Q. 34

# For real gases the relation between p, V and T is given by van der Waals equation:

P + an^{2}(V - nb) / V^{2}= nRT

Where‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas.

‘a’ is the measure of magnitude of intermolecular attraction.

(i) Arrange the following gases in the increasing order of ‘b’. Give reason. O_{2}, CO_{2}, H_{2}, He

(ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason. CH_{4}, O_{2}, H_{2}

Answer :

(i) As the Vander Waals constants,‘b’ is approximately equal to the total volume of the molecules of a gas. Hence, the increasing order of ‘b’ is as follows:

He < H_{2}< O_{2}< CO_{2}

This is because the parameter ‘b′ is proportional to the proper volume of a single particle, and the volume of CO_{2} is maximum followed by O_{2} which is followed by H_{2} and He. Or in other words the value of ‘b’ is directly proportional to the size of gas molecules. Hence we get the order shown above.

The volume of CO_{2} is maximum; this is because CO_{2} contains 3 atoms, whereas O_{2} has greater volume than H_{2} because Oxygen atom has 2 shells whereas H has only one shell. Hence, H is smaller in size than O. And He is having the lowest volume because it is single atom and has only one shell like H, but unlike He, H-H is having greater volume than He.

(ii) The value of‘a’ for any gas depends on the strength of inter molecular attraction. Molecules having the weakest forces of attraction has the smallest value of ‘a’ whereas the molecules having the strongest force of attraction has the largest a values.

Hence, as the surface area of CH_{4} is highest so, it has highest Vander Waal’s force of attraction so, has highest value of ‘a’, followed by O_{2} and H_{2}.

So the decreasing order is found to be:

CH_{4}> O_{2}> H_{2}

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