Deviation From Ideal Gas Behaviour

A gas which obeys the gas laws and the gas equation PV = nRT strictly at all temperatures and pressures is said to be an ideal gas. The molecules of ideal gases are assumed to be volume less points with no attractive forces between one another. But no real gas strictly obeys the gas equation at all temperatures and pressures. Deviations from ideal behaviour are observed particularly at high pressures or low temperatures. The deviation from ideal behaviour is expressed by introducing a factor Z known as compressibility factor in the ideal gas equation. Z may be expressed as $Z=\frac{PV}{nRT}$

$\bullet$In case of ideal gas, PV = nRT $\therefore$ Z = 1

$\bullet$In case of real gas, PV $\ne$ nRt $\therefore$ Z $\ne$ 1

Thus in case of real gases Z can be < 1 or > 1

(i) When Z < 1, it is a negative deviation. It shows that the gas is more compressible than expected from ideal behaviour.

(ii) When Z > 1, it is a positive deviation. It shows that the gas is less compressible than expected from ideal behaviour.

Causes of deviation from ideal behaviour

The causes of deviations from ideal behaviour may be due to the following two assumptions of kinetic theory of gases. There are

$\bullet$The volume occupied by gas molecules is negligibly small as compared to the volume occupied by the gas.

$\bullet$The forces of attraction between gas molecules are negligible.

The first assumption is valid only at low pressures and high temperature, when the volume occupied by the gas molecules is negligible as compared to the total volume of the gas. But at low temperature or at high pressure, the molecules being in compressible the volumes of molecules are no more negligible as compared to the total volume of the gas.

The second assumption is not valid when the pressure is high and temperature is low. But at high pressure or low temperature when the total volume of gas is small, the forces of attraction become appreciable and cannot be ignored.

Van Der Waal's Equation

The general gas equation PV = nRT is valid for ideal gases only Van der Waal is 1873 modified the gas equation by introducing two correction terms, are for volume and the other for pressure to make the equation applicable to real gases as well.

Volume correction

Let the correction term be v

$\therefore$ Ideal volume v_i = (V – v)

Now v $\propto$ n or v = nb

[n = no. of moles of real gas; b = constant of proportionality called Van der Waal's constant]

$\therefore$ V_i = V – nb

b = 4$\times$volume of a single molecule.

Pressure Correction

Let the correction term be P

$\therefore$ Ideal pressure P_i = (P + p)

Now, $P\propto {\left(\frac{n}{V}\right)}^2$= $\frac{a{n}^2}{V^2}$

Where a is constant of proportionality called another Van der Waal's constant.

Hence ideal pressure

P_i = $\left(P+\frac{a{n}^2}{V^2}\right)$

Here, n = Number of moles of real gas

V = Volume of the gas

a = A constant whose value depends upon the nature of the gas

Substituting the values of ideal volume and ideal pressure, the modified equation is obtained as

$\left(P+\frac{a{n}^2}{V^2}\right)(v-nb)=nRT$

Illustration 1. 1 mole of SO₂ occupies a volume of 350 ml at 300K and 50 atm pressure. Calculate the compressibility factor of the gas.

Solution: P = 50 atm V = 350 ml = 0.350 litre

n = 1 mole

T = 300L $\therefore$ Z = $\frac{PV}{nRT}$

$\therefore$ Z =$\frac{50\times 0.350}{1\times 0.082\times 300}=0.711$

Thus SO₂ is more compressible than expected from ideal behaviour.

Vander Waals equation, different forms

$\bullet$At low pressures: 'V' is large and 'b' is negligible in comparison with V. The Vander Waals equation reduces to:

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

PV + $\frac{a}{V}$ = RT

PV = RT - $\frac{a}{V}$ or PV < RT

This accounts for the dip in PV vs P isotherm at low pressures.

$\bullet$At fairly high pressures

$\frac{a}{V^2}$may be neglected in comparison with P. The Vander Waals equation becomes

P (V – b) = RT

PV – Pb = RT

PV = RT + Pb or PV > RT

This accounts for the rising parts of the PV vs P isotherm at high pressures.

$\bullet$At very low pressures: V becomes so large that both b and $\frac{a}{V^2}$ become negligible and the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal behaviour at very low pressures.

$\bullet$Hydrogen and Helium: These are two lightest gases known. Their molecules have very small masses. The attractive forces between such molecules will be extensively small. So $\frac{a}{V^2}$ is negligible even at ordinary temperatures. Thus PV > RT. Thus Vander Waals equation explains quantitatively the observed behaviour of real gases and so is an improvement over the ideal gas equation.

Vander Waals equation accounts for the behaviour of real gases. At low pressures, the gas equation can be written as,

$\left(P+\frac{a}{V_m^2}\right)\left(V_m\right)=RT$ or Z = $\frac{V_m}{RT}=1-\frac{a}{V_{m}RT}$

Where Z is known as compressibility factor. Its value at low pressure is less than 1 and it decreases with increase of P. For a given value of V_m, Z has more value at higher temperature.

At high pressures, the gas equation can be written as

P (V_m – b) = RT

Z = $\frac{P{V}_m}{RT}$ = 1 + $\frac{Pb}{RT}$

Here, the compressibility factor increases with increase of pressure at constant temperature and it decreases with increase of temperature at constant pressure. For the gases H₂ and He, the above behaviour is observed even at low pressures, since for these gases, the value of 'a' is extremely small.

Illustration 2. One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found are 1.072 & 1.375 respectively at initial and final states. Calculate the final volume.

Solution: P₁V₁ = Z₁nRT₁ and P₂V₂ = Z₂nRT₂

$\frac{P_2{V}_2}{T_2}\times \frac{T_1}{P_1{V}_1}=\frac{Z_2}{Z_1}$

or V₂ =$\frac{Z_2}{Z_1}\times \frac{T_2}{T_1}\times \frac{P_1{V}_1}{P_2}$ = $\frac{1.375}{1.072}\times \frac{273}{473}\times \frac{300\times 1}{600}$ = 370.1 ml

Some other important definitions

Relative Humidity (RH)

At a given temperature it is given by equation

RH = $\frac{partialpressureofwaterinair}{vapourpressureofwater}$.

Boyle's Temperature (T_b)

Temperature at which real gas obeys the gas laws over a wide range of pressure is called Boyle's Temperature. Gases which are easily liquefied have a high Boyle's temperature [T_b(O₂)] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle's temperature [T_b(He) = 26K].

Boyle's temperature T_b = $\frac{a}{Rb}=\frac{1}{2}{T}_i$

where T_i is called Inversion Temperature and a, b are called van der Waals constant.

Critical Constants

$\bullet$Critical Temperature (T_c): It (T_c) is the maximum temperature at which a gas can be liquefied i.e. the temperature above which a gas can't exist as liquid.

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

$\bullet$Critical Pressure (P_c): It is the minimum pressure required to cause liquefaction at T_c

$P_c=\frac{a}{27b^2}$

$\bullet$Critical Volume: It is the volume occupied by one mol of a gas at T_c and P_c

V_c = 3b

Molar heat capacity of ideal gases: Specific heat c, of a substance is defined as the amount of heat required to raise the temperature of is defined as the amount of heat required to raise the temperature of 1 g of substance through 1⁰C, the unit of specific heat is calorie g^-1 K^-1. (1 cal is defined as the amount of heat required to raise the temperature of 1 g of water through 1⁰C)

Molar heat capacity C, is defined as the amount of heat required to raise the temperature of 1 mole of a gas trough 1⁰C. Thus,

Molar heat capacity = Sp. Heat $\times$ molecular wt. Of the gas

For gases there are two values of molar heats, i.e., molar heat at constant pressure and molar heat at constant molar heat at constant volume respectively denoted by C_p and C_v. C_p is greater than C_v and C_p-R = 2 cal mol^-1 K^-1.

From the ratio of C_p and C_v, we get the idea of atomicity of gas.

For monatomic gas C_p = 5 cal and C_v =3 cal

$\therefore$$\lambda =\frac{4}{3}=1.67\left(\gamma ispoissonapossratio=\frac{C_p}{C_v}\right)$for diatomic gas C_p = 7 cal and C_v = 5 cal $\gamma =\frac{7}{5}=1.40$For polyatomic gas C_p = 8 cal and C_v= cal $\gamma =\frac{8}{6}=1.33$also C_p = C_p$\times$m, $C_v=C_v\times m$where, C_p and C_v are specific heat and m, is molecular weight.Illustration 3.Calculate Vander Waals constants for ethylene T_C= 282.8 k P_C = 50 atmSolution:b = $\frac{1}{8}\frac{R{T}_C}{P_C}=\frac{1}{8}\times \frac{0.082\times 282.8}{50}$ = 0.057 litres/molea = $\frac{27}{64}{R}^2\times \frac{T_C^2}{P_C}=\frac{27}{64}\times {(0.082)}^2\times {(282.8)}^2\times \frac{1}{50}$ = 4.47 lit² atm mole^𠄲Gas EudiometryThe relationship amongst gases, when they react with one another, is governed by two laws, namely Gay-Lussac law and Avogadro’s law.Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one end and the other closed end of which is provided with platinum terminals for the passage of electricity through the mixture of gases. Such a tube is known as Eudiometer tube and hence the name Eudiometry also used for Gas analysis.During Gas analysis, the Eudiometer tube filled with mercury is inverted over a trough containing mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which displaces an equivalent amount of mercury. Next a known excess of oxygen is introduced and the electric spark is passed, whereby the combustible material gets oxidised. The volumes of carbon dioxide, water vapour or other gaseous products of combustion are next determined by absorbing them in suitable reagents. For example, the volume of CO₂ is determined by absorption in KOH solution and that of excess of oxygen in an alkaline solution of pyrogallol. Water vapour produced during the reaction can be determined by noting contraction in volume caused due to cooling, as by cooling the steam formed during combustion forms liquid (water) which occupies a negligible volume as compared to the volumes of the gases considered. The excess of oxygen left after the combustion is also determined by difference if other gases formed during combustion have already been determined. From the data thus collected a number of useful conclusions regarding reactions amongst gases can be drawn.$\bullet$Volume-volume relationship amongst Gases or simple Gaseous reactions.$\bullet$ omposition of Gaseous mixtures.$\bullet$Molecular formulae of Gases.$\bullet$Molecular formulae of Gaseous Hydrocarbons.The various reagentsused for absorbing different gases are O₃ $\to$ turpentine oilO2alkaline pyrogallolNO $\to$ FeSO₄ solutionCO₂,SO₂ $\to$alkali solution (NaOH, KOH, Ca(OH)₂, HOCH₂CH₂NH₂, etc.)NH₃ $\to$ acid solution or CuSO₄ solutionCl₂ $\to$ waterEquation for combustion of hydrocarbons.C_xH_y + $\left(x+\frac{y}{4}\right)$O₂ $\to$ xCO₂ + $\frac{y}{2}$H₂OGeneral Assumptions: In all problems, it is assumed that the sparking occurs at room temperature. This implies that water formed would be in liquid state andthat nitrogen gas is inert towards oxidation. Illustration 4.A gaseous hydrocarbon requires 6 times its own volume of O₂ for complete oxidation and produces 4 times its volume of CO₂. What is its formula?Solution:The balanced equation for combustionC_xH­_y + $\left(x+\frac{y}{4}\right)$O₂ $\to$ xCO₂ + $\frac{y}{2}$H₂O1 volume $\left(x+\frac{y}{4}\right)$ volume $\therefore$$x+\frac{y}{4}$ = 6 (by equation)

or 4x + y = 24 …(1)

Again x = 4 since evolved CO₂ is 4 times that of hydrocarbon

$\therefore$ 16 + y = 24 or y = 8 $\therefore$ formula of hydrocarbon C₄H₈