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For an ideal mixture, the Helmholtz free energy is a sum over species. At fixed $T,V$ and all other populations, the molecular partition function $q_i$ does not depend on $N_i$. Differentiating the contribution written using each species' own energy reference givesThe desired additive energy term requires a consistent energy reference in molecular partition functions. Let ${\epsilon }^{0,i}$ be the energy per molecule of species $i$ in its reference state, measured relative to a common chemical energy zero, and let $q_i$ sum excitations above that reference. ThenThereforeOne convenient choice is each species' rovibrational ground state, so the molecular reference includes its zero-point energy. A different consistent reference is possible, but the corresponding partition function must change. The offset cannot arise from differentiation of a formula that has already omitted it, and it must not be counted twice. These chemical potentials are per particle; molar expressions instead use the molar gas constant.

Here $p$ is absolute pressure, $V$ is volume, $n$ is amount of substance in moles, $R$ is the molar gas constant and $T$ is absolute thermodynamic temperature in kelvin. This is the ideal gas law; consistent SI units give pressure in pascals, volume in cubic metres and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Let $G$ be Gibbs free energy, $H$ enthalpy and $T$ thermodynamic temperature for a fixed amount and composition of material, with the derivative taken at constant pressure. Apply the Gibbs–Helmholtz equation to the standard molar reaction quantities, obtained by summing product minus reactant contributions with their stoichiometric coefficients:Define $K(T)$ as the dimensionless thermodynamic equilibrium constant expressed using activities in specified standard states. Since ${\Delta }_r{G}^{\circ }=-RT\ln K$, where $R$ is the molar gas constant,If the reaction enthalpy is constant over the interval, integration givesThis is the requested Van't Hoff equation. For an exothermic reaction, increasing temperature decreases $K$. Although the name “isochore” is often used in this context, the displayed enthalpy form follows from standard Gibbs energies and the stated constant-pressure Gibbs–Helmholtz identity; the analogous ideal-gas concentration form involves internal energy when the temperature dependence of standard concentration is retained.

Rearrange the Arrhenius relation using the molar activation energy and the molar gas constant:The exponent is dimensionless. There is no Avogadro factor in it because both $E_a$ and $R$ are expressed per mole; the molecular concentration convention only determines the units of the prefactor.

The Arrhenius equation isHere $E_a$ is the molar activation energy, $R$ the molar gas constant, $T$ the absolute temperature, and $A$ the pre-exponential factor, taken approximately constant over the measured range. For this bimolecular elementary reaction, $A$ and $k_1$ both have units $\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$.

Take representative samples of a measured volume, identify the impurity by an appropriate reaction, and express the blank-corrected amount per volume of ward air. Excess carbon dioxide, moisture, organic contamination, Ammonia and suspended material are the principal groups to investigate; combustion or defective drains may introduce additional gases. Smell alone measures none of them reliably.

Record occupancy, the position and height of sampling, open doors or windows, heating and lighting, temperature and sampling time. Compare indoor and outdoor samples collected under comparable conditions. Avoid breathing into the collecting vessel. A sample near an open window may miss conditions at the beds, and a sample during cleaning may contain unusually disturbed dust. Explain the sampling conditions with the result.

For carbon dioxide, “carbonic acid” in period terminology, collect a known volume in a calibrated vessel and introduce a measured portion of standardized barium hydroxide solution. Shake to promote absorption. Carbon dioxide consumes the alkaline reagent and forms barium carbonate:Determine the residual alkali in the clear liquid, excluding the carbonate precipitate from an acid determination that would redissolve it. Compare with an equal unexposed reagent portion. An 1864 report describes 7–10 litre globes and checks against gravimetric absorption, so this is an available period method rather than an instrument projected backwards.^[ref]

If $c$ is the hydrogen ion equivalent concentration of the titrating acid, and $v_b$ and $v_s$ are the respective blank and exposed-solution titration volumes, then, for the entire absorption charge,The factor two arises because one carbon dioxide molecule consumes two alkali equivalents. If only an aliquot is titrated, multiply by the ratio of total solution volume to aliquot volume. Here $P$, $T$ and $V_{\mathrm{air}}$ describe the same measured gas sample, with consistent wet- or dry-gas conventions; $R$ is the molar gas constant. Report $10^6{x}_{\mathrm{C}{\mathrm{O}}_2}$ as parts per million by volume. Other absorbed acidic gases are potential interferences. The calculation is a stoichiometric and ideal gas law interpretation, not a direct measure of infectiousness.

For water vapour, compare dry- and wet-bulb thermometers to estimate relative humidity, or pass a known volume through a suitable preweighed drying absorber and determine its mass gain. The latter gives collected water mass divided by sampled air volume, with a blank and appropriate control of other absorbable vapours. Humidity is relevant to ward conditions, but is not by itself evidence that the air contains a particular disease agent.

For Ammonia, moist red litmus becoming blue is a screening reaction, not a selective quantitative analysis. Absorb a measured volume of air into a known amount of dilute acid; the reduction in residual acidity measures absorbed volatile bases. Other alkaline vapours interfere, so do not call the result pure Ammonia without further discrimination. A modern comparison is the validated acid-treated sorbent collection and ion chromatography of the institutional air-sampling method; it too requires calibrated flow, blanks and checks for sample loss.^[ref]

For suspended dust, collect material on a preweighed filter, condition and reweigh it, and divide the blank-corrected mass increase by sampled volume. Microscopy can describe fibres, cells and other particles, but cannot turn an unspecified speck into an identified pathogen. A modern gravimetric sampling method illustrates the importance of controlled filter handling and air-volume measurement.^[ref]

Period investigators also assessed oxidizable organic contamination through the consumption of an oxidizing reagent such as potassium permanganate. Such a result is an operational measure of chemical consumption, dependent on the procedure; it does not reveal a unique substance or count living organisms. An 1865 review already warned that inorganic reducing substances also decolourized the reagent, making supposed quantitative organic estimates unreliable.^[ref] Similarly, a lead-salt darkening reaction may support the presence of hydrogen sulfide where drains are implicated, but this gas need not be present in every ward. Investigate carbon monoxide if combustion is suspected: it is odourless and cannot be excluded because the room smells fresh.^[ref]

A useful report therefore separates measured carbon dioxide, humidity, chemical contamination and particles, states detection limits and interferences, and avoids declaring ward air safe from a single result. Infectious aerosols require a separate assessment; settling plates or a few microscopic fields do not quantitatively characterize everything that patients inhale. Current infection-control guidance likewise warns about the limitations of nonspecific environmental sampling.^[ref]

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

For the numbered engine, denote the low and high pressures by $p_1=p_4=p_l$ and $p_2=p_3=p_h$. Let $R_g$ denote the molar gas constant. The ideal gas law makes the constant-pressure branches straight rays in a temperature-volume diagram:The high-pressure ray is steeper. For a constant heat capacity ratio $\gamma$, each reversible ideal-gas adiabat obeysConsequently $1\to 2$ curves upward and left during compression; $2\to 3$ goes upward and right along the steeper ray; $3\to 4$ curves downward and right during expansion; and $4\to 1$ goes downward and left along the shallower ray.

The plot is an original calculation for $\gamma =7/5$, $p_h/p_l=10$ and $V_3/V_2=2$; its state directions follow the original paper. The dashed extensions identify the two constant-pressure rays, not additional stages of the cycle.

Use distinct symbols for the dimensions of mass, length, time, temperature and amount of substance. The molar gas constant has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}{\mathsf{\Theta }}^{-1}{\mathsf{N}}^{-1}$, while the given $M$ is a molar mass with dimensions $\mathsf{M}{\mathsf{N}}^{-1}$. Matching dimensions in the proposed power law giveswith the amount-of-substance equation repeating $\alpha +\gamma =0$. ThereforeDimensional analysis fixes these exponents but not a dimensionless multiplier. For example, an open-tube mode number and the gas's heat-capacity ratio enter the actual resonance frequency without changing the dimensions.

Assume an ideal gas dominated by carbon dioxide, constant mean molar heat capacity $C_p$, and a dry convective adiabat connecting the stated emitting layer to the ground. Conservation of entropy gives $C_{p}d\ln T=Rd\ln p$, where $R$ is the universal molar gas constant. ThusThe estimate extrapolates across a pressure ratio of $450$. It neglects radiatively stable layers, variable heat capacity, departures from an ideal gas and changes in composition; these approximations limit its accuracy. A brightness temperature observed from space cannot generally be equated to a unique material level, but this calculation uses the layer interpretation explicitly supplied by the question.

Notation. The source uses $p_s$ for both the hypothetical dry surface pressure and the actual surface pressure after evaporation. Distinguish them as $P_d$ and $P$, respectively, and interpret the specified pressure thickness $\Delta p$ as fixed. This conserves the original hydrogen inventory and permits $P$ to vary. Reading both occurrences as one fixed number would be inconsistent with adding water vapor to a closed atmospheric column mass budget.

Let $E(T)$ denote the saturation pressure of water, let $x$ be the mole fraction of water vapor in the mixed layer, and let $M_d,M_v$ be the molar masses of hydrogen and water. Surface phase equilibrium and the ideal gas law giveBecause the layer has constant temperature, its relative humidity isThus every level strictly above the surface is subsaturated: hydrostatic balance makes $p<P$, whereas the saturation pressure stays constant. The result does not require a dilute water vapor approximation.

The Clausius-Clapeyron relation, neglecting the specific volume of liquid water and treating the water vapor as an ideal gas, givesHere $T_t,E_t$ are the triple point values, $L_v$ is the specific latent heat, and $R$ is the molar gas constant. All temperatures in these expressions are absolute. As throughout the question, this is a constant-latent heat, ideal gas model, not an extrapolation valid through the real critical point of water.

For a well-mixed ideal gas mixture, its mean molar mass and water vapor mass fraction areBy hydrostatic balance, the mixed layer has total mass per unit area $\Delta p/g$, of which $w_v\Delta p/g$ is water vapor. The remaining atmosphere contains no water vapor. Consequently conservation of hydrogen mass givesEquivalently the requested quadratic equation isFor each $T_s$, calculate $E$, take the positive root of this quadratic equation, and retain it only if $0\le x<1$ and $P>\Delta p$. The last condition leaves a nonempty dry layer. In particular the dry constituent's surface partial pressure is not generally $P_d$: $P_d$ measures its column weight, and the hydrogen and water vapor have different molar masses.

Use a rock strength envelope: at each depth plot the smaller differential stress needed for frictional failure and for sustained ductile material creep at a specified strain rate. Shallow rock generally strengthens with depth because effective confining pressure increases; sufficiently deep hot rock weakens because thermally activated material creep becomes efficient.

At shallow depth, Byerlee's law or a Mohr-Coulomb failure criterion gives a relation of the formwhere $C$ is cohesion, $\mu$ a coefficient of friction, ${\sigma }_n$ the compressive normal stress and $P_f$ the pore pressure. Increasing overburden generally increases effective stress, hence the frictional strength. Conversion from fault shear stress to differential stress depends on the stress regime and fault orientation; a sketch should not label a universal numerical slope. High pore pressure or weak fault gouge can lower the frictional strength substantially.

A simplified dislocation creep relation isHere $Q$ is activation energy, $V^{\ast }$ activation volume, $R$ the molar gas constant and $T$ absolute temperature; $A$ and $n$ depend on composition and deformation mechanism. At fixed strain rate, increasing temperature usually decreases the required stress strongly. The competing effects of pressure, water weakening of rocks, grain size and composition also matter. The intersection with frictional failure gives a brittle-ductile transition, not a change from solid to liquid.

For the Indian continental interior, a useful first sketch has an upper-crustal strength peak, a weaker lower continental crust, and renewed strength in lithospheric mantle beneath the Moho. Quartz-bearing rocks can undergo ductile material creep at conditions where colder, drier olivine-rich mantle retains appreciable strength. This creates a possible two-peak, or “jelly sandwich”, rock strength envelope. With further depth the mantle also weakens. A cold craton can retain strong lithospheric mantle to much greater depths than a tectonically heated continental region.

For the central Pacific, show thin mafic oceanic crust above a relatively thick layer of strong lithospheric mantle. There need not be a conspicuous weak lower-crustal layer. Much of the integrated strength can reside below the Moho, with a broad strength peak followed by ductile weakening towards the asthenosphere. Cooling and age control the thickness: young hot oceanic lithosphere would give a much shallower and weaker profile than an old central-Pacific plate.

The plotted Moho depths of about 38 km and 7 km, and all plotted stresses, are illustrative assumptions. “India” alone does not specify a unique geothermal gradient, water content or strain rate, and the Pacific includes plates of different ages. The robust comparison is the possible crust–mantle layering in a continent versus the thin oceanic crust and mantle-dominated strength of an old oceanic plate. The Moho is a compositional boundary; neither it nor the deepest small plotted strength is automatically the seismic lithosphere-asthenosphere boundary.

Left: ordinary frictional strengthening with depth followed by thermally activated material creep. Right: a very weak shallow faulted layer above a narrow strong region near 10 km, followed by essentially the same weakening material creep branch. These are the shapes actually plotted in Figure 2. The right curve is not a strong upper crust over a uniformly weak lower crust; neither diagram displays a second mantle-strength peak.

For the left curve, increasing overburden increases the effective normal load on faults. A schematic Coulomb friction law iswhere $C$ is cohesion, $\mu$ the coefficient of friction, ${\sigma }_n$ the compressive normal stress and $P_f$ the pore pressure. With ordinary rock friction and approximately hydrostatic pressure, ${\sigma }_n-P_f$ increases roughly linearly with depth. The stress needed for frictional sliding therefore increases, as in the straight shallow branch. Byerlee's law summarizes the relatively high friction found in many laboratory rocks.^[ref]

At higher temperature, dislocation creep or another ductile mechanism accommodates the imposed deformation at lower stress. For example,Here $\dot{ϵ}$ is strain rate, $Q$ an activation energy, $R$ the molar gas constant and $T$ absolute temperature. At fixed strain rate, increasing $T$ lowers the required stress. The operative envelope is approximately $\min ({\sigma }_f,{\sigma }_d)$, expressed in a consistent stress convention; the peak marks the brittle-ductile transition. This accounts for the left curve's peak near 100 MPa and its decline toward small strength by about 20 km.

For the right curve, normal high-friction sliding with depth-independent fluid-pressure ratio cannot explain the nearly vertical weak shallow branch. Plausible weakening includes high pore pressure tracking the overburden, low-friction fault gouge, or a connected network of pre-existing weak faults. Roughly constant small ${\sigma }_n-P_f$ yields roughly constant small frictional strength. The abrupt increase near the base of this layer requires a change of conditions: for example, the weak fluid-rich fault network terminates or becomes discontinuous, leaving stronger intact or less altered rock below. A change in mineralogy or deformation localization can also give a contrast. Simply saying “there are fluids” does not explain the return to high strength. Below the strong region, thermal material creep again lowers the strength.

The figure alone does not identify the weakening mechanism or a particular lithological boundary. In particular, a 10 km strength maximum is not evidence for a 10 km deep Mohorovičić discontinuity. The strong-fault/weak-fault evidence and the distinction between fault strength and surrounding crustal strength are discussed in Stress orientations at intermediate angles to the San Andreas Fault, California by Jeanne L. Hardebeck and Andrew J. Michael.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

For an original idealized mathematical model, let $n_{{\mathrm{O}}_2}(t)$ be the amount of substance of molecular oxygen consumed since the initial reading and $n_{{\mathrm{CO}}_2}(t)$ the carbon dioxide produced. Suppose a fraction $f$ of that carbon dioxide is removed from the gas, all other gas inventories remain fixed, and temperature $T$ and gas pressure $P$ stay constant. The ideal gas law gives the positive gas contractionwhere $R$ is the molar gas constant and $V$ the gas volume. With complete absorption, $f=1$, the contraction reflects molecular oxygen uptake regardless of the amount of carbon dioxide made. For a uniform pipette bore of area $A$, an inward boundary displacement $x$ gives $D=Ax$. A graduated pipette can instead supply the displaced volume directly.

Let $D_s$ denote sample contraction and $D_b$ the matched blank contraction. Define $D_{\mathrm{corr}}=D_s-D_b$ with both taken relative to their own initial readings. Under the preceding assumptions,The slope of corrected volume against time estimates uptake over the chosen interval. The approximate sign acknowledges that a liquid boundary also responds to hydrostatic and capillary pressure, while the gas can exchange water vapour. A fixed-volume apparatus would instead show a pressure decrease; its ideal conversion is $n_{{\mathrm{O}}_2}\simeq V(P(0)-P(t))/(RT)$ after complete absorption and appropriate blank correction.

The ideal gas law also exposes drift without any biology: for fixed gas inventory, $V\propto T/P$. A one-percent rise in absolute temperature at fixed pressure produces a one-percent expansion. Matching the initial gas spaces makes such common changes easier to subtract. Subtracting unmatched blanks can leave a physical signal that looks biological.

Relative to the specified solid, $Q_{\mathrm{sp}}<K_{\mathrm{sp}}$ favors dissolution if that solid is available; $Q_{\mathrm{sp}}>K_{\mathrm{sp}}$ favors precipitation; equality marks saturation. These comparisons identify a driving direction favored by thermodynamics, not a reaction rate.^[ref]

A proof outline comes from the reaction Gibbs free energy for dissolution:Here $R$ is the molar gas constant and $T>0$ the absolute temperature. If $Q_{\mathrm{sp}}>K_{\mathrm{sp}}$, the logarithm is positive, making the reverse transfer into solid favorable. At equality the bulk driving difference vanishes. Equality can also occur at the boundary where the last solid disappears, but it is not a universal constraint on every solid-free chemical solution.

For an ideal dilute chemical solution, moving one mole of a species with signed ion charge number $z$ from outside to inside changes its electrochemical potential byHere $R$ is the molar gas constant, $T$ the uniform absolute temperature, $F$ the Faraday constant, and $c$ the concentration. This assumes the same reference chemical potential on both sides. At thermodynamic equilibrium, $\Delta \tilde{\mu }=0$, henceThe Nernst equilibrium potential uses thermodynamic activities in the general case; replacing them by concentrations is an ideal-solution approximation.

Choose an original passive model: only ${\mathrm{K}}^{+}$ permeates ($z=+1$), other ions cannot cross, there are no pumps or imposed electric currents, and there is no net water transfer. Stipulate reservoir concentrations $c_{\mathrm{in}}=140\mathrm{mM}$ and $c_{\mathrm{out}}=5\mathrm{mM}$, with $T=310\mathrm{K}$, $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. Treat reservoir concentrations as effectively unchanged by charging. ThenThe concentration ratio is dimensionless; $J/C=\mathrm{V}$ supplies the voltage unit. At $V_m=0$, $\Delta \tilde{\mu }>0$ for inward transport, so outward transport is favored. At $V_m=E_{\mathrm{K}}$ it vanishes. A more negative $V_m$ favors inward transport. These sign checks explain the result physically.

Keeping the other chosen inputs fixed, doubling $c_{\mathrm{out}}$ to $10\mathrm{mM}$ raises $E_{\mathrm{K}}$ by $(RT/F)\ln 2\approx 18.52\mathrm{mV}$. Equal inside and outside concentrations give $E_{\mathrm{K}}=0$; reversing their ratio reverses its sign. In finite reservoirs, transport changes both concentrations; the exact final thermodynamic equilibrium must also satisfy material conservation. The fixed-reservoir calculation is useful when only a small fraction of the available ions moves.

For transfer of the same species $X$ from a gaseous reference state into a solvent $S$, define the molar solvation free energy byHere ${\mu }^{\circ }$ is a standard chemical potential. Negative solvation free energy means that transfer into the stated solvent lowers the Gibbs free energy relative to the stated gaseous reference. Include solute-solvent interactions, making room in the solvent, and its reorganization; counting only a favorable pair interaction omits part of the transfer.^[ref]

State the temperature, gaseous reference pressure, dissolved reference concentration or other standard state convention, and species identity. Reference-state changes alter the quoted solvation free energy. For example, with an ideal dilute thermodynamic activity $a=c/c^{\circ }$, replacing $c^{\circ }$ by $\alpha c^{\circ }$ while preserving the physical chemical potential requires ${\mu }^{\circ }$ to increase by $RT\ln \alpha$. Thus the corresponding solvation free energy increases by $RT\ln \alpha$ if the gaseous reference stays fixed. This follows directly from $\mu ={\mu }^{\circ }+RT\ln a$, with $R$ the molar gas constant.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.