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I have having difficulty in understanding the liquefaction of gases.In the graph of isotherms of $\ce{CO2}$ I cannot understand how can gas coexist as liquid and vapour under area of $\mathrm{XCEBY}$.

enter image description here

I have read the following statements in my textbook:

  1. At $\pu{21.5 ^\circ C}$, $\ce{CO2}$ remains a gas up to point $\mathrm B$. At point B a liquid of specific volume appears.In fact,the gas and liquid $\ce{CO2}$ coexist along the horizontal line $\mathrm{BC}$. The decrease of volume represents condensation of more and more $\ce{CO2}$ gas till point $\mathrm C$ is reached.

  2. A careful study of isotherms reveal that it is possible to change gas to liquid by following a path in which a single phase is present throughout the process.For example,move vertically from $\mathrm{A}$ to $\mathrm{F}$ by increasing the temperature of the gas.Now move from point $\mathrm{F}$ to point $\mathrm{G}$ by increasing pressure at constant temperature.Move from point $\mathrm{G}$ to point $\mathrm{H}$ by lowering the temperature.As soon as we cross point $\mathrm H$ we get a liquid.

(i)Why $\ce{CO2}$ coexist as liquid and vapour in the area of $\mathrm{XCEBY}$?(instead of converting directly into liquid by condensation of all the molecules simultaneously)

(ii)Why under some range of pressure and temperature $\ce{CO2}$ converts directly into liquid without going through the two-phase region?

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    $\begingroup$ In order to change from vapor to liquid at constant temperature and pressure, you need to remove heat. The amount of liquid formed depends on the amount of heat removed. If you only remove a tiny amount of heat, you can't suddenly change all the vapor to liquid. $\endgroup$ – Chet Miller Sep 21 '18 at 14:23
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    $\begingroup$ But the pressure is constant (p2) along the 21.5 C isotherm in the 2 phase region. $\endgroup$ – Chet Miller Sep 21 '18 at 16:34
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    $\begingroup$ chemistry.stackexchange.com/questions/101900/… $\endgroup$ – Mithoron Sep 21 '18 at 22:33
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    $\begingroup$ I have a feeling, you're getting too far, not knowing most elementary things. 1) If you put some water into pressure cooker, it won't suddenly boil off. 2) There's no boiling at all in this region, because you have supercritical fluid. $\endgroup$ – Mithoron Sep 21 '18 at 22:41
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    $\begingroup$ @Mithoron Do you really think water in a pressure cooker is supercritical (i.e., above the critical temperature)? $\endgroup$ – Chet Miller Sep 22 '18 at 0:11
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  1. Vapor-liquid coexistence.

This is possible because changing the system's molar volume $V_\mathrm m$ does not affect the Gibbs free energy of either phase, so equality of the Gibbs free energy (that is, phase coexistence,) can be maintained for a range of $V_\mathrm m$; this is the coexistence region $\mathrm{XCEBY}$.

On the other hand, changing $p$ and $T$ will affect the Gibbs free energy, and if instead we plot this phase diagram on the $p,T$-plane, the entire coexistence region $\mathrm{XCEBY}$ will reduce to a single curve, the liquid-vapor coexistence curve.

In other words: Fix $T$. Only one value of $p$ results in phase coexistence; phase coexistence is a curve on the $p,T$-plane. A range of $V_\mathrm m$ results in phase coexistence; phase coexistence is a region on the $p,V_\mathrm m$-plane.

  1. Phase transitions without coexistence.

This is made possible because the liquid and vapor phases share continuous translational symmetry. In contrast, a phase transition from the liquid phase to a solid, crystalline phase requires breaking this continuous symmetry, resulting instead in the discrete translational symmetry possessed by a crystalline lattice. This shared symmetry permits the possibility of continuously going from one phase to the other without an explicit phase transition.

This possibility is realized past the critical temperature, the temperature represented by the critical point $\mathrm E$. Past the critical isotherm identified by line segment $\mathrm{HE}$ is the supercritical fluid "phase", which shares properties of both liquid and vapor. Isotherms like $\mathrm{GF}$ do not represent a liquid-vapor transition, because the species remains as a supercritical fluid throughout. A liquid-vapor transition that bypasses the coexistence region, like $\mathrm{AFGHD}$, is a continuous transformation from liquid to supercritical fluid to vapor, rather than a discrete phase transition.

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  • $\begingroup$ It seems beyond my scope of understanding from what I have studied upto now,but thanks for the answer. $\endgroup$ – pranjal verma Sep 21 '18 at 16:26
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    $\begingroup$ @pranjalverma, I'm happy to help, and might be able to answer any questions you have regarding my answer. In general, however, I would recommend that you wait longer before accepting an answer, especially if it's not easy for you to understand, because someone else might provide a more accessible answer. $\endgroup$ – a-cyclohexane-molecule Sep 21 '18 at 16:37
  • $\begingroup$ Your answer was indeed useful as I will understand all the stuff after reading carefully about thermodynamics. $\endgroup$ – pranjal verma Sep 22 '18 at 9:14

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