Intermediate Chemistry - Mono Mole

October 30, 2019September 25, 2024

Critical constants

The critical constants of a real gas describe the conditions for liquefaction of the gas.

Thomas Andrews, an Irish chemist, conducted experiments on the liquefaction of gases in the 1860s and developed the concept of critical constants, which is summarised as follows:-

Consider a gas, Y, confined by a piston in a cylinder. The diagram below shows the pressure of the gas in the cylinder as the piston is pushed to vary the volume at various temperatures with T₁> T₂ > T₃ > T₄ > T₅. Each coloured line, called an isotherm, describes the change in pressure versus the change in molar volume of the gas at a specific temperature.

The isotherms at higher temperatures have smooth curves, while those at lower temperatures consist of horizontal portions. As we compress the gas at point A at temperature T₅, its pressure increases until just to the left of point B, where the gas starts to condense into its liquid form (it exists in both the liquid and gaseous phases with a defined surface separating the two). The system is now at equilibrium, where $Y(g)\rightleftharpoons Y(l)$ . The fact that the gas condenses into a liquid implies that it is a real gas (an ideal gas is devoid of intermolecular interactions).

As we manually compress the system further, more liquid is formed, as the system reacts according to Le Chatelier’s principle: the system counteracts the increase in pressure (decrease in volume) by shifting the position of the equilibrium of $Y(g)\rightleftharpoons Y(l)$ to the right to reduce the pressure, resulting in the pressure of the gas being constant. Hence, the curve from B to C is horizontal and is called a tie line. The pressure corresponding to the tie line is known as the vapour pressure of the liquid atT₅. At point C, all the gas is condensed into liquid with the piston touching the surface of the liquid.

The volume hardly changes when we compress the system further from C to D, as the liquid state of a substance is relatively incompressible. If we repeat the entire compression process at higher temperatures, we find that the width of the tie line shortens, e.g. for T₄, where it reduces to a maximum point (marked X) at T₃. We call this point the critical point of the substance and the isotherm that passes through it, the critical isotherm.

The temperature that produces the critical isotherm is called the critical temperature, T_c, e.g. T_cfor CO₂ is 31.04⁰C. The pressure and molar volume at the critical point is known as the critical pressure, p_c, and critical molar volume, V_c, respectively. p_c, V_c, and T_c are collectively called the critical constants of a substance.

The gas that is described by isotherms at temperatures above T_ccannot be compressed into a liquid regardless of the pressure applied. Due to the high pressure environment at temperatures above T_c, the gas has a density that is closer to that of a liquid and is called a supercritical fluid even though it is by definition a gas.

The density of a supercritical fluid and hence its solubility can be optimised for a particular application by adjusting its pressure. Supercritical CO₂, for example, is used to dissolve and extract caffeine from coffee beans and as a dry-cleaning solvent.

Generally, the regions shaded yellow, blue, pink and grey in the above graph represent the gas phase, the liquid-gas phase, the liquid phase and the supercritical fluid phase respectively.

Lastly, it is important to note that at points B, C, X and all other points on the boundary separating the liquid-gas zone from the other zones, the substance exists only in one phase. For example, at the critical point, the meniscus between the liquid and gas (supercritical fluid) phases disappears, as there is no difference in densities of the liquid and gas (supercritical fluid).

Next article: van der Waal’s equation of state

Previous article: Compression factor

Content page of real gases

Content page of intermediate chemistry

Main content page

October 30, 2019December 5, 2024

van der Waal’s equation of state

Johannes van der Waals, a Dutch physicist, introduced an equation in 1873 called the van der Waals equation to mathematically describe the physical state of a real gas.

This equation is:

$\left [ p+a\left ( \frac{n}{V} \right )^2 \right ]\left ( V-nb \right )=nRT\; \; \; \; \; \; \; \; 3$

where a and b are called the van der Waals coefficients, which are temperature-independent constants specific to each gas.

The equation looks similar to the ideal gas equation except for the pressure and volume factors of $a\left ( \frac{n}{V} \right )^2$ and nb, respectively. To understand how van der Waals’ equation is derived, we need to know how the ideal gas law is conceptualised.

The ideal gas equation, pV = nRT, is derived from the combined experiments of Robert Boyle and Jacques Charles, and a principle developed by Amedeo Avogadro. This means that, for a fixed amount of gas at a relatively high temperature, the product of the experimentally observed values of p and V in the equation is a constant. The equation works reasonably well at low pressures where gases approach ideality. Under such conditions, gas particles can be assumed to be point masses with zero volume and, therefore, have no intermolecular forces of attraction or repulsion between them. In other words, the equation only works when the observed variables of p and V are ‘ideal’:

$p_{ideal}V_{ideal}=nRT\; \; \; \; \; \; \; \; 4$

In reality, gases have finite volumes and repel one another, especially at high pressures. The experimentally observed values of pressures and volumes are actually p_real and V_real. Therefore, for eq4 to work, we have to modify it.

Due to the presence of repulsive forces between real gas molecules at high pressures, the observed volume of a real gas is larger than the volume of its ideal counterpart, that is, V_real > V_ideal. The diagram below shows the volume expansion (from yellow to blue) if the point masses have finite volumes.

Letting the difference in molar volume of a real gas and an ideal gas be b, we have $\frac{V_{real}}{n}-\frac{V_{ideal}}{n}=b$ or

$V_{ideal}=V_{real}-nb\; \; \; \; \; \; \; \; 5$

Substituting eq5 in eq4 and rearranging yields

$V_{real}=\frac{nRT}{p_{ideal}}+nb\; \; \; \; \; \; \; \; 6$

Eq6 predicts that the molar volume of a gas at zero Kelvin is b. Experiments indeed reveal that b has approximately the same value as the molar volume of the solid form or liquid form of the gas near that temperature. On the other hand, the ideal gas law, $V_{ideal}=\frac{nRT}{p_{ideal}}$ , wrongly predicts that the volume of a gas at 0 K is zero.

Next, the pressure that a real gas exerts on the wall of a container is lower than the pressure of its ideal counterpart (p_ideal > p_real), as the molecules of a real gas experience intermolecular forces of attraction that decelerates their motion when they are about to impact the wall of the container. This means that we need to add a pressure shortfall, Δp_real, to the observed real pressure to reflect the ideal pressure that fits eq4:

$p_{ideal}=p_{real}+\Delta p_{real}\; \; \; \; \; \; \; \; 7$

If we perceive the intermolecular force of attraction (i.e., the force that one molecule of gas experiences near the wall of the container) as the attraction between that molecule and the bulk of the gas, the pressure shortfall due to that particular molecule is proportional to the density of the gas:

$\Delta p_{real,\, one\, molecule}=k\left ( \frac{n}{V_{real}} \right )$

Furthermore, the total number of molecules N near the wall of the container is again proportional to the density of the gas:

$N=k'\left ( \frac{n}{V_{real}} \right )$

The total pressure shortfall is therefore:

$\Delta p_{real}=N\Delta p_{real,\, one\, molecule}=k'\left ( \frac{n}{V_{real}} \right )k\left ( \frac{n}{V_{real}} \right )=a\left ( \frac{n}{V_{real}} \right )^2$

where a = k’k is a proportionality constant specific to a particular gas.

Eq7 becomes

$p_{ideal}=p_{real}+a\left ( \frac{n}{V_{real}} \right )^2\; \; \; \; \; \; \; \; 8$

Substituting eq5 and eq8 in eq4, we get the explicit form of eq3:

$\left [ p_{real}+a\left ( \frac{n}{V_{real}} \right )^2 \right ]\left ( V_{real}-nb \right )=nRT\; \; \; \; \; \; \; \; 9$

The pressure and volumes in eq9 now represent experimentally observed real pressure and real volumes, respectively. We have essentially converted an equation that only works for ideal gases into one that is applicable for real gases. For simplicity, we can omit the subscripts, which gives eq3.

Using eq3, the van der Waals equation can also be expressed in the molar form by substituting it with the molar volume, $V_m=\frac{V}{n}$ , to give:

$\left ( p+\frac{a}{V_m^{\: 2}} \right )\left ( V_m-b \right )=RT$

which rearranges to:

$p=\frac{RT}{V_m-b}-\frac{a}{V_m^{\: 2}}\; \; \; \; \; \; \; \; 10$

At high molar volumes (i.e. low pressures), V_m – b ≈ V_m and the second term on the RHS of eq10 becomes very small, with eq10 reducing to the ideal gas equation: $p\approx \frac{RT}{V_m}$ .

The van der Waals equation is an improvement over eq2, as it is able to describe the physical state of a real gas in terms of the van der Waals coefficients a and b, which relate to the strength of the forces of attraction between molecules and the size of the molecules, respectively.

Since the critical isotherm of a gas has an inflexion point at the critical point, we can determine the van der Waals coefficients for a gas by finding the 1^st and 2^nd derivatives of eq10, letting these be equal to zero and solving for the critical constants:

$\frac{dp}{dV_m}=-\frac{RT}{\left ( V_m-b \right )^2}+\frac{2a}{V_m^{\: 3}}=0\; \; \; \; \; \; \; \; 11$

$\frac{dp}{dV_m}=\frac{2RT}{\left ( V_m-b \right )^3}-\frac{6a}{V_m^{\:4}}=0\; \; \; \; \; \; \; \; 12$

From eq11 and eq12, $a=\frac{RTV_c^{\: 3}}{2\left ( V_c-b \right )^2}$ and $a=\frac{RTV_c^{\: 4}}{3\left ( V_c-b \right )^3}$ respectively. Combining both values of a,

$V_c=3b\; \; \; \; \; \; \; \; 13$

Substitute eq13 back in eq11 where V_m is now V_c,

$T_c=\frac{8a}{27Rb}\; \; \; \; \; \; \; \; 14$

Substitute eq13 and eq14 back in eq10,

$p_c=\frac{a}{27b^2}\; \; \; \; \; \; \; \; 15$

Hence, by measuring the pressure, volume and temperature at the critical point of a gas (i.e. the critical constants), we can calculate the van der Waals coefficients for that gas. The table below shows the van der Waals coefficients for some common gases:

	a / atm dm⁶mol^-2	b / 10^-2 dm³mol^-1
H₂	0.2420	2.65
N₂	1.352	3.87
O₂	1.364	3.19
CO₂	3.610	4.29

Next article: Virial equation of state

Previous article: Critical constants