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On the straight sloping leg, $p=p_0(3-V/V_0)$ and $dp=-p_{0}dV/V_0$. Put $v=V/V_0$, which increases from $1$ to $2$. The reversible heat transfer isThus heat is absorbed for $1<v<15/8$ and emitted for $15/8<v<2$, with zero instantaneous heat at $v=15/8$. The switching point has $(p,V)=(9p_0/8,15V_0/8)$. In particular the net heat on this leg, $\frac{3}{2}{p}_0{V}_0$, is not the total positive heat on it.

For a reversible change, the first law of thermodynamics is $TdS=dU+pdV$. Using the monatomic internal energy,Along the constant-volume leg, $Q_{AB}=\frac{3}{2}{V}_0(2p_0-p_0)=\frac{3}{2}{p}_0{V}_0>0$, so heat is absorbed. Along the constant-pressure return leg, $Q_{CA}=\frac{5}{2}{p}_0(V_0-2V_0)=-\frac{5}{2}{p}_0{V}_0$, so heat is emitted. The work signs are zero on $AB$ and negative, or work on the gas, on $CA$.

The total work by the gas is the enclosed triangular area, $W=\frac{1}{2}{p}_0{V}_0$. The positive heat on the sloping leg isAdd $Q_{AB}=48p_0{V}_0/32$, excluding the emitting part of $BC$ and all of $CA$. HenceAs an independent conservation of energy check, the heat emitted on late $BC$ is $p_0{V}_0/32$ and on $CA$ is $80p_0{V}_0/32$; thus $Q_{\mathrm{in}}-Q_{\mathrm{out}}=16p_0{V}_0/32=W$. Using only the net heat on $BC$ would give an incorrect thermal efficiency.

Macroscopically, an adiabatic process exchanges no heat with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore isentropic, since $dS=\delta Q_{\mathrm{rev}}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a heat bath; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a thermodynamic cycle. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the Carnot cycle. Reversibility and zero net entropy change giveThe thermal efficiency is the net work output divided by the absorbed heat:ConsequentlyThe Carnot theorem shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

The stagnant layer transports heat by thermal conduction, so its upward heat flux isAt the maximum-flux interface from part iii,Equating $F_s$ to $F_{\max }$ givesThe stagnant layer has positive thickness only ifFor an ice-covered freshwater lake, take $T_s\simeq 0{}^{\circ }\mathrm{C}$ and $T_m\simeq 4{}^{\circ }\mathrm{C}$. The largest interior temperature compatible with this penetrative convection in an ice-covered lake model is therefore

A phase diagram partitions a space of thermodynamic variables into regions occupied by different equilibrium thermodynamic phases and records the phase coexistence curves between them. For a pure, one-component substance, a pressure–temperature phase diagram gives one point for each pair $(T,p)$. In a single-thermodynamic phase region both controls can vary independently. On a two-thermodynamic phase phase coexistence curve, they obey one relation; at a triple point, both are fixed. These statements assume bulk thermodynamic equilibrium, fixed composition and no additional applied fields.^[ref]

At fixed temperature and pressure, the stable bulk thermodynamic phase minimizes Gibbs free energy for a fixed amount of substance. For a pure substance this is a comparison of chemical potentials on a common normalization. Two candidate thermodynamic phases coexist where ${\mu }_{\alpha }(T,p)={\mu }_{\beta }(T,p)$ and neither is exceeded in stability by a third thermodynamic phase. Differentiating their equality gives the Clapeyron equation. Thus a phase coexistence curve is a locus of equal stability, rather than a trajectory that every heated sample must follow.^[ref]

The following original cartoon uses dimensionless positive coordinates $\theta =T/T_{\ast }$ and $\pi =p/p_{\ast }$, with arbitrary reference scales $T_{\ast },p_{\ast }$. It represents no particular substance and contains no measured data. Stipulate a triple point at $(1,1)$, a liquid-gas critical point at $(3,5)$, and these drawing rules:They specify sublimation, liquid–gas phase coexistence and melting boundaries for the picture. They are not an equation of state, do not predict latent heat or mass density, and do not supply a quantitative critical theory. In particular, the melting slope is merely an illustrative choice.

For example, at fixed $\pi =3$ the melting boundary is at $\theta =1.2$ and the liquid–gas boundary is at $\theta =2$. Heating from $(\theta ,\pi )=(1.6,3)$ to $(3.6,3)$ therefore starts in liquid, crosses phase coexistence at $(2,3)$, and ends in gas. In an equilibrium conversion at that crossing, added heat can change the phase fraction at fixed temperature through latent heat. The pressure–temperature point alone does not specify how much of each thermodynamic phase is present: for that one needs an additional amount or volume constraint.

Now connect exactly the same endpoints by raising $\pi$ from $3$ to $6$ at $\theta =1.6$, moving to $\theta =3.6$ at $\pi =6$, then lowering $\pi$ to $3$. The first leg stays on the liquid side because ${\theta }_{\mathrm{melt}}$ ranges only from $1.2$ to $1.5$ and ${\pi }_{\mathrm{vap}}(1.6)=2.2<3$. The second leg lies above the entire liquid–gas boundary, whose largest drawn pressure coordinate is $5$; the third lies beyond its largest temperature coordinate, $3$. No leg crosses a first-order phase transition boundary. The middle leg passes through a supercritical fluid region, with $\theta >3$ and $\pi >5$, and the final leg returns to a gaslike state continuously.

Avoiding a first-order phase transition does not mean that no heat or work is exchanged. Along the detour, mass density, enthalpy and heat capacity may change substantially but continuously. The liquid-gas critical point itself has singular behavior, so the detour goes around it rather than through it. This connected fluid region is the striking lesson of a phase diagram: a familiar change from liquid to gas need not involve boiling across a sharp boundary.

An equilibrium phase diagram also does not predict how rapidly a conversion occurs. Metastability and a barrier described by classical nucleation theory can delay the observed phase transition. Nor should a binary-composition phase diagram be read as the same picture: its coexistence regions and lever rule encode composition and phase fractions, and can occupy an area even though pure-substance two-thermodynamic phase coexistence occupies a curve in the pressure–temperature plane.

In the published chlorine experiment, negative ions were accelerated and mass-selected, then overlapped with a tunable laser beam. A downstream electric field deflected charged ions; neutral chemical atoms continued to a plate, whose secondary Electrons were counted. Scanning the laser frequency produced an onset in this neutral-atom signal. The apparatus therefore inferred an electron affinity from a threshold, rather than measuring the heat of attachment.^[ref]

Reversing the laser direction and averaging the two thresholds corrected the first-order Doppler shift. A Fabry–Pérot interferometer and reference spectral lines calibrated the frequency. Finite laser bandwidth, unresolved quantum state splitting and shifts of the calibration lines limited the inference; the authors included these effects in fitting and the error estimate. The level diagram shows only the extracted gap, not a measured signal curve.^[ref]

Photosynthesis captures light energy and transforms it into chemical forms that a biological cell can use. In plants, algae and cyanobacteria, this supports carbon fixation: incorporation of carbon from carbon dioxide into chemical compounds used to build living material. Capturing light and incorporating carbon are distinct processes coupled through metabolism. See OpenStax Biology 2e, 8.1: overview of photosynthesis.

For oxygenic photosynthesis, a simplified net equation writes a six-carbon sugar, glucose, as the representative product:This is net stoichiometry, not a one-step mechanism. A worked bookkeeping check counts six carbon chemical atoms, twelve hydrogen chemical atoms, and eighteen oxygen chemical atoms on each side: the product side has six oxygen chemical atoms in the sugar and twelve in molecular oxygen. The light label supplies energy, not additional chemical atoms.

Crucially, the released molecular oxygen comes from water, not carbon dioxide. Cancelling water made and consumed in different steps hides this distinction in the net equation. The Calvin cycle supplies a net three-carbon triose phosphate, such as glyceraldehyde 3-phosphate, for subsequent sugar synthesis; it does not directly export glucose. See OpenStax Biology 2e, 8.2: light-dependent reactions and OpenStax Biology 2e, 8.3: using light energy to make organic molecules.

The useful achievement is connecting incoming light to both usable energy and new biological material. ATP is an immediate energy currency; NADPH supplies Electrons for synthesis; carbon dioxide supplies the carbon incorporated into carbohydrates. These roles are complementary. Conservation of energy still holds: photosynthesis transforms energy, never creates it, and some absorbed energy is dissipated as heat. A plant also carries out cellular respiration, including in biological cells that perform photosynthesis. Gas exchange reflects both processes. Oxygenic photosynthesis remains central to interpreting terrestrial molecular oxygen as an atmospheric biosignature gas. See OpenStax Biology 2e, 8.3: photosynthesis and respiration.

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.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

Radiometric absorptance is the fraction of incident radiant flux converted to heat in the specified receiver: $P_{\mathrm{abs}}=a{P}_{\mathrm{in}}$, with $0\le a\le 1$ for a passive receiver. Specify the entrance boundary, wavelength, incidence angle and receiver temperature. Radiation reflected or transmitted out of that receiver is not absorbed power. A cavity can improve radiometric absorptance by giving light repeated opportunities for wave absorption; a dark coating alone does not establish $a=1$.

For broadband light, spectral radiant flux matters: $P_{\mathrm{abs}}=\int a(\lambda ){\Phi }_{e,\lambda }(\lambda )d\lambda$. A single effective radiometric absorptance therefore depends on the incident optical spectrum as well as on the receiver. If the stated incident boundary lies before a window of known power transmission ${\tau }_w$, the receiver absorbs $a{\tau }_w{P}_{\mathrm{before}}$ when other optical losses are absent. Window loss and receiver radiometric absorptance are separate corrections.