--%>

Describe physical adsorption and its characteristics.

When the forces of attraction existing between adsorbate and adsorbent are van der Waal's forces, the adsorption is called physical adsorption. This type of adsorption is also known as physisorption or van der Waal's adsorption. Since the forces existing between adsorbent and adsorbate are very weak, therefore, this type of adsorption can be easily reversed by heating or by decreasing the pressure.

Characteristics of Physisorption

Some of the important characteristics of physisorption are as follows:

(i) Deficient of specificity: since the van der Waal forces are universal, a given surface of adsorbent does not show any preference for any specific gas. It can adsorb all the gases but to a different extent.

(ii) Reversible nature: physical adsorption of a gas by the solid is reversible and thus equilibrium is reached rapidly

Solid + Gas  1387_Physical adsorption.png  gas/solid + Heat

Thus, according to Le-chatelier's principle,

(a) Increase of pressure pushes the equilibrium in forward direction leading to more adsorption of gas and decrease of pressure cause desorption to occur.

(b) Since process is exothermic, therefore, lowering of temperature favours more adsorption and increase of temperature leads to desorption.

(iii) Surface area of adsorbent: the extent of adsorption increase with the increase of surface area of adsorbent. Thus, finely divided metals and rough surfaces are good adsorbents.

(iv) Nature of adsorbate: the amount of gas adsorbed by solid depends on nature of gas. In general, easily liquefiable gases (i.e gases with higher critical temperature) are readily as van der Waal forces are stronger near the critical temperature.

(v) Enthalpy of adsorption: the enthalpy of adsorption is low (20-40 kJ mol-1). This is because of weak nature of van der Waal's forces.

(vi) State of adsorbate: molecular state of adsorbate remains unaltered.

(vii) Activation energy: physical adsorption does not involve any chemical reaction and therefore, it requires very low activation energy.

   Related Questions in Chemistry

  • Q : Molarity of HCl solution 20 ml of HCL

    20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is:  (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.

  • Q : Problem on decinormal strength Can

    Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml  of its aqueous solution to provide decinormal strength: (i) 1g  (ii)2g  (iii) 10g  (iv) 20g<

  • Q : Normality of sulphuric acid Help me to

    Help me to go through this problem. Normality of sulphuric acid is: (a) 2N (b) 4N (c) N/2 (d) N/4

  • Q : How haloalkanes are prepared from

    Alkyl halides can be prepared from alkanes through substitution and from alkenes through addition of halogen acids or through allylic substitution.    From alkanesWhen alkanes are treated with halogens, chlo

  • Q : Vapour pressure of methanol in water

    Give me answer of this question. An aqueous solution of methanol in water has vapour pressure: (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water (d) Less than that of water

  • Q : Define Virial Equation The constant of

    The constant of vander Waal's equation can be related to the coefficients of the virial equation.  Vander Waal's equation provides a good overall description of the real gas PVT behaviour. Now let us

  • Q : Molar mass of solute The boiling point

    The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1  (b)

  • Q : Explain the preparation of phenols. The

    The methods used for the preparation of phenols are given below:    From aryl sulphonic acids

  • Q : Problem on normality Help me to solve

    Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N

  • Q : Molar mass of compound The freezing

    The freezing point of a solution having 4.8 g of a compound in 60 g of benzene is 4.48. Determine the molar mass of the compound (Kf = 5.1 Km-1) , (freezing point of  benzene = 5.5oC)          &n