## Critical Constant and Liquefaction of Gases

**Critical constants** like temperature, pressure, and volume of gas determine the condition and formula of liquefaction of ideal and real gases. Liquefaction of gases is an important property in physical chemistry and used for the transportation of gas. Condition of liquefaction and critical constant of real gases define by Andrews’s isotherms and determination in terms of van der walls constant.

A gas can liquify by lowering the temp and increasing pressure. Thus most of the real gas can be liquefied at natural pressure by the cool suitable lowering of the temp.

But gas cannot be liquefied unless its temperature below a certain value depending upon the nature of the gas whatever high pressure may be applied.

The temperature at which the gas can be liquified is called its critical temperature. Therefore Liquefaction has done when the temperature below the critical constant of gases.

### Critical Temperature, Pressure, Volume

The essential formula for the cool or liquefaction is to maintain critical temperature, pressure, and volume of the gas molecules. The critical temperature, pressure, and volume simply represent T_{C}, P_{C}, and V_{C} respectively.

The** Critical temperature** is the maximum temperature at which a gas can be liquefied and the temperature above which a liquid cannot exist.

**Critical pressure** is the maximum pressure required to cause liquefaction at the critical temperature.

**Critical volume** is the volume occupied by one mole of gas molecules at critical temperate and critical pressure.

### Liquefaction Condition of Gas Andrews Isotherms

In 1869, Thomas Andrews carried out an experiment in which P – V relations for liquefication of carbon dioxide gas at different temp.

- High temperatures the isotherms for liquefication of CO
_{2}look like those of an Ideal gas. - Low temp, the nature of the curves has altogether different appearances. For curve a 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 to a liquid with much higher density. This conversion takes place at constant pressure.
- Point C, liquefaction of CO
_{2}is complete and thus the CD is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that AB represents the gaseous state, BC, liquid, and vapor in equilibrium, and CD shows the liquid state only. - Still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The pressure corresponding to this portion is higher than at lower temperatures.
- At temperatures, T
_{C}the horizontal portion is reduced to a mere point. At temperatures higher then T_{C}there is no indication of qualification at all.

For every gas can have a limit of temperature above which they can not be liquefied, no matter what the pressure is.

### Continuity of State in Chemistry

It appears from the Amagat curve at T there is a discontinuity or break during the transformation of gas to liquid.

The continuity of the states from the gas to liquid can be explained from the following Andrews isotherm ABCD at T_{1}.

When A is heated with a specific heat to B at constant volume along with AB. Then the gas gradually cool or temp decreases at constant P along with BC, the volume and reduced considerably. Again cool or decreasing temp at constant volume until point D is reached. Nowhere in the process liquefaction would appear.

At D, the system is a highly compressed gas. But from this temperature curve, this critical point is the representation of the liquefaction state of gases. Hence there is hardly a distinction between the liquid state 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 Phenomena of Real Gases

Van der Waals equation for 1-mole real gas

(P + a/V^{2})(V – b) = RT

This equation has three roots in volume for given values of a, b, P, and T. It found that either all the three roots are real or one is real and the other two are imaginary.

- At higher temperatures and higher volume regions, the isotherms look much like the isotherms from Ideal gas law.
- At the temperature lower than Tc the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V
_{1}, V_{2}, and V_{3}at pressure.

The section AB and ED of the Van der Waals curve at T_{1}can be realized experimentally. ED represents supersaturated or super cool vapor and AB represents super-heated liquid at this temp. Both of these states are meta-stable.

These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium. - The section BCD of the Van der Waals isotherms cannot be realized experimentally. In this region the slope of the P – V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease in pressure. The line BCD represents the metastable state.

### Determination of Critical Constants

Again with the increase of temperature the minimum and maximum points come close to each other and at the critical constant (C) both maximum and minimum coalesce. The slope and curvature both become zero at this point.

Mathematical condition of the critical constant

(dP/dV)_{T} = 0

(d^{2}P/dV^{2})_{T} = 0

### Critical Temperature Pressure and Volume Formula

When P = P_{c}, V = V_{c}, and T = T_{c}, we can calculate the critical temperature, pressure, and volume formula of the gases from the Van der walls equation. For Details calculation see the above video on this page.

Therefore the measured value of critical constants

V_{C} = 3b

T_{C} = 8a/27Rb

P_{C} = a/27b^{2}

Problem

Calculate Van der Waals constant for methane gas if critical temperature = 280.8 K and critical pressure = 50 atm.

Answer

a = 0.057 lit mol^{-1} and b = 4.47 lit^{2} atm mol^{-2}

### Compressibility Factor of Gas

Compressibility factor formula for the one-mole ideal gas

Z = PV/RT

Therefore the critical Coffeicient

Z_{C} = RT_{C}/P_{C}V_{C} = 8/3 = 1.66

### Critical Constant and Van der Waals Constants

a and b for real gas can be determined from critical constants formula. V_{c} in the expression is avoided due to difficulty in its determination.

From the critical constants formula of the gas

b = V_{C}/3

a = 27 R^{2} T_{C}^{2}/64P_{C}

Problem

The critical constants for water are 647 K, 22.09 MPa, and 0.0566 dm^{3} mol^{-1}. What is the value of a and b?

Solution

T_{C} = 647 K

P_{C} = 22.09 Mpa = 22.09 × 10^{3} kPa

V_{C} = 0.0566 dm^{3} mol^{-1}

b = V_{C}/3 = (0.0566 dm^{3} mol^{-1})/3

∴ b = 0.0189 dm^{3} mol^{-1}

From the critical constants formula of real gas

a = 3 P_{C} V_{C}^{2}

= 3 (22.09 × 10^{3}) × (0.0566)^{2}

∴ a = 213.3 kPa mol^{-2}

Problem

Argon has T_{C} = – 122°C, P_{C} = 48 atm. What is the radius of the argon atom?

Answer

Radius of argon atom = 1.47 × 10^{-8} cm.

### Critical Temperature of the Ideal Gas

For an ideal gas a = 0, since there exist no forces of attraction between the molecules. Thus the critical temperature of the ideal gas equal to zero.

Cool up to below the critical temperature is the essential condition for liquefaction of ideal gases. Therefore the ideal gas can not be liquefied at critical temperature or zero Kelvin because practically this temp does not exist in our universe.