Equations of state and gases

A vessel and four numbers

A stationary vessel contains n = 0.100 mol of a hypothetical gas in V = 2.50 L = 0.00250 m³ at a uniform absolute temperature T = 300 K. Assume the gas obeys the ideal gas model. What pressure should an absolute pressure sensor read? If a second vessel contains twice the amount at the same volume and temperature, what changes?

The selected system is the gas inside the vessel. A sealed vessel is closed to matter transfer; its wall may still transfer energy. An equilibrium state has no macroscopic tendency to change under the allowed exchanges. Uniform pressure and temperature here describe the bulk gas at rest, with gravity, wall layers and other spatial variations neglected. These assumptions make a single p, V and T meaningful.

An equation of state, abbreviated EOS, relates properties of an equilibrium state. The ideal gas equation pV = nRT gives p = 99.8 kPa for the first vessel and 199.5 kPa for the second. The increase follows the number of particles per volume. It does not tell us how the second vessel was filled or how much energy that operation required.

On a first reading, follow the vessel, the ideal gas comparisons, the molecular pressure argument, the mixture and the compression factor Z. Then inspect how the assigned real-fluid model differs from the ideal curve. Sections marked “Further” provide a return path through virial truncation, derivatives, critical conditions and phase branches. Published preparation is “Physical systems, states, boundaries, and environments”, “Dimensions, scaling, and similarity”, and “Conservation of mass, momentum, and energy”. The needed thermodynamic terms are introduced locally; the systems and temperature lessons are not prerequisites that must already be available.

Name amount, pressure and absolute temperature

Pressure p is isotropic normal force per area for the bulk equilibrium fluid; 1 Pa = 1 N/m². The pressure in an EOS is absolute pressure, referenced to vacuum. Gauge pressure is a difference from a specified surrounding pressure: p_gauge = p_absolute − p_ambient. At an assigned ambient absolute pressure of 101 325 Pa, the opening gas has p_gauge ≈ −1.55 kPa while p_absolute ≈ 99.8 kPa. A negative gauge reading is compatible with positive absolute pressure.

Temperature T in these gas equations uses the absolute thermodynamic scale in kelvin, K. Celsius temperature t is related by T/K = t/°C + 273.15. Temperature ratios in gas laws require the absolute scale: doubling a Celsius reading generally does not double T. Thermal equilibrium concerns the absence of a thermal driving difference under thermal contact; the later temperature and zeroth-law subject develops thermometers and the scale.

Amount of substance n counts specified entities in mol. The particle count is N = nN_A, where N_A = 6.022 140 76 × 10²³ mol⁻¹ exactly. The Boltzmann constant k_B = 1.380 649 × 10⁻²³ J/K is exact, and R = N_Ak_B ≈ 8.314 462 618 J/(mol K). A mole is not a mass unit: mass m = nM depends on molar mass M in kg/mol.

Molar volume v_m = V/n has units m³/mol. Mass density ρ = m/V has units kg/m³, and the specific gas constant R_s = R/M has units J/(kg K). Thus pV = nRT, pv_m = RT, p = ρR_sT and pV = Nk_BT express the same ideal relation with different counting conventions. Keep the universal R with moles and R_s with mass. For the assigned M = 0.0280 kg/mol, the opening gas has m = 0.00280 kg, ρ = 1.12 kg/m³ and v_m = 0.0250 m³/mol.

A state relation is not a process

For a uniform single phase with fixed chemical identity and composition, a thermal EOS can be written p = p(T,v_m). Fixing two independent intensive coordinates on a specified branch determines the third locally. This is a conditional statement: near singular points the chosen coordinates can lose independence, and several candidate branches can share T and p. A phase or branch must then be specified or selected by equilibrium conditions.

V and n are extensive properties: replicating the same homogeneous bulk state multiplies both. T, p and v_m stay unchanged. The ideal equation respects this scaling because multiplying V and n by the same factor leaves n/V fixed. Doubling n while holding V fixed is a different operation and increases the ideal pressure. A statement such as “pressure doubles when the gas amount doubles” needs the fixed V and T conditions.

An isotherm is a curve of constant T. At fixed n and T, the ideal relation p ∝ 1/V gives Boyle’s inverse-volume relation. At fixed n and V, p ∝ T. At fixed n and p, V ∝ T. These compare equilibrium states. A real rapid compression can have temperature gradients and a changing temperature; the isotherm alone does not describe that event.

The table gives four separate supplied endpoints. The double-inventory row could belong to a filling experiment, but no inflow temperature, wall exchange or energy history is supplied. Heat and work are transfers during a process; internal energy is a state property. The thermal EOS by itself supplies neither the process history nor all energy properties. The first-law, calorimetry and thermodynamic-potential subjects develop that additional structure.

Why molecular motion produces pressure

The classical dilute-gas picture uses many particles whose sizes and interactions make negligible corrections to the bulk pressure at the state of interest. Translational motion follows Newtonian mechanics; wall encounters transfer momentum. Equilibrium random velocities are isotropic, so their mean squared components are equal. Collisions can redistribute momentum and establish equilibrium even when interaction contributions to the equilibrium EOS are negligible.

For a particle of mass m_1 striking a stationary wall elastically, the wall-normal momentum change has magnitude 2m_1|c_x|. Counting the rate of encounters for many particles gives p = (N/V)m_1⟨c_x²⟩. Isotropy gives ⟨c_x²⟩ = ⟨c²⟩/3, hence p = ρ⟨c²⟩/3. Here c is the random velocity relative to the bulk flow, angle brackets denote an equilibrium average, and c_rms = √⟨c²⟩. A directed jet’s bulk speed cannot simply be inserted as c_rms.

Classical equilibrium translational equipartition gives ½m_1⟨c²⟩ = 3k_BT/2. Combining the two relations yields pV = Nk_BT = nRT and c_rms = √(3RT/M). For the assigned gas at 300 K and M = 0.0280 kg/mol, c_rms ≈ 517 m/s. The stationary vessel has no bulk flow even though individual particles move rapidly.

The translational result is not a universal formula for total internal energy. A classical monatomic ideal gas with only translational energy has U = 3nRT/2 up to a reference constant; molecules can also store rotational, vibrational or electronic energy. Those modes change caloric properties without automatically changing the ideal translational pressure relation. At low enough temperature or high enough density, quantum statistics, interactions, condensation or changes of chemical species can invalidate this model. “Low pressure” needs a temperature, composition and required accuracy; there is no universal cutoff in Pa for every gas.

An ideal mixture: count each component

Now consider a separate nonreacting ideal mixture at the opening volume and temperature, containing n_A = 0.0600 mol and n_B = 0.0400 mol. Its total amount is 0.100 mol. Each component contributes its own wall-momentum flux in the ideal model, so p_i = n_iRT/V and p = Σp_i. This is Dalton’s ideal-mixture pressure relation.

The mole fraction y_i = n_i/n is dimensionless and sums to one. Therefore p_i = y_ip. Here y_A = 0.600, y_B = 0.400, p_A ≈ 59.9 kPa and p_B ≈ 39.9 kPa. Both components occupy the same vessel volume; dividing the container into literal volumes 0.6V and 0.4V is not how this pressure calculation is made.

The convention p_i = y_ip can also define a partial pressure for a real mixture; what is specific to the ideal calculation is p_i = n_iRT/V and its independent wall-pressure interpretation. For a real mixture, composition affects interactions and the EOS. Pure-component correction factors cannot generally be averaged to produce a valid mixture EOS, and p_i = y_ip does not substitute for the chemical-equilibrium quantities needed for a nonideal mixture. Reaction can also change the amounts of the species being counted. Mixture thermodynamics and chemical equilibrium retain those developments; the present numerical mixture has neither interactions nor reaction.

Measure departure with the compression factor

The dimensionless compression factor, also called the compressibility factor, is Z = pV/(nRT) = pv_m/(RT). An ideal gas has Z = 1. At the same T and molar volume, Z is the ratio of the actual EOS pressure to the ideal pressure. At the same T and p, it is the ratio of the actual molar volume to RT/p. Name the comparison held fixed before interpreting a departure.

Attractions can reduce pressure at a given T and v_m, whereas short-range repulsion and excluded-volume effects can increase it. This provides a useful interpretation of some Z departures, not an exact decomposition for every gas and regime. Z < 1 does not mean that repulsion is absent, and Z = 1 at a state can result from cancelling effects. A single Z value also does not determine derivatives, internal energy or all nearby properties.

For a stable dilute classical gas phase at fixed suitable temperature and composition, the ideal relation is a low-density limit. Moving toward high density, a phase boundary, a critical region or a changing chemical composition requires renewed model checks. A very small numerical Z departure may be adequate for one pressure estimate and inadequate for a sensitive derivative. Choose an accuracy goal and compare with a validated EOS or suitable property data over the stated range.

An EOS is a closure relation: it completes a larger calculation by relating variables that conservation alone leaves undetermined. A compressible-flow problem also needs energy information and transport assumptions. The compressible-flows subject treats those balances; inserting p = ρR_sT does not by itself supply a sound speed, a shock solution or a temperature history.

Further: a controlled virial expansion

For a pure gas, a density virial expansion takes the form Z = 1 + B(T)/v_m + C(T)/v_m² + ⋯. The second virial coefficient B has units m³/mol; C has units m⁶/mol². Every term in Z is dimensionless. The coefficients encode material-specific interaction information and depend on temperature. The expansion is useful in a sufficiently dilute single gas phase where the retained terms and remainder meet the required accuracy.

A pressure expansion Z = 1 + B_p(T)p + C_p(T)p² + ⋯ uses different coefficients: B_p has units Pa⁻¹ and C_p has units Pa⁻². For consistent expansions, B_p = B/(RT) and C_p = (C − B²)/(RT)². Do not insert a density-series coefficient directly into a pressure series. Nor is small |B/v_m| enough to prove accuracy when B nearly cancels and higher terms remain.

To make the truncation question checkable, use the separate assigned van der Waals model introduced next. Define reduced volume v_r = v_m/v_c and reduced temperature T_r = T/T_c, where v_c and T_c are this model’s critical molar volume and temperature, derived below. At T_r = 1.20 its exact Z is 1/[1 − 1/(3v_r)] − 9/(8T_rv_r). Expanding its first term gives B/v_c = −0.6041667 and C/v_c² = 1/9. Here “exact” means exact within this assigned EOS, not agreement with measured gas data.

The B-only approximation at v_r = 2 underestimates this model’s Z by about 4.56%; adding C reduces the underestimate to about 0.760%. At v_r = 20 those errors are about 0.0291% and 0.000485%. The signed relative error is (Z_approx − Z_model)/Z_model. For this particular geometric series, the remainder after C is [1/(3v_r)]³/[1 − 1/(3v_r)], valid when |1/(3v_r)| < 1. This supplied remainder establishes the comparison; it is not a general error formula for empirical virial equations.

At the model’s Boyle temperature T_B = a/(Rb), B = 0. Nevertheless Z = 1.03333 at v_r = 2. The initial density correction vanishes while higher terms persist. Thus “at the Boyle temperature the gas is ideal at all densities” is false even in this simple model.

An assigned fluid with two corrections

The molar van der Waals equation is p = RT/(v_m − b) − a/v_m², with positive constants a and b and domain v_m > b. The first term reduces the volume available to molecular centers in this model; the second represents a mean attraction correction to pressure. These are approximate mechanisms. They do not resolve detailed molecular structure, all dense-liquid properties, solids or a real material’s precise critical behavior.

For the rest of the model comparisons, assign a = 0.360 Pa m⁶/mol² and b = 4.00 × 10⁻⁵ m³/mol. This is a separate hypothetical fluid, not a fit to the opening gas or any named measured substance. Its calculations test consequences of an explicit equation. The displayed decimal precision records numerical evaluation of that equation and carries no experimental accuracy claim.

At fixed T, v_m approaching b from above makes the repulsion term diverge. The point v_m = b and smaller volumes are outside the model domain. At large v_m, both corrections become small relative to RT/v_m and Z tends to one. Expanding gives B(T) = b − a/(RT), C(T) = b² and further powers of b. The excluded-volume series converges for |b/v_m| < 1, which includes the stated physical domain; convergence alone does not guarantee a short truncation is accurate near b.

A cubic EOS can yield more than one homogeneous volume at a supplied T and p. Solving the polynomial is only the first step. Reject volumes outside v_m > b, examine the local slope and compare permitted equilibrium states. Choosing the largest positive root automatically would discard a possible liquid branch; choosing every real root as an equilibrium state would retain unstable candidates.

Further: critical and reduced variables

For this model, the liquid-like and vapor-like extrema of a subcritical isotherm meet where (∂p/∂v_m)_T = 0 and (∂²p/∂v_m²)_T = 0. Differentiating p gives −RT/(v_m − b)² + 2a/v_m³ and 2RT/(v_m − b)³ − 6a/v_m⁴. Solving the two equations gives v_c = 3b, T_c = 8a/(27Rb) and p_c = a/(27b²). These are the model critical values.

For the assigned parameters, v_c = 0.120 L/mol, T_c ≈ 320.73 K and p_c ≈ 8.3333 MPa. The critical compression factor is Z_c = p_cv_c/(RT_c) = 3/8. It is a prediction of this EOS, not a universal measured critical value for real fluids.

Reduced variables divide by those critical scales: v_r = v_m/v_c, T_r = T/T_c and p_r = p/p_c. Substitution gives p_r = 8T_r/(3v_r − 1) − 3/v_r², with v_r > 1/3. Thus all members of this two-parameter model share the same reduced curves. Approximate corresponding-state ideas can be useful for related real fluids, but different molecular structure and additional parameters prevent this model curve from being an exact universal law.

At T_r = 1.20 the displayed curve decreases monotonically. At T_r = 1 the horizontal inflection is at (v_r,p_r) = (1,1). At T_r = 0.90 the homogeneous equation has a loop and, only for pressures between its two extrema, three physical roots. The overview plot uses 0.5 ≤ v_r ≤ 5 and 0 ≤ p_r ≤ 8. It does not show the forbidden-volume boundary or the full divergence near v_r = 1/3. A separate focused plot resolves the subcritical branches.

Further: derivatives and local mechanical stability

The isothermal compressibility κ_T = −(1/v_m)(∂v_m/∂p)_T measures fractional volume decrease per pressure increase, in Pa⁻¹. The thermal expansion coefficient α_p = (1/v_m)(∂v_m/∂T)_p measures fractional volume increase per temperature increase at constant pressure, in K⁻¹. Amount and composition are fixed. The subscripts say what the partial derivative holds constant; they are part of the physical definition.

Implicit differentiation of a smooth single-phase EOS gives κ_T = −1/[v_m(∂p/∂v_m)_T] and α_p = −(∂p/∂T)_v/[v_m(∂p/∂v_m)_T]. These formulas require a locally invertible volume relation. For an ideal gas they yield κ_T = 1/p and α_p = 1/T. The compression factor Z and compressibility κ_T are different quantities with different units.

At the assigned state T_r = 1.20, v_r = 2, the reduced pressure is 1.170 and the reduced slope is −0.402. In ordinary units T ≈ 384.87 K, v_m = 0.240 L/mol and p = 9.750 MPa. The derivatives give κ_T ≈ 1.493 × 10⁻⁷ Pa⁻¹ and α_p ≈ 6.205 × 10⁻³ K⁻¹. The ideal formula at the same T and p gives κ_T ≈ 1.026 × 10⁻⁷ Pa⁻¹ and α_p ≈ 2.598 × 10⁻³ K⁻¹; that ideal comparison has a different molar volume.

For a homogeneous simple fluid under isothermal mechanical variations, a negative p-versus-v_m slope gives positive κ_T, a necessary local mechanical-stability condition. A positive slope gives negative κ_T and a mechanically unstable homogeneous candidate. Zero slope marks a spinodal limit of this mean-field model, where this inverse-slope response becomes singular. It does not supply a finite measured response or a relaxation time.

Positive κ_T does not establish global equilibrium: a separated pair of phases may have lower free energy. It also does not establish stability against every kind of perturbation. Thermal stability requires suitable caloric information, such as positive heat capacity in the relevant conditions; the thermal EOS alone does not determine that information. Here branch labels concern the stated fixed-temperature volume and phase comparisons.

Further: which subcritical branch is equilibrium?

At T_r = 0.90 the assigned isotherm has zero-slope volumes v_r ≈ 0.718597 and 1.528505. Its pressure minimum and maximum are p_r ≈ 0.419843 and 0.724013. Between these extrema three homogeneous roots can occur at the same T and p. The middle positive-slope branch fails the local mechanical test. The two negative-slope branches still need a global comparison.

Allow a macroscopic one-component sample to separate into two fluid phases at the same T and p, with interface contributions neglected. Equilibrium also requires equal molar Gibbs free energy, the free-energy quantity that decides material transfer at fixed T and p. A consistent Helmholtz free energy F has p = −(∂F/∂V) at fixed T and n. At fixed T, integrating this EOS determines its volume-dependent part, which is enough for the present comparison even though an arbitrary function of T remains.

With F_m = F/n and a_r = F_m/(p_cv_c), this model gives a_r(v_r) = −(8T_r/3)ln(3v_r − 1) − 3/v_r plus a function of T_r only. Since g_r = G_m/(p_cv_c) = a_r + p_rv_r, equal endpoint pressure and equal g_r are equivalent to ∫ from v_L to v_V [p_r(v_r) − p_s] dv_r = 0. This is Maxwell’s equal-area construction. It selects a common tangent to the fixed-temperature free-energy curve rather than treating the positive-slope segment as a traversed equilibrium path.

At T_r = 0.90 the construction gives p_s ≈ 0.646998, v_L ≈ 0.603402 and v_V ≈ 2.348842. Thus T ≈ 288.65 K, p_s ≈ 5.39165 MPa, v_m,L ≈ 0.0724082 L/mol and v_m,V ≈ 0.281861 L/mol. Between these endpoint molar volumes, the equilibrium mean volume changes through the phase amounts while T and p stay fixed. A mean molar volume in that interval is not the volume of a uniform phase on the loop.

The negative-slope homogeneous segments from v_L to the liquid spinodal and from the vapor spinodal to v_V are metastable against phase separation in this isothermal model: they pass the local slope test but do not minimize the allowed macroscopic free energy. The positive-slope segment is mechanically unstable. Outside the coexistence interval, the displayed outer branches are the isothermally stable homogeneous fluid states within this model’s allowed fluid states. No solid phase is supplied.

Metastable does not specify a lifetime. Nucleation barriers, interfaces, impurities and fluctuation dynamics require additional models. This construction neglects interfacial cost and gives no transition rate. Nor does it claim that a tiny disturbance always destroys every metastable state. Full phase-equilibrium, free-energy and statistical-mechanics subjects develop these ideas beyond this single-EOS example.

Further: what a thermal EOS leaves undetermined

For a thermodynamically consistent simple one-component fluid, equilibrium relations imply (∂U_m/∂v_m)_T = T(∂p/∂T)_v − p, where U_m is molar internal energy in J/mol. This identity relates a volume dependence of energy to the thermal EOS; its derivation uses the first and second laws and is developed with thermodynamic potentials. It is an advanced consistency statement here, not a new definition of heat.

For the ideal EOS the right side vanishes, so U_m can depend on T but not v_m within the consistent ideal model. Its temperature function still needs caloric information: the pressure equation does not decide whether rotational or vibrational modes contribute to heat capacity.

For the assigned van der Waals equation, (∂p/∂T)_v = R/(v_m − b), giving T(∂p/∂T)_v − p = a/v_m². Integrating at fixed T gives U_m(T,v_m) = u₀(T) − a/v_m. The arbitrary temperature function u₀(T) remains undetermined. Choosing it requires additional physics or property information; this lesson assigns no heat capacity.

This illustrates two useful checks. An independently proposed energy model must satisfy the consistency identity if it is to represent the same equilibrium fluid. Conversely, matching p at one state, or even throughout a thermal EOS, does not determine every caloric property. Entropy, heat capacities, sound speed and energy changes along an unspecified process should not be read from a pressure plot alone.

Choose and use a gas model

Begin with the physical system, chemical species, amount or composition, absolute temperature, absolute pressure, volume range and required accuracy. Decide whether a uniform equilibrium phase is an adequate description. During fast flow or rapid relaxation, an equilibrium EOS is often used locally only when local equilibration is adequate; a single vessel-wide equilibrium state need not exist. Transport, shocks and reaction require their own balances and timescale arguments.

Use the ideal model where its approximations meet the task. For modest departures in a dilute gas phase, a measured or theoretically established virial coefficient and a controlled truncation can be useful. Cubic equations provide compact broader-range models, with physical-domain and branch selection checks. High-accuracy work uses fluid-specific formulations and documented validity ranges, often based on a consistent free-energy representation and evaluated property information. An assigned teaching EOS is not a substitute for that validation.

Check units, counting basis, pressure reference, phase and allowed branch before trusting a number. Then check the residual of the equation and whether a derivative or omitted term changes the conclusion. Near criticality, phase separation, strong association, dissociation, quantum degeneracy or other unmodeled physics, a formally smooth curve can still be the wrong description. An extrapolated computer result is a prediction of the extrapolated model, not new evidence.

The opening pressure question is now part of a larger method. State the conditions, choose an EOS, calculate the property it actually constrains, and inspect the model’s limits. Return to temperature and the zeroth law for the thermal scale, the first law and calorimetry for energy, potentials and phase equilibrium for global selection, and statistical mechanics for microscopic derivations and critical behavior. Those neighboring subjects keep their full development.

Worked examples

Try it yourself

1. A fixed-n, fixed-V ideal gas changes from 300 K to 360 K. What is the pressure ratio? Could the Celsius ratio be used?

   Show the answer and reasoning

   p₂/p₁ = 360/300 = 1.20; Celsius ratios cannot be used.

   The proportionality uses absolute temperature. Here the separate endpoint pressure is about 119.7 kPa if the first is 99.8 kPa.

2. Compare doubling n alone at fixed V,T with doubling both n and V at fixed T.

   Show the answer and reasoning

   The first doubles ideal pressure; the second leaves it unchanged.

   Pressure depends on n/V. Bulk replication preserves molar volume; adding inventory to the same volume does not.

3. What gas constant belongs in p = ρ R_s T for assigned M = 0.0280 kg/mol?

   Show the answer and reasoning

   R_s = R/M ≈ 296.945 J/(kg K).

   ρ is mass per volume, so the gas constant must be per mass. Inserting universal R without converting the counting basis gives incorrect units.

4. In the separate ideal mixture, what fraction of the total pressure is contributed by 0.0400 mol out of 0.1000 mol?

   Show the answer and reasoning

   0.400, giving about 39.9 kPa.

   The ideal-mixture relation p_i = y_i p uses the common T and total vessel volume. It is not a real-mixture interaction model.

5. A state has Z = 1. Does this prove κ_T = 1/p or that all gas properties are ideal?

   Show the answer and reasoning

   No.

   Z fixes one value of pv_m/(RT). Compressibility involves a neighboring-state derivative, and caloric properties require additional information. Corrections can cancel at one state.

6. State the units of B and C in Z = 1 + B/v_m + C/v_m² and of B_p in Z = 1 + B_p p + ⋯.

   Show the answer and reasoning

   B: m³/mol; C: m⁶/mol²; B_p: Pa⁻¹.

   Every correction to dimensionless Z must be dimensionless. Density-series and pressure-series coefficients have different definitions.

7. At the assigned Boyle temperature, B = 0. Does the supplied Z = 1.03333 at v_r = 2 contradict that fact?

   Show the answer and reasoning

   No; higher-density terms remain.

   B = 0 removes only the first density correction. It does not make the full EOS ideal at finite density.

8. What do v_c = 3b and Z_c = 3/8 establish for the assigned EOS? Are they universal real-fluid measurements?

   Show the answer and reasoning

   They are model predictions, not universal measured values.

   They follow from the EOS and its simultaneous first and second volume-derivative conditions. Real-fluid validation is separate.

9. Why is a positive slope (∂p/∂v_m)_T problematic for a homogeneous simple-fluid state?

   Show the answer and reasoning

   It gives κ_T < 0 and mechanical instability under the stated isothermal variations.

   Compression would lower its pressure rather than provide restoring resistance. A negative slope is necessary locally but does not establish global phase equilibrium.

10. At T_r = 0.90, is v_r = 0.718597 the same boundary as v_L = 0.603402?

    Show the answer and reasoning

    No. The first is the liquid spinodal; the second is the liquid coexistence endpoint.

    The spinodal is a local zero-slope limit. Coexistence is selected by equal pressure and molar Gibbs free energy. The intervening homogeneous branch is metastable in this isothermal model.

11. Does the equilibrium plateau between v_L and v_V describe one uniform phase with every intermediate density?

    Show the answer and reasoning

    No; the mean volume changes by changing phase amounts.

    The endpoint phases retain their respective molar volumes at the common T and p. Intermediate mean volumes are two-phase states under the stated macroscopic assumptions.

12. Can p(T,v_m) alone assign a numerical U_m(T,v_m), heat capacity and rapid-compression temperature history?

    Show the answer and reasoning

    No.

    A consistent thermal EOS constrains the volume dependence of U_m but leaves temperature-dependent caloric information. A process also needs energy balances and boundary conditions.

What to carry forward

• A thermal EOS relates equilibrium state properties; its counting basis, pressure reference, domain and phase conditions must be explicit.
• Ideal pressure follows dilute classical isotropic molecular translation; internal energy modes require separate caloric information.
• Z measures a dimensionless EOS departure; κ_T is a derivative response with units Pa⁻¹.
• Virial truncations need units, a density range and an error argument; B = 0 does not erase higher terms.
• A cubic root is a candidate state. Local mechanical stability and global permitted phase equilibrium require different checks.
• Assigned model curves and numerical precision do not establish measured-fluid accuracy, transition rates or a complete process history.