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# Critical Constants

## Critical Constants and Liquefaction of Gases

Critical constants like critical temperature, pressure, and volume of gas determine the condition and formula of liquefaction of real and ideal gases. Liquefaction of gases is an important property in physical chemistry which uses for the transportation of natural gas. Most of the gases in our environment are liquefy at ordinary pressure but the suitable lowering of cooling the temperature. But many gases like hydrogen, nitrogen, oxygen, methane, etc can not be liquefied at ordinary temperature however high pressure may be used. Hence the condition of liquefaction and critical temperature, pressure, and volume of real gas molecule define by Andrews’s isotherms and determine in terms of van der Waals constant.

According to the kinetic theory, ideal gas can not be liquefied in critical temperature because the gas molecules are considered as point masses with no intermolecular attraction. Real gases also, cannot be liquefied unless its temperature below a certain value depending upon the properties of gases. The temperature at which the gas can be liquefied is called its critical temperature for learning chemistry or physical chemistry.

### Liquefaction Condition and Andrews Isotherms

In 1869, Thomas Andrews carried out an experiment with a small quantity of carbon dioxide in a glass tube enclosed and sealed by the element mercury. He studied P – V relations for carbon dioxide at different temperatures or the condition of liquefaction at the critical temperature. At high temperatures or above the critical temperature the isotherms for liquefaction of carbon dioxide follow ideal gas law. At low temperatures or below the critical temperature, the nature of the curves has altogether different appearances. As the pressures increase, the volume of the gas decreases in curve A to B. At the point, B liquefaction commences and the volume decreases rapidly as the gas converted into the liquid with much higher density.

Point C, liquefaction of carbon dioxide is complete. The CD of the curve evidence for the fact. Therefore, AB represents the gaseous state, BC represents liquid or vapor in equilibrium, and CD shows the liquid state only. Still higher temperatures at T2, we get a similar type of curve like ABCD. At temperatures TC, the horizontal portion is reduced to a mere point, called the critical point or state of the gases. Therefore, every gas can have a limit of temperature above which they can not be liquefied.

### Critical Temperature Pressure Volume

The critical temperature, pressure, and volume simply represent TC, PC, and VC respectively, and above which the gases can not be liquefied. Therefore, the Critical temperature is the maximum temperature at which the gas can be liquefied and the temperature above which the liquid cannot exist. Critical pressure is the maximum pressure required for the liquefaction of gases at the critical temperature. Critical volume is the volume occupied by one gram mole of gaseous substances at critical temperate and pressure.

### Continuity of State in Chemistry

An examination of the PV-curve at the temperature below the critical temperature may be discontinuous or break down during the transformation of gas to liquid. The continuity of the states from the gas to liquid can be explained from the above Andrews isotherm ABCD at temperature, T1. Suppose the gas is heated with the specific heat at constant volume along with AB. Then the gas gradually cooling at constant pressure along with BC, the volume will be reduced considerably. On reaching D the process liquefaction would appear.

At point D, the system contains highly compressed gas. But from the Andrews curve, the critical temperature point is the representation of the liquefaction point of the gases. Hence there is hardly a distinction between the liquid and the gaseous state. There is no line of separation between the two phases. This is known as the principle of continuity of the state.

### Critical Temperature Pressure Volume

With the increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both become zero at this point. Therefore, these conditions, we can calculate the critical temperature, pressure, and volume formula for real gases from the Van der walls equation.

Hence, the measured value of VC, TC, and PC
VC = 3b
TC = 8a/27Rb
PC = a/27b2

Question: Calculate Van der Waals constant for the gas when critical temperature and pressure = 280.8 K and = 50 atm respectively.

Answer: a = 0.057 lit mol-1 and b = 4.47 lit2 atm mol-2

### Compressibility Factor of Real Gases

Compressibility factor formula at this state of the gas, Zc = PCVC/RTC = 3/8 = 0.375 and critical coefficient value = 8/3 = 2.66. If we compare these values with the experimental values, we found that the agreement is very poor. Because Van der Waals equation at the critical state not very accurate.

Therefore, the experimental value of the compressibility factor for non-polar or slightly polar bonding molecules like helium, neon, argon, oxygen, and methane close to 0.29. For the molecule having polarity or polarization like chlorine, carbon disulfide, chloroform, and ethylene close to 0.26 or 0.27. For hydrogen bonding molecules like ammonia, water, methyl alcohol close to 0.22 to 0.24.

### Critical Constants and Van der Waals Constant

Van der Waals constant for real gas can be determined from the critical constants formula (temperature and pressure) and volume in the expression is avoided due to difficulty of determination. From the critical constants like temperature, pressure, and volume formula of Van der Waals constants, b = VC/3 and a = 27 R2 TC2/64PC.

Problem: The critical constants for water are 647 K, 22.09 MPa, and 0.0566 dm3 mol-1. What is the value of a and b?

Solution:TC = 647 K, PC = 22.09 Mpa = 22.09 × 103 kPa, VC = 0.0566 dm3 mol-1.

b = VC/3 = (0.0566 dm3 mol-1)/3
∴ b = 0.0189 dm3 mol-1
From the critical constants formula of real gas
a = 3 PC VC2
= 3 (22.09 × 103) × (0.0566)2
∴ a = 213.3 kPa mol-2

Question: An atom  in the molecule has TC = – 122°C, PC = 48 atm. What is the radius of the atom?