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The change in overpotential per decade of current density in the activation-controlled regime of a Tafel equation. In a single-step Butler-Volmer equation with constant transfer coefficient, the anodic slope is $b_a=RT\ln 10/({\alpha }_{a}nF)$, where $T$ is temperature, $R$ is the molar gas constant, $n$ is the number of transferred electrons and $F$ is the Faraday constant. The magnitude of the cathodic slope is $b_c=RT\ln 10/({\alpha }_{c}nF)$. Inferring one from the other requires the model's relation between the two coefficients; the equality ${\alpha }_a+{\alpha }_c=1$ applies to this effective single-step model and need not describe a multistep reaction.

The thermodynamic activity is the dimensionless quantity defined by ${\mu }_i={\mu }_i^{\circ }+RT\ln a_i$, where $R$ is the molar gas constant. For an ideal dilute solution with a specified concentration reference, $a_i=c_i/c^{\circ }$; departures from ideality are represented by an activity coefficient multiplying this ratio.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

The Van der Waals equation is a simple real gas equation of state that can also describe a liquid-gas critical point. With particle count $N$ fixed, write the volume per particle as $v=V/N$:or equivalently $(p+a{N}^2/V^2)(V-Nb)=N{k}_{B}T$. Here $p$ is pressure, $T$ is absolute temperature, and $k_B$ is the Boltzmann constant. In SI, $v$ and $b$ have units ${\mathrm{m}}^3$ per particle, $a$ has units $\mathrm{Pa}{\mathrm{m}}^6=\mathrm{J}{\mathrm{m}}^3$, and $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$; the particle count is dimensionless. The pole at $v=b$ excludes smaller volume values from this model.

The $b$ correction represents excluded volume; the attraction term lowers pressure at fixed temperature and particle number density $1/v$. This is a mean-field theory, replacing detailed intermolecular structure by two parameters. At $v\gg b$ with $a/v^2\ll k_{B}T/v$, it approaches the ideal gas law $pv=k_{B}T$. It supplies a useful qualitative route from a dilute gas to phase coexistence, without a claim of quantitative accuracy for every fluid.^[ref]

Do not mix this particle convention with the molar volume convention elsewhere in this article. With amount of substance $n_{\mathrm{mol}}=N/N_A$, Avogadro constant $N_A$, and molar gas constant $R=N_A{k}_B$, one has $v_m=N_{A}v$, $b_m=N_{A}b$, and $a_m=N_A^{2}a$, so $p=RT/(v_m-b_m)-a_m/v_m^2$. The molar parameters have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, respectively.

Johannes Diderik van der Waals introduced this equation of state in 1873, connecting molecular attraction with departures from the dilute ideal gas law. Its historical significance is the unified treatment of gases and liquids, rather than a universally exact fit.^[ref]

A hypothetical Van der Waals equation model illustrates positive and negative Joule-Thomson coefficient regions and a finite throttle that first warms, then cools. Use the molar convention already distinguished in Van der Waals equation:Here $v_{\mathrm{m}}$ is molar volume, $R$ the molar gas constant, $u_{\mathrm{m}}$ internal energy per mole, and $C_{V,\mathrm{m}}$ the stipulated constant molar heat capacity at fixed volume. The second equation is an additional caloric assumption: an equation of state alone does not specify the temperature dependence of internal energy. It satisfies the equilibrium identity $(\partial u_{\mathrm{m}}/\partial v_{\mathrm{m}}{)}_T=T(\partial p/\partial T{)}_{v_{\mathrm{m}}}-p=a_{\mathrm{m}}/v_{\mathrm{m}}^2$. The arbitrary energy reference is set to zero.

Define dimensionless variables $r=v_{\mathrm{m}}/b_{\mathrm{m}}$, $\theta =R{b}_{\mathrm{m}}T/a_{\mathrm{m}}$, $\pi =b_{\mathrm{m}}^{2}p/a_{\mathrm{m}}$ and $\eta =b_{\mathrm{m}}{h}_{\mathrm{m}}/a_{\mathrm{m}}$, where $h_{\mathrm{m}}$ is molar enthalpy. These differ from the critical-value scaling in Van der Waals law of corresponding states. The throttle calculation reduces toThus a prescribed inlet fixes $\eta$, and solving the first and last equations at the outlet pressure gives the final temperature. For an inlet $({\theta }_1,{\pi }_1)=(0.8,0.12)$ and outlet ${\pi }_2=0.025$, the model gives ${\theta }_2\simeq 0.752287$, a decrease of $0.047713$. The inlet $(2.2,0.12)$ instead gives ${\theta }_2\simeq 2.205786$, an increase of $0.005786$. The purple inlet $(1.45,0.30)$ initially warms but ends at ${\theta }_2\simeq 1.437585$ for the same outlet pressure. These are reduced toy values, not kelvin readings or experimental evidence. No physical gas parameters were fitted.

For a short inversion derivation, implicit differentiation of the Van der Waals equation at fixed pressure givesInsert this into $T(\partial v_{\mathrm{m}}/\partial T{)}_p=v_{\mathrm{m}}$ and substitute back into the equation of state. The Joule-Thomson inversion curve isPositive inversion pressure requires $r>3/2$. Its maximum pressure is $\pi =1/3$ at $r=3$, while the upper dilute limit is ${\theta }_{\mathrm{inv}}\to 2$. Using Van der Waals critical point values, the latter is $T_{\mathrm{inv},\max }/T_c=27/4$. The inversion locus depends on the equation of state; the chosen molar heat capacity additionally determines how much temperature changes along each finite throttle.^[ref]

The displayed region has $\theta >8/27$, above the model's critical temperature, and uses only $r>1$, $p>0$, $(\partial p/\partial v_{\mathrm{m}}{)}_T<0$ and positive $C_{p,\mathrm{m}}$. Subcritical calculations require attention to phase coexistence, metastability and the Maxwell construction; an unstable homogeneous continuation is not an equilibrium throttle endpoint. Even on a stable branch, this mean-field theory with constant $C_{V,\mathrm{m}}$ is a qualitative model, not an accurate inversion map for a particular real gas. Phase transitions, changing composition, appreciable heat leakage or bulk kinetic energy changes require a different property calculation or steady-flow energy balance. At phase coexistence, the smooth single-thermodynamic phase coefficient formula cannot replace a balance involving the phase fraction.

An ideal gas law measurement obtains amount of substance without knowing the chemical formula in advance:Here $p$ is vapor pressure, $V$ is its occupied volume, $T$ is absolute temperature, $R$ is the molar gas constant, and $m$ is the mass of vapor occupying that volume at those conditions.

Santa Monica College's teaching experiment uses a $125\mathrm{mL}$ Erlenmeyer flask with a pinhole foil cap, a $600\mathrm{mL}$ water bath on a hot plate, a thermometer, and a barometer. Excess volatile liquid displaces air. A dissolved dye loses its colour and deposits when the liquid evaporates; three more minutes of heating follow. Cooling condenses retained vapor. Subtracting the dry assembly mass from the cooled assembly mass supplies $m$. The flask volume comes from brim-full water mass and mass density at its measured temperature. Controls include a dry exterior, excluding bath water, and three trials.^[ref]

The inference assumes vapor pressure equals atmospheric pressure, vapor and bath reach thermal equilibrium, the retained vapor mass is recovered on cooling, and the vapor obeys the ideal gas law. Residual air, thermal expansion of the flask, and vapor temperature below bath temperature are explicit error questions in the teaching experiment. Repeating trials assesses repeatability; it does not remove a shared bias.^[ref]

The virial expansion describes the first departures from the ideal gas law as a power series in number density. For a homogeneous, single-species gas in three dimensions at thermal equilibrium, write $n=N/V$, with $V$ the volume, and define the dimensionless compressibility factor $Z$ byHere $p$ is absolute pressure, $T$ is absolute thermodynamic temperature, and $k_B$ is the Boltzmann constant. The second virial coefficient determines the initial slope away from $Z=1$; higher coefficients describe further corrections. The symbol $Z$ here is an equation of state ratio, not a canonical partition function. ^[ref]

The left panel sketches the measurement principle; the right panel uses deliberately chosen polynomials, $Z=1+bx+6x^2$ with $x=n{\ell }^3$ and $b=\pm 2$, where $\ell$ is a reference length. These curves are not measured data. Their initial slopes have opposite signs, but both bend because of the quadratic term. Using a finite secant from $x=0$ to $x=h$ gives $(Z(h)-1)/h=b+6h$, rather than $b$: reducing the fitting range reduces this particular truncation bias. This algebra explains why measuring one nonideal pressure is insufficient to extract a reliable second virial coefficient.

Waxman and Hastings's 1971 Burnett experiment used sample and expansion chambers in a thermostatic bath, with a valve between them and a pressure transducer linked to a piston gauge. They measured successive isothermal expansions: evacuate the expansion chamber, open the valve, record the equilibrated pressure, isolate and evacuate that chamber again, then repeat. This avoids separately measuring the absolute gas amount and chamber volumes. For carbon dioxide at $50{}^{\circ }\mathrm{C}$, their measurements below $35$ bar gave a molar second virial coefficient of approximately $-102.2\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. ^[ref]

An idealized reduction makes the signal explicit. Let $q=(V_A+V_B)/V_A$ be the calibrated chamber volume ratio, $p_i$ the pre-expansion pressure, and $p_f$ the post-expansion pressure. Particle conservation during one valve opening gives $n_f=n_i/q$. At the same temperature, division of the two equations of state yieldsThus the intercept determines $q$, while the dilute-limit slope determines the second virial coefficient. In the actual experiment, helium runs calibrated the cell constant and chamber deformation corrections were included. Agreement of the carbon dioxide low-pressure intercept with the helium calibration, and the absence of an anomalous low-pressure upturn, checked for gas adsorption on chamber walls. ^[ref]

For a limited dilute-gas fit, hold temperature fixed, allow thermal equilibrium after every expansion, calibrate absolute pressure and the chamber ratio, and check leaks and sample composition. Repeat the analysis with a smaller upper pressure cutoff and an added curvature term; compare the resulting second virial coefficients with their uncertainties. Too wide a range mixes in higher coefficients; too narrow a range makes the nonideal signal hard to distinguish from instrument error. The displayed chamber equation neglects deformation, dead spaces and adsorption, so it is a teaching reduction rather than the complete experimental analysis.

Units matter. With number density in ${\mathrm{m}}^{-3}$, $B_2$ has units ${\mathrm{m}}^3$ and $B_3$ has units ${\mathrm{m}}^6$. For molar density $c=n/N_{\mathrm{A}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$, the same equation of state isHere $N_{\mathrm{A}}$ is the Avogadro constant and $R=N_{\mathrm{A}}{k}_B$ is the molar gas constant; the molar coefficients have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and ${\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$. A coefficient tabulated per mole must therefore multiply molar density, not number density.

For the classical formulas below, use a nondegenerate gas with thermal de Broglie wavelength ${\lambda }_{\mathrm{th}}$ satisfying $n{\lambda }_{\mathrm{th}}^3\ll 1$, negligible quantum corrections to relative motion, and a stable, sufficiently short-ranged pair potential. Diluteness must also control the omitted interaction terms: $|B_{2}n|\ll 1$ alone is not enough when positive and negative contributions cancel. A finite truncation is a local low-number density approximation, not an equation of state for a dense liquid or across a phase transition. ^{[ref] [ref]}

Water potential expresses the chemical potential of water relative to a stated pure-water reference in units of pressure. For a nearly incompressible liquid at fixed temperature, use a constant reference molar volume ${\bar{V}}_w$ and writeThe reference is pure water at the same temperature and a chosen pressure $P_0$. Both sides must use the same reference. Neglecting height differences and interactions with solid surfaces, the approximation isHere $a_w$ is the dimensionless thermodynamic activity of water, $R$ the molar gas constant, and $T$ the absolute temperature. The natural logarithm is therefore applied to a dimensionless quantity. The pressure contribution ${\Psi }_p=P-P_0$ increases with pressure; the solute potential ${\Psi }_s$ is negative when $a_w<1$. Across an available passive water route, net water transfer favors higher to lower water potential. The familiar plant description adds effects of height and interactions with solid surfaces when they matter; they are omitted in the two liquid compartments here.^[ref]

Adding solute and raising pressure can thus cancel in water potential while each contribution remains nonzero. Equality of water potential does not imply equality of pressure or of every solute concentration. In a supported plant cell, the cell wall permits a positive turgor pressure to balance a negative solute potential; plasmolysis illustrates the opposite loss of support after water leaves.