Q1(d). Continuity of state: Andrews' isotherms of CO₂
Explain “continuity of state” with the help of isotherms of CO₂. How is it affirmed by van der Waals' equation?
Answer in brief
Above the critical point (31.04°C, 72.9 atm for CO₂) the gas passes into the liquid with no visible change of phase, so gas and liquid are two extremes of one fluid state. The van der Waals equation gives both branches from one continuous S-shaped isotherm below T_c, merging at T_c.
Key points
- on compression the gas pressure rises until the vapour pressure is reached. Liquid then appears and the volume falls at constant pressure (red tie line) while both phases are present. When all is liquid the isotherm rises steeply.
- The horizontal part shrinks as T rises and vanishes at the critical point C. At T_c liquid and gas have the same density; the meniscus disappears.
- no liquefaction at any pressure; the isotherms approach Boyle's hyperbolas at high T.
- Take gas at a point on the right of the dome. Heat it above T_c at constant volume, compress it above p_c, then cool it at constant volume to below T_c. It ends as liquid, yet at no stage two phases were present and no…
The full model answer
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Andrews' isotherms of CO₂
- Below T_c: on compression the gas pressure rises until the vapour pressure is reached. Liquid then appears and the volume falls at constant pressure (red tie line) while both phases are present. When all is liquid the isotherm rises steeply.
- The horizontal part shrinks as T rises and vanishes at the critical point C. At T_c liquid and gas have the same density; the meniscus disappears.
- Above T_c (304.2 K for CO₂): no liquefaction at any pressure; the isotherms approach Boyle's hyperbolas at high T.
Continuity of state
- Take gas at a point on the right of the dome. Heat it above T_c at constant volume, compress it above p_c, then cool it at constant volume to below T_c. It ends as liquid, yet at no stage two phases were present and no meniscus was seen.
- So the change gas → liquid can be made continuously: gas and liquid are not different states but two forms of one fluid state. This is the principle of continuity of state (James Thomson).
How van der Waals' equation affirms it
- (p+aVₘ²)(Vₘ-b)=RT is a cubic in Vₘ: Vₘ³-(b+RT/p)Vₘ²+a/pVₘ-ab/p=0
- Below T_c: three real roots. The curve is one continuous S-shaped loop (dashed) linking the liquid branch to the gas branch. The tie line cuts it so that the two loop areas are equal (Maxwell). The loop portions are partly realised as superheated liquid and supercooled vapour.
- At T_c: the three roots merge at C, a point of inflexion with (∂ p/∂ V)_T=0 and (∂²p/∂ V²)_T=0. This gives V_c=3b, p_c=a/27b², T_c=8a/27Rb. Above T_c: one real root.
- A single equation describes gas, liquid and their passage into one another; the liquid branch is the continuation of the gas isotherm. This is the theoretical affirmation of continuity of state.
| CO₂ (a=3.610 atm dm⁶ mol⁻², b=4.29×10⁻² dm³ mol⁻¹) | T_c / K | p_c / atm | V_c / cm³ mol⁻¹ | Z_c |
| from van der Waals constants | 303.8 | 72.6 | 128.7 | 0.375 |
| experiment | 304.2 | 72.9 | 94.0 | 0.274 |
Agreement in T_c and p_c is close; V_c and Z_c show the limits of the equation.
Application. supercritical CO₂ (above 31°C and 73 atm) has liquid-like density and gas-like viscosity, and is used to extract caffeine from coffee and essential oils without organic solvents.
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All parts of Question 1
- 1(a)(i) Bond angles of CH₄, NH₃ and H₂O5 marks
- 1(a)(ii) Dipole moments of BF₃ and PF₃5 marks
- 1(b) Residual entropy of CO and N₂O10 marks
- 1(c)(i) Derivation of the Bragg equation5 marks
- 1(c)(ii) Spacing of the (110) planes of a cubic cell5 marks
- 1(d) Continuity of state: Andrews' isotherms of CO₂10 marks
- 1(e) Specific conductance of saturated BaSO₄10 marks