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Concept 04 / 07States of Matter

Liquefaction of Gases and Critical Constants

Andrews' isotherms for carbon dioxide, critical temperature, pressure and volume, the critical point, continuity of states, critical constants from the van der Waals equation, the relation Zc = 3/8 and methods of liquefying gases.

Figure 1
Andrews' experiments (1869). Thomas Andrews measured PP-VV isotherms of CO2\mathrm{CO_{2}} at several temperatures. At high temperature (48∘C48^{\circ}\mathrm{C}) the isotherm resembles Boyle's hyperbola: CO2\mathrm{CO_{2}} stays a gas at any pressure. At 31.1∘C31.1^{\circ}\mathrm{C} the isotherm has a point of inflection with a horizontal tangent: the critical point. Below 31.1∘C31.1^{\circ}\mathrm{C} (for example 21.5∘C21.5^{\circ}\mathrm{C}) compression first reduces the volume of the gas; then liquid appears and the volume falls at constant pressure (the flat portion, where liquid and vapour coexist at the vapour pressure); finally, when all is liquid, the curve rises steeply because liquids are hardly compressible. Critical constants. Critical temperature TcT_{c}: the highest temperature at which a gas can be liquefied by pressure alone. Above TcT_{c}, no pressure produces a separate liquid phase. (CO2\mathrm{CO_{2}} 304 K304\,\mathrm{K}, NH3\mathrm{NH_{3}} 406 K406\,\mathrm{K}, O2\mathrm{O_{2}} 155 K155\,\mathrm{K}, N2\mathrm{N_{2}} 126 K126\,\mathrm{K}, H2\mathrm{H_{2}} 33 K33\,\mathrm{K}, He\mathrm{He} 5.2 K5.2\,\mathrm{K}.) Critical pressure PcP_{c}: the pressure needed to liquefy the gas at TcT_{c}. Critical volume VcV_{c}: the molar volume at TcT_{c} and PcP_{c}. At the critical point the densities of liquid and vapour become equal and the meniscus disappears. Continuity of states. A gas can be converted to a liquid without ever showing two phases, by going round the critical point: heat above TcT_{c}, compress, then cool. So there is no sharp distinction between gas and liquid above TcT_{c}; the substance is a supercritical fluid. The word vapour is used for a gas below its critical temperature, which can be liquefied by pressure alone. From the van der Waals equation. Requiring the isotherm at TcT_{c} to have a horizontal inflection (dPdV=0\frac{dP}{dV} = 0 and d2PdV2=0\frac{d^{2}P}{dV^{2}} = 0) gives Vc=3bV_{c} = 3b, Pc=a27b2P_{c} = \frac{a}{27b^{2}}, Tc=8a27RbT_{c} = \frac{8a}{27Rb}. Hence Zc=PcVcRTc=38=0.375Z_{c} = \frac{P_{c}V_{c}}{RT_{c}} = \frac{3}{8} = 0.375 for every van der Waals gas (real gases give about 0.230.23 to 0.310.31). Also Tc=827TBT_{c} = \frac{8}{27}T_{B}, and a=3PcVc2a = 3P_{c}V_{c}^{2}, b=Vc3b = \frac{V_{c}}{3}. Below TcT_{c} the van der Waals equation predicts an unphysical S-shaped loop; the real isotherm replaces it with a flat line placed so that the areas above and below are equal (Maxwell's construction; see figure). Liquefaction methods. A gas must first be cooled below TcT_{c}. Gases with high TcT_{c} (NH3\mathrm{NH_{3}}, SO2\mathrm{SO_{2}}, Cl2\mathrm{Cl_{2}}, CO2\mathrm{CO_{2}}) liquefy by pressure at or near room temperature. Permanent gases (O2\mathrm{O_{2}}, N2\mathrm{N_{2}}, H2\mathrm{H_{2}}, He\mathrm{He}) need cooling first, by adiabatic expansion or Joule-Thomson expansion (Linde's process). Joule-Thomson expansion cools a gas only below its inversion temperature Ti=2aRbT_{i} = \frac{2a}{Rb}; H2\mathrm{H_{2}} and He\mathrm{He} must be precooled because their TiT_{i} is low. Key Takeaways: - TcT_{c} is the maximum temperature for liquefaction by pressure. - Vc=3bV_{c} = 3b, Pc=a27b2P_{c} = \frac{a}{27b^{2}}, Tc=8a27RbT_{c} = \frac{8a}{27Rb}, Zc=38Z_{c} = \frac{3}{8}. - Ti:TB:Tc=2:1:827T_{i} : T_{B} : T_{c} = 2 : 1 : \frac{8}{27} in units of aRb\frac{a}{Rb}. - Higher aa means higher TcT_{c} and easier liquefaction. - Pitfall: a larger bb lowers TcT_{c}, so liquefiability is judged by ab\frac{a}{b}, not by aa alone. Brain-Teaser 1: For CO2\mathrm{CO_{2}}, a=3.59 L2 atm mol−2a = 3.59\,\mathrm{L^{2}\,atm\,mol^{-2}} and b=0.0427 L mol−1b = 0.0427\,\mathrm{L\,mol^{-1}}. Estimate TcT_{c} and PcP_{c}. Solution: Tc=8×3.5927×0.0821×0.0427=28.720.0947=303 KT_{c} = \frac{8 \times 3.59}{27 \times 0.0821 \times 0.0427} = \frac{28.72}{0.0947} = 303\,\mathrm{K}, close to the measured 304 K304\,\mathrm{K}. Pc=3.5927×0.04272=3.590.0492=72.9 atmP_{c} = \frac{3.59}{27 \times 0.0427^{2}} = \frac{3.59}{0.0492} = 72.9\,\mathrm{atm}, close to the measured 73 atm73\,\mathrm{atm}. Brain-Teaser 2: The critical temperatures of gases A and B are 400 K400\,\mathrm{K} and 150 K150\,\mathrm{K}. Which has the larger ab\frac{a}{b}, and which can be liquefied at room temperature by pressure alone? Solution: Tc∝abT_{c} \propto \frac{a}{b}, so A has the larger ratio, meaning stronger attractions relative to size. Room temperature (300 K300\,\mathrm{K}) is below TcT_{c} of A but above that of B, so only A can be liquefied by pressure alone. Brain-Teaser 3: A cylinder of CO2\mathrm{CO_{2}} at 25∘C25^{\circ}\mathrm{C} shows a constant pressure of about 64 atm64\,\mathrm{atm} as gas is withdrawn, until suddenly the pressure starts falling. Explain. Solution: At 25∘C25^{\circ}\mathrm{C} (below Tc=31∘CT_{c} = 31^{\circ}\mathrm{C}) the cylinder holds liquid CO2\mathrm{CO_{2}} in equilibrium with its vapour. The pressure equals the vapour pressure, which depends only on temperature. As gas is withdrawn, more liquid evaporates and the pressure stays constant. When the last liquid is gone, only gas remains and the pressure falls as Boyle's law predicts.