathrm{mmol}$. B formation already exceeds the initial amount of substance of C by a factor of three, while the total C inventory remains $0.20\mathrm{mmol}$. Adding the two steps cancels C and AC, giving the net A-to-B chemical reaction; counting their constituents proves these balances. Conserved total catalyst does not mean constant free catalyst concentration.

Dissolved contact is useful because reactants can encounter the catalytic chemical species throughout the mixed liquid, without first reaching a macroscopic solid surface. Recovering a dissolved catalyst from the same chemical solution can instead require a separation. Neither ease of contact nor phase classification determines an activation energy or an equilibrium constant; those depend on the relevant chemical kinetics and endpoint thermodynamics.

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

The ideal gas law is the equation of state relating the absolute pressure $P$, volume $V$, particle count $N$, and absolute thermodynamic temperature $T$ of a classical ideal gas:Here $k_B$ is the Boltzmann constant, $N_{\mathrm{A}}$ is the Avogadro constant, $R$ is the molar gas constant, and $n_{\mathrm{mol}}$ is the amount of substance in moles. The distinct symbol $n_{\mathrm{mol}}$ avoids confusing amount of substance with the number density $n=N/V$, measured in ${\mathrm{m}}^{-3}$: the same law can be written $P=n{k}_{B}T$. Use absolute pressure, relative to a vacuum, rather than an unconverted gauge pressure. ^[ref]

The exact SI definitions of the Avogadro constant and Boltzmann constant giveThis numerical value is exact; the applicability of the ideal gas model to a real gas is an approximation. ^{[ref] [ref]}

For $n_{\mathrm{mol}}=1\mathrm{mol}$, $T=300\mathrm{K}$, and $V=0.025{\mathrm{m}}^3$, direct substitution givesAt fixed temperature and particle count, doubling the volume to $0.050{\mathrm{m}}^3$ halves the pressure to approximately $49.887\mathrm{kPa}$: this is Boyle's law. At fixed volume and particle count, doubling the temperature from $300\mathrm{K}$ to $600\mathrm{K}$ doubles the pressure to approximately $199.547\mathrm{kPa}$. These ratios use kelvins, not readings in the degree Celsius scale: $300\mathrm{K}$ is $26.85{}^{\circ }\mathrm{C}$, and $600\mathrm{K}$ is $326.85{}^{\circ }\mathrm{C}$.

The curved slices show Boyle's law at three temperatures; the straight slice shows proportionality to absolute temperature at one fixed volume. They are relations among equilibrium states, rather than a prescribed process history. Compression could hold the temperature fixed through heat transfer, or raise it during adiabatic compression. To calculate a process's heat or work, supplement the ideal gas law with the process constraint, the first law of thermodynamics, and the relevant internal energy or heat capacity information. The graph's extrapolation toward $T=0$ is mathematical; it does not assert that a real gas remains classical or gaseous there.

For a stellar mixture with fixed mean molecular weight, the ideal gas law can be written $P=\mathcal{R}\rho T/\mu$, where $\rho$ is mass density and $\mathcal{R}$ is the stellar gas constant. It assumes negligible interaction corrections and a nondegenerate gas; excitation or ionization can change the particle count and thermodynamic response.

In his 1834 memoir on the motive power of heat, Émile Clapeyron combined gas laws into a relation between pressure, volume, and temperature while analysing heat engines (printed page 164). The equation above uses modern SI notation and absolute thermodynamic temperature; it is not a transcription of his historical temperature convention. ^[ref]