Critical constants of gases definition

Critical constants of real gas formula

Liquefaction of gases is an important physical property. It used for transportation of gas. But the condition and formula of liquefication of gas identified by critical constants.

A gas can cool or 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. Gas can only be liquefied when the temperature below the critical temperature.

Conditions for liquefaction of gases

The essential condition for the cool or liquefaction is to maintain critical temp, pressure, and volume of the gas molecules. The critical temperature, pressure, and volume simply represent TC, PC, and VC respectively.

A simple definition of critical constant

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 of CO2-Andrews isotherms

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

Critical constants of gases definition
Critical constants
  1. High temperatures the isotherms for liquefication of CO2 look like those of an Ideal gas.
  2. 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.
  3. 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.
  4. Point C, liquefaction of CO2 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.
  5. 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.
  6. At temperatures, TC the horizontal portion is reduced to a mere point. At temperatures higher then TC 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 T1.

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 liquid would appear.

At D, the system is a highly compressed gas. But from the curve, this point is the representation of the liquid state. 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.

Van der Waals equation critical phenomena

Van der Waals equation for 1-mole real gas

\left (P+\frac{a}{V^{2}} \right )\left ( V-nb \right )=RTor, V^{3}-\left (b+\frac{RT}{P} \right )V^{2}-\frac{aV}{P}-\frac{ab}{P}=0

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.

Van der Waals equation critical constant of gas formula
Critical phenomena
  1. At higher temperatures and higher volume regions, the isotherms look much like the isotherms from Ideal gas law.
  2. 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 V1, V2, and V3 at pressure.
    The section AB and ED of the Van der Waals curve at T1 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.
  3. 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

\left (\frac{\mathrm{d}P }{\mathrm{d} V} \right ) _{T}=0

\left (\frac{\partial^2 P}{\partial V^2} \right )_{T}=0

From Van der Waals equation

\left (P+\frac{a}{V^{2}} \right )\left ( V-b) \right )=RT

or, P= \frac{RT}{\left ( V-b \right )}-\frac{a}{V^{2}}

Differentiating with respect to the volume at constant temperature gives the slope of the curve.

∴ Slope

\left (\frac{\partial P}{\partial V} \right )_{T}=-\frac{RT}{\left ( V-b \right )^{2}}+\frac{2a}{V^{3}}

Curvature

\left (\frac{\partial^2 P}{\partial V^2} \right )_{T}=\frac{2RT}{\left (V-b \right )^{3}}-\frac{6a}{V^{4}}

The critical volume of real gases

When P = Pc, V = Vc, and T = Tc.

-\frac{RT_{C}}{\left ( V_{C}-b \right )^{2}}+\frac{2a}{V_{C}^{3}}=0

\frac{RT_{C}}{\left ( V_{C}-b \right )^{2}}=\frac{2a}{V_{C}^{3}}

-\frac{2RT_{C}}{\left ( V_{C}-b \right )^{3}}+\frac{6a}{V_{C}^{4}}=0

or,-\frac{2RT_{C}}{\left ( V_{C}-b \right )^{3}}=\frac{6a}{V_{C}^{4}}

\therefore \frac{\left ( V_{C}-b \right )}{2}=\frac{V_{C}}{3}

∴ VC = 3b

The critical temperature of the gases

\frac{RT_{C}}{\left ( V_{C}-b \right )^{2}}=\frac{2a}{V_{C}^{3}}

Again, VC =3b.

\therefore \frac{RT_{C}}{4b^{2}}=\frac{2a}{27b^{3}}

or, T_{C}=\frac{8a}{27Rb}

Critical pressure of real gas formula

VDW equation at the critical state,

P_{C}=\frac{RT_{C}}{V_{C}-b}-\frac{a}{V_{C}^{2}}

Again VC = 3b and TC= 8a/27Rb

\therefore P_{C}=\frac{a}{27b^{2}}

Problem
Calculate Van der Waals constants 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 lit2 atm mol-2

Compressibility factor of the natural gas formula

Compressibility factor formula for the one-mole ideal gas

Z = PV/RT

At the critical point

Z_{C}=\frac{RT_{C}}{P_{C}{V_{C}}}

=\frac{R\times \frac{a}{27Rb}}{\frac{a}{27b^{2}}\times 3b}=0.375

Critical coefficient

=\frac{RT_{C}}{P_{C}V_{C}}=\frac{8}{3}=1.66

Critical constants in terms of Van der Waals constants

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

From the critical volume formula of the gas

b=\frac{V_{C}}{3}

\frac{P_{C}V_{C}}{RT_{C}}=\frac{3}{8}\therefore V_{C}=\frac{3}{8}\times \frac{RT_{C}}{P_{C}}or,b=\frac{RT_{C}}{8P_{C}}

a=P_{C}\times 27b^{2}=3\times P_{C}\times \left ( 3b \right )^{2}=3P_{C}\times \left (\frac{3RT_{C}}{8P_{C}} \right )^{2}
\therefore a=\frac{27R^{2}T_{C}^{2}}{64P_{C}}

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 constant of gas formula
a = 3 PC VC2
= 3 (22.09 × 103) × (0.0566)2
∴ a = 213.3 kPa mol-2

Problem
Argon has TC = – 122°C, PC = 48 atm. What is the radius of the argon atom?

Answer
Radius of argon atom = 1.47 × 10-8 cm.

The 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 temp is the essential condition for an ideal gas to be liquified. Thus the ideal gas can not be liquefied because zero kelvin temp practically does not exist in the universe.