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 = PV / nRT   

• In case of ideal gas, PV = nRT ∴ Z = 1  

• In case of real gas, PV ≠ nRt ∴ Z ≠ 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  

• The volume occupied by gas molecules is negligibly small as compared to the volume occupied by the gas.  

• 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  

∴ Ideal volume vi = (V – v)  

Now v ∝ n or v = nb                   

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

∴Vi = V – nb  

b = 4 × volume of a single molecule.  

Pressure Correction  

Let the correction term be P  

∴ Ideal pressure Pi = (P + p)  

Now,  = P ∝ (n/V)2 = an2 / V2 

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

Hence ideal pressure  

                        Pi = (P + an2 / V2)  

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  

                        (P + an2 / V2) (V–nb) = nRT

Illustration 14. 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  Z = PV / nRT   

                        ∴ Z = 50 × 0.350 / 1 × 0.082 × 300 = 0.711   

                        Thus SOis more compressible than expected from ideal behaviour.    

Exercise 15.Out of NH3 and N2 which will have  

      (a) Larger value of a  

      (b) Larger value of b    

Exercise 16:- 2 moles of NH3 occupied a volume of 5 litres at 27°C. Calculate the pressure if the gas obeyed Vander Waals equation. Given a = 4.17 litre2 atm mole–2, b = 0.037 litre/mole.    

Exercise 17.Calculate the percentage of free volume available in 1 mole gaseous water at 1 atm and  Density of liquid  is 0.958g / mL.  

Vander Waals equation, different forms  

At low pressures: ‘V’ is large and ‘b’ is negligible in comparison with V. The Vander Waals equation reduces to:             (P + a / V2) V = RT;          PV + a/ V = RT             PV = RT – a/V  or PV < RT             This accounts for the dip in PV vs P isotherm at low pressures.

  352_Deviation of gases.JPG Deviation of gases from ideal behaviour with pressure.

At fairly high pressures           a/V2  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.

  487_Gas at different temperature.JPG The plot of Z vs P for N2 gas at different temperature is shown here.

At very low pressures: V becomes so large that both b and a/V2 become negligible and the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal behaviour at very low pressures.

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 a/V2 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,  

            (P + a/v2m) (Vm)  = RT or Z = Vm / RT = 1 – a/VmRT  

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 Vm, Z has more value at higher temperature.  

At high pressures, the gas equation can be written as  

P (Vm – b) = RT  

Z = PVm / RT = 1 + 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 15. 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  

                      P2V2 / T× T1 / P1V1 = Z2 / Z1  

                        or V2 =  Z2 / Z1 × T1 / T1 × P1V1 / P2  = 1.375 / 1.072 × 273 / 473  × 300 × 1 / 600 = 370.1 ml    

Illustration 16. The behaviour of a real gas is usually depicted by plotting compressibility factor Z versus P at a constant temperature. At high temperature and high pressure, Z is usually more than one. This fact can be explained by van der Waal’s equation when  

                        (A) the constant ‘a’ is negligible and not ‘b’     

                        (B) the constant ‘b’ is negligible and not ‘a’  

                        (C) both constants ‘a’ and ‘b’ are negligible     

                        (D) both the constants ‘a’ and ‘b’ are not negligible.    

Solution:         (P + n2a / V2) (V – nb) = nRT  

At low pressures, ‘b’ can be ignored as the volume of the gas is very high. At high temperatures ‘a’ can be ignored as the pressure of the gas is high.  

                        ∴   P (V–b) = RT  

                        PV – Pb = RT => PV = RT + Pb  

                        PV / RT = Z = 1 + Pb / RT  

                        Hence, (A) is correct.    

Exercise 18.The compressibility factor for CO2 at 273K and 100atm pressure is 0.2005. Calculate the volume occupied by 0.2 mole of CO2 gas at 100 atm assuming ideal behaviour    

Some other important definitions  

Relative Humidity (RH)  

At a given temperature it is given by equation  

RH = partial pressure of water in air / vapour pressure of water  

Boyle’s Temperature (Tb)  

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 [Tb(O2)] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle’s temperature [Tb(He) = 26K].  

Boyle’s temperature Tb = a / Rb = 1/2 T1  

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

Critical Constants  

• Critical Temperature (Tc): It (Tc) 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 = 8a / 27Rb

• Critical Pressure (Pc): It is the minimum pressure required to cause liquefaction at Tc  

            Pc = a/27b2  

• Critical Volume: It is the volume occupied by one mol of a gas at Tc and Pc  

            Vc = 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  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 Cp and Cv. Cp is greater than Cv and Cp-R = 2 cal mol-1 K-1.  

From the ratio of Cp and Cv, we get the idea of atomicity of gas.  

For monatomic gas Cp = 5 cal and Cv =3 cal  

                  ∴ λ = 4/3 = 1.67             (γ is poisson's ratio = Cp / Cv)    

for diatomic gas Cp = 7 cal and Cv = 5 cal

                     γ 7/5 = 1.40

For polyatomic gas Cp = 8 cal and Cv= cal  

           γ   = 8/6 = 1.33

also Cp = Cpm,  

where, Cp and Cv are specific heat and m, is molecular weight.   

Illustration 17. Calculate Vander Waals constants for ethylene  

                        TC = 282.8 k; PC = 50 atm  

Solution: b = 1/8 RTc / Pc = 1/8 × 0.082 × 282.8 / 50 = 0.057 litres/mole  

                              a = 27 / 64 R2 × T2C / Pc = 27/64 × (0.082)2 × (282.8)2 × 1/50 = 4.47 lit2 atm mole–2

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