Pure Substances: Phases, p-v-T Behaviour, Ideal and van der Waals Gases

This chapter answers Section XE4.2 “Properties of Pure Substances” of XE4 Thermodynamics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Thermodynamic properties of pure substances in solid, liquid and vapor phases; P-v-T behaviour of simple compressible substances, Concept of triple point and critical point; Ideal and real gases, ideal gas equation of state and van der Waals equation of state.” It teaches phase behaviour on the p–v, T–v and p–T diagrams, saturation and quality, superheated vapour and compressed liquid, the triple and critical points, the ideal-gas law and the compressibility factor, and the van der Waals equation with its constants, reduced form and critical compressibility 3/8. The numericals are quality from specific volume, mass of an ideal gas, a real-gas mass from Z, van der Waals constants from critical data, and the reduced van der Waals pressure. Every property value a numerical needs is given in its stem.

1. Phases and the p–v–T surface

A pure substance has a fixed chemical composition throughout (water, nitrogen, or air as long as it stays a single gas phase). It may exist in several phases: solid, liquid and vapour (gas). In a simple compressible substance two independent intensive properties fix the state, so all equilibrium states form a surface in p–v–T space, with p = f(v, T). On the surface, single-phase regions (solid, liquid, vapour) are separated by two-phase regions (solid–liquid, solid–vapour, liquid–vapour) and the three phases coexist along the triple line, which appears as a point on the p–T diagram.

DiagramWhat a point or line showsWhere the two-phase region appears
p–TEach line is a phase boundary: fusion, vaporisation (saturation), sublimation; they meet at the triple pointCollapsed onto the lines (the pressure fixes T_sat)
p–vIsotherms; the saturated-liquid and saturated-vapour curves meet at the critical pointUnder the dome; isotherms are horizontal there
T–vIsobars; the dome is bounded by the saturated-liquid and saturated-vapour linesUnder the dome; isobars are horizontal there

The triple point is the unique state at which solid, liquid and vapour coexist in equilibrium. For water it is 273.16 K at 0.6117 kPa. By Gibbs’ phase rule F = C − P + 2, a pure substance (C = 1) with three phases (P = 3) has F = 0 degrees of freedom: T and p are fixed and cannot be varied. The critical point is the end of the liquid–vapour line. For water Tc = 647.1 K (about 374 °C) and pc = 22.06 MPa. Above pc and Tc there is no distinct liquid or vapour, only a supercritical fluid; at the critical point the saturated-liquid and saturated-vapour states are identical and the latent heat of vaporisation h_fg is zero.

⚠️ Water is the exception
For most substances the solid is denser than the liquid, so the fusion line on the p–T diagram slopes upward. Water expands on freezing (ice is less dense), so its fusion line slopes downward: raising the pressure on ice can melt it. Do not generalise this slope to other substances.

2. Saturation, quality and the vapour dome

At a given pressure, boiling occurs at the saturation temperature T_sat(p); at a given temperature, the corresponding saturation pressure p_sat(T). In the two-phase region T and p are not independent, so a third property is needed to fix the state. A saturated liquid is liquid at T_sat (about to vaporise); a saturated vapour is vapour at T_sat (about to condense); between them is a saturated mixture. The quality (dryness fraction) is x = m_vapour / m_total, with x = 0 for saturated liquid and x = 1 for saturated vapour; it is defined only inside the dome.

Any extensive-per-mass property Y of a mixture is the mass-weighted average of the phases: v = v_f + x v_fg, u = u_f + x u_fg, h = h_f + x h_fg, s = s_f + x s_fg, where v_fg = v_g − v_f and h_fg = h_g − h_f. Solving, x = (v − v_f)/(v_g − v_f). Here subscript f means saturated liquid and g saturated vapour at the given p or T.

Worked example. A rigid vessel of 1.2 m³ holds 4 kg of water at 200 kPa. Take, from the table at 200 kPa, v_f = 0.001061 m³/kg and v_g = 0.8857 m³/kg. Then v = 1.2/4 = 0.300 m³/kg, which lies between v_f and v_g, so the water is a mixture: x = (0.300 − 0.001061)/(0.8857 − 0.001061) = 0.2989/0.8846 = 0.338. Vapour mass = 0.338 × 4 = 1.35 kg. Although the vapour is only 34% by mass, it fills almost the entire volume, since v_g is about 835 times v_f.

⚠️ Quality is a mass fraction
x compares masses, not volumes, and is meaningless for a compressed liquid or superheated vapour. Compare v with v_f and v_g first: v < v_f means compressed liquid, v_f < v < v_g a mixture, v > v_g superheated vapour (at the same p or T).

3. Superheated vapour, compressed liquid and heating at constant volume

A superheated vapour has T > T_sat(p) (equivalently p < p_sat(T)); here p and T are independent, and the degree of superheat is T − T_sat. A compressed (subcooled) liquid has T < T_sat(p) (equivalently p > p_sat(T)). Liquid properties depend weakly on pressure, so for a compressed liquid the usual approximations are v ≈ v_f(T), u ≈ u_f(T) and h ≈ h_f(T), using the saturated-liquid value at the same temperature (not the same pressure). The enthalpy approximation can be refined by h ≈ h_f(T) + v_f (p − p_sat), a small correction at moderate pressure.

Heating a closed rigid vessel. The state moves along a vertical line (v = constant) on the T–v or p–v diagram. If the overall v is less than the critical specific volume v_c, the liquid expands until it fills the vessel: the final state is saturated liquid. If v > v_c, the liquid evaporates entirely: the final state is saturated vapour; for v = v_c the mixture passes through the critical point. For water v_c ≈ 0.0031 m³/kg. A sealed bottle of water that is nearly full therefore ends as liquid, and a nearly empty one ends as vapour.

RegionCondition at given p (or T)Independent properties
Compressed liquidT < T_sat or v < v_fp and T
Saturated mixturev_f < v < v_gp or T, plus x (or v)
Superheated vapourT > T_sat or v > v_gp and T

4. Ideal gas, specific gas constant and compressibility factor

An ideal gas obeys pV = nR_uT, or pv = RT per unit mass, where R_u = 8.314 J mol⁻¹ K⁻¹ is the universal gas constant and the specific gas constant is R = R_u/M, with M the molar mass. For air, M ≈ 28.97 kg/kmol gives R ≈ 287 J/kg·K. In molar form pV = N R_u T with N in kmol uses R_u = 8.314 kJ/kmol·K. Ideal-gas behaviour is the limit of real behaviour at low pressure and high temperature (low density, with molecules far apart and negligible interaction). Its internal energy and enthalpy depend on temperature only.

A real gas deviates from the ideal law by the compressibility factor Z = pv/(RT) = pV/(mRT). Z = 1 for an ideal gas; Z < 1 means attraction dominates (the gas is more compressible than ideal), Z > 1 means repulsion (finite molecular size) dominates. For a real gas the volume is v = Z RT/p, so mass m = pV/(Z R T). As p → 0, Z → 1 for every gas. The law of corresponding states says that Z is nearly the same function of the reduced properties p_r = p/p_c and T_r = T/T_c for all gases, which gives the generalised compressibility chart; at the critical point Z_c is about 0.27 to 0.29 for most gases.

Worked example. Air (R = 287 J/kg·K, ideal gas) fills 0.5 m³ at 300 kPa and 300 K: m = pV/(RT) = (300 000 × 0.5)/(287 × 300) = 150 000/86 100 = 1.74 kg. If a gas at 5 MPa and 400 K has Z = 0.92 and R = 0.287 kJ/kg·K, then 0.1 m³ of it has mass 5000 × 0.1/(0.92 × 0.287 × 400) = 4.73 kg, which is 8.7% more than the ideal-gas value of 4.35 kg.

⚠️ Temperature and units
Use absolute pressure and absolute temperature in pv = RT. With p in kPa and R in kJ/kg·K the product pv is in kJ/kg; mixing kPa with J/kg·K gives an error of 1000.

5. The van der Waals equation of state

The van der Waals equation corrects the ideal law for two molecular effects: (p + a/v²)(v − b) = RT (per unit amount, with v the molar or specific volume and a, b matched to that basis). The constant b is the volume excluded by the finite size of the molecules, so the free volume is v − b. The term a/v² is the correction to the pressure for intermolecular attraction, which reduces the wall force of the molecules. It reduces to pv = RT when a → 0 and b → 0.

Critical constants. At the critical point the critical isotherm has a horizontal tangent and an inflection: (∂p/∂v)_T = 0 and (∂²p/∂v²)_T = 0 at (T_c, v_c). Writing p = RT/(v − b) − a/v² and solving the two conditions gives v_c = 3b, T_c = 8a/(27Rb), p_c = a/(27b²). Inverting: b = RT_c/(8p_c) and a = 27R²T_c²/(64p_c). Then Z_c = p_c v_c/(RT_c) = (a/27b²)(3b)/(R × 8a/27Rb) = 3/8 = 0.375 for every van der Waals gas, whereas real gases show about 0.27 to 0.29. The equation is therefore qualitative near the critical point.

Reduced form. With p_r = p/p_c, T_r = T/T_c, v_r = v/v_c, the equation becomes (p_r + 3/v_r²)(3v_r − 1) = 8T_r, which contains no substance-specific constant: this is the law of corresponding states in exact form for a van der Waals fluid. Worked example. For T_r = 1.2 and v_r = 2: p_r = 8T_r/(3v_r − 1) − 3/v_r² = 9.6/5 − 0.75 = 1.17. For CO₂ with T_c = 304.2 K and p_c = 7.38 MPa, R_u = 8.314 J mol⁻¹ K⁻¹: b = 8.314 × 304.2/(8 × 7.38 × 10⁶) = 4.28 × 10⁻⁵ m³/mol and a = 27 × 8.314² × 304.2²/(64 × 7.38 × 10⁶) = 0.366 Pa·m⁶/mol².

⚠️ Below T_c the cubic has three roots
Below the critical temperature the van der Waals isotherm has a loop (a swallow-tail in v). The middle root is physically unstable; the real isotherm replaces the loop by a horizontal line at the saturation pressure, drawn so that the two areas between the line and the loop are equal (Maxwell construction). Above T_c there is one real root for v.

Key takeaways

  • The triple point (water: 273.16 K, 0.6117 kPa) has zero degrees of freedom; the critical point (water: 647.1 K, 22.06 MPa) is where h_fg = 0 and liquid and vapour become indistinguishable.
  • In the two-phase region p and T are tied; fix the state with quality x = (v − v_f)/(v_g − v_f), a mass fraction, and use h = h_f + x h_fg for the other properties.
  • Compare v (or T) with the saturated values first to classify the region; a compressed liquid is approximated by the saturated-liquid values at the same temperature.
  • The ideal-gas law pv = RT with R = R_u/M holds at low p and high T; Z = pv/(RT) measures the deviation, and m = pV/(Z R T) for a real gas.
  • Van der Waals: (p + a/v²)(v − b) = RT, with b = RT_c/(8p_c), a = 27R²T_c²/(64p_c), v_c = 3b and Z_c = 3/8; the reduced form (p_r + 3/v_r²)(3v_r − 1) = 8T_r has no substance constants.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which statement about the critical point of a pure substance is correct?

    1. Solid, liquid and vapour coexist, so the substance has no free intensive property
    2. The liquid can be boiled at constant pressure only above the critical temperature
    3. Saturated liquid and vapour states coincide, and latent heat of vaporisation is zero
    4. The latent heat of vaporisation reaches its maximum value, since the vapour is densest here
    Show answer

    Answer: C — Saturated liquid and vapour states coincide, and latent heat of vaporisation is zero

    At the critical point the two phases become identical, so v_fg = 0 and h_fg = 0; latent heat decreases steadily along the saturation line towards zero rather than peaking here. Coexistence of all three phases describes the triple point, not the critical point. Above the critical pressure no boiling occurs at all.

  2. Apply Gibbs’ phase rule F = C − P + 2 to pure water (one component). How many intensive variables can be varied independently when ice, liquid water and water vapour coexist in equilibrium?

    1. Zero, the temperature and pressure are both fixed
    2. Two, the temperature and the pressure
    3. One, either the temperature or the pressure
    4. Three, the temperature, the pressure and the quality
    Show answer

    Answer: A — Zero, the temperature and pressure are both fixed

    With C = 1 and P = 3, F = 1 − 3 + 2 = 0, so the triple point is invariant: T and p are fixed (273.16 K and 0.6117 kPa for water). One degree of freedom would apply to two coexisting phases, where p fixes T_sat, and two to a single phase.

  3. Water exists at 200 kPa and a temperature below the saturation temperature at 200 kPa. For a thermodynamic calculation, its specific volume and specific enthalpy are best approximated by the values of

    1. Saturated vapour at 200 kPa, since the pressure is the same
    2. Saturated liquid at 200 kPa, since the pressure is the same
    3. A mixture at the actual temperature, with quality fixed by v
    4. Saturated liquid at the actual temperature of the water
    Show answer

    Answer: D — Saturated liquid at the actual temperature of the water

    The water is a compressed (subcooled) liquid. Liquid properties depend mainly on temperature, so v ≈ v_f(T) and h ≈ h_f(T) at the actual temperature. Using the saturated values at 200 kPa would use the saturation temperature, which is higher than the true temperature and so overstates the enthalpy. The quality is undefined outside the vapour dome.

  4. A pure substance at a given pressure has a specific volume greater than v_g at that pressure. Which statement holds?

    1. It is a saturated mixture with quality greater than one
    2. It is superheated vapour, and p and T are independent properties for it
    3. It is saturated vapour, and p alone fixes its temperature
    4. It is a compressed liquid, because its density is lower than that of saturated vapour
    Show answer

    Answer: B — It is superheated vapour, and p and T are independent properties for it

    v > v_g at the given pressure means the temperature exceeds T_sat, so the substance is superheated vapour, whose state needs both p and T. A quality above one has no meaning, and a liquid has a specific volume below v_f, not above v_g.

  5. A rigid sealed vessel contains water as a saturated liquid–vapour mixture with overall specific volume 0.0012 m³/kg. The critical specific volume of water is about 0.0031 m³/kg. The vessel is heated slowly. What is the final state when the contents reach saturation again at a higher pressure?

    1. Saturated vapour, because all liquid eventually evaporates
    2. A mixture of constant quality, since the volume is fixed
    3. Saturated liquid, as the liquid level rises until it fills the vessel
    4. The critical point, because every heated sealed vessel passes through it
    Show answer

    Answer: C — Saturated liquid, as the liquid level rises until it fills the vessel

    The state moves along a vertical line v = 0.0012 m³/kg, which is below v_c, so it meets the saturated-liquid line: the liquid expands and fills the vessel. It would reach saturated vapour only if v were above v_c, and the critical point only if v equalled v_c. The quality does not stay constant, since the saturation values v_f and v_g change with temperature.

  6. A real gas has Z = pv/(RT) greater than one at a given p and T. Compared with an ideal gas at the same p and T, its specific volume is

    1. Larger, because repulsion from finite molecular size dominates
    2. Smaller, because intermolecular attraction dominates at this state
    3. Equal, because Z is only a small correction applied to the pressure
    4. Smaller, because Z multiplies the pressure in the ideal-gas law
    Show answer

    Answer: A — Larger, because repulsion from finite molecular size dominates

    v = Z RT/p, so Z > 1 gives a larger volume than the ideal value at the same p and T; this happens when repulsion dominates (high pressure or high temperature). Z < 1 is the attraction-dominated case with the smaller volume.

  7. The law of corresponding states, which underlies the generalised compressibility chart, states that

    1. all gases have the same Z at the same absolute pressure and temperature
    2. every gas has a compressibility factor of exactly 1 at its critical point
    3. Z is fixed by the molar mass of the gas alone at any state
    4. gases at equal reduced pressure and temperature have nearly the same Z
    Show answer

    Answer: D — gases at equal reduced pressure and temperature have nearly the same Z

    With p_r = p/p_c and T_r = T/T_c, Z is approximately a universal function Z(p_r, T_r). Equal absolute p and T do not give equal Z, since the critical constants differ between gases. At the critical point Z_c is about 0.27 to 0.29 for most gases, not 1.

  8. In the van der Waals equation (p + a/v²)(v − b) = RT, the term a/v² and the constant b account respectively for

    1. molecular volume and intermolecular attraction
    2. intermolecular attraction and the finite volume of the molecules
    3. thermal motion and the compressibility of the gas
    4. the pressure of the surroundings and the volume of the container
    Show answer

    Answer: B — intermolecular attraction and the finite volume of the molecules

    The a/v² term adds to the measured pressure to restore the pressure that would exist without attraction between molecules; b is the volume the molecules exclude, so only v − b is free. The reversed pairing swaps the two effects.

  9. For a van der Waals gas, the critical compressibility factor Z_c = p_c v_c/(RT_c) is

    1. 1, the value for an ideal gas at every state
    2. 8/27, because a = 27R²T_c²/(64p_c) enters the ratio
    3. 3/8 for all van der Waals gases, whatever a and b
    4. 0.27, the value measured for many real gases at the critical point
    Show answer

    Answer: C — 3/8 for all van der Waals gases, whatever a and b

    From v_c = 3b, T_c = 8a/(27Rb) and p_c = a/(27b²): Z_c = (a/27b²)(3b)/(R × 8a/(27Rb)) = 3/8 = 0.375 for all van der Waals fluids. The measured value for real gases is lower (about 0.27 to 0.29), which shows that the equation is only qualitative near the critical point.

  10. Which of the following statements about phase behaviour are correct?

    1. Above the critical pressure and temperature a pure substance cannot be separated into distinct liquid and vapour phases.
    2. In the two-phase region of a pure substance the temperature fixes the pressure.
    3. The fusion line of water on the p–T diagram has a positive slope because ice is denser than liquid water.
    4. Quality is defined as the ratio of the volume of vapour to the total volume of the mixture.
    Show answer

    Answer: A — Above the critical pressure and temperature a pure substance cannot be separated into distinct liquid and vapour phases.; B — In the two-phase region of a pure substance the temperature fixes the pressure.

    Beyond the critical point there is only a supercritical fluid, and under the dome p = p_sat(T), so both are correct. Ice is less dense than water, so the fusion line slopes downward; quality is a mass ratio, not a volume ratio.

  11. Which of the following statements about ideal and real gases are correct?

    1. The specific gas constant R = R_u/M has a different value for each gas.
    2. As the pressure tends to zero, Z tends to one for every real gas.
    3. The specific gas constant of a given gas increases with temperature.
    4. At its critical point every real gas has Z exactly equal to one.
    Show answer

    Answer: A — The specific gas constant R = R_u/M has a different value for each gas.; B — As the pressure tends to zero, Z tends to one for every real gas.

    R = R_u/M depends on the molar mass only, so it differs between gases but does not change with T, and Z → 1 as p → 0 because molecules then interact negligibly. At the critical point Z_c is about 0.27 to 0.29 for most gases, far below one.

  12. For a van der Waals fluid with critical constants T_c, p_c and v_c, which of the following are correct?

    1. The critical specific volume equals three times the constant b.
    2. The constant b equals RT_c/(8p_c).
    3. The reduced equation (p_r + 3/v_r²)(3v_r − 1) = 8T_r contains the constants a and b.
    4. At T_r = 1 and v_r = 1 the reduced equation gives p_r = 1.
    Show answer

    Answer: A — The critical specific volume equals three times the constant b.; B — The constant b equals RT_c/(8p_c).; D — At T_r = 1 and v_r = 1 the reduced equation gives p_r = 1.

    v_c = 3b and b = RT_c/(8p_c) follow from the critical-point conditions. The reduced form has no substance constants: a and b cancel when p, T and v are scaled by their critical values. At the critical point itself, (p_r + 3)(2) = 8 gives p_r = 1.

  13. Air may be treated as an ideal gas with R = 287 J/kg·K. A rigid tank of volume 0.5 m³ holds air at an absolute pressure of 300 kPa and a temperature of 300 K. The mass of air in the tank, in kg, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.74

    m = pV/(RT) = (300 000 × 0.5)/(287 × 300) = 150 000/86 100 = 1.742 kg ≈ 1.74 kg. Using 300 kPa as 300 Pa or R in kJ would shift the answer by a factor of 1000.

  14. A rigid vessel of volume 1.2 m³ contains 4 kg of water as a saturated mixture at 200 kPa. At this pressure take v_f = 0.001061 m³/kg and v_g = 0.8857 m³/kg. The quality (dryness fraction) of the mixture, to three decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.338

    v = V/m = 1.2/4 = 0.300 m³/kg. x = (v − v_f)/(v_g − v_f) = (0.300 − 0.001061)/(0.8857 − 0.001061) = 0.298939/0.884639 = 0.3379 ≈ 0.338. Using V/m without dividing by the mass, or v/v_g = 0.339, gives a slightly different or wrong figure.

  15. A gas with specific gas constant R = 0.287 kJ/kg·K is held in a 0.1 m³ vessel at 5 MPa and 400 K, where its compressibility factor is Z = 0.92. The mass of gas in the vessel, in kg, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 4.73

    m = pV/(Z R T) = (5000 kPa × 0.1 m³)/(0.92 × 0.287 × 400) = 500/105.62 = 4.734 kg ≈ 4.73 kg. The ideal-gas value is 4.355 kg; Z < 1 means the gas packs more mass into the vessel.

  16. A gas has critical temperature T_c = 304.2 K and critical pressure p_c = 7.38 MPa. Treat it as a van der Waals gas with R_u = 8.314 J mol⁻¹ K⁻¹. The van der Waals constant b, in cm³/mol, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 42.8

    b = R_u T_c/(8 p_c) = 8.314 × 304.2/(8 × 7.38 × 10⁶) = 2529.1/59.04 × 10⁻⁶ = 4.284 × 10⁻⁵ m³/mol = 42.84 cm³/mol ≈ 42.8 cm³/mol. The critical molar volume is 3b = 128.5 cm³/mol.

  17. A fluid obeys the van der Waals equation. It is at a reduced temperature T_r = 1.2 and a reduced specific volume v_r = 2. Using the reduced form (p_r + 3/v_r²)(3v_r − 1) = 8T_r, the reduced pressure p_r, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.17

    p_r = 8T_r/(3v_r − 1) − 3/v_r² = 9.6/5 − 3/4 = 1.92 − 0.75 = 1.17. Dropping the 3/v_r² term gives 1.92, the wrong answer that treats the fluid as a gas with excluded volume only.