Question

a and b are van der waals constants for gases. Chlorine is more easily liquefied than ethane because:
A. a and b for $C{l_2}$ >a and b for ${C_2}{H_6}$
B. a and b for $C{l_2}$ < a and b for ${C_2}{H_6}$
C. a for $C{l_2}$ < b for ${C_2}{H_6}$ but b for $C{l_2}$>b for ${C_2}{H_6}$
D. a for $C{l_2}$ >a for ${C_2}{H_6}$ but b for $C{l_2}$ < b for ${C_2}{H_6}$

Answer 1

Hint: Van der waals modified the ideal gas equation by suggesting that the gas molecules were not point masses but behave like rigid spheres having a certain diameter and that there exist intermolecular forces of attraction between them. The two correction terms introduced by van der waals are upon volume and pressure.

Complete step by step solution:
let us understand the significance of the constants, ‘a’ is a measure of Van der waals force of attraction existing between the molecules of a given gas.
The greater the value of a, the greater is the strength of the van der waals forces. The greater the value of a, the greater is the ease in which gas can be liquefied, ’b’ is just the incompressible volume per mole of any gas, which is directly proportional to the size of gas molecules.
Applying the concept to the above question, we know that chlorine is more liquefied than ethane therefore it simply implies that the value of a for chlorine is higher than the value of a in ethane. Constant ‘b’ is a correction for finite molecular size and hence heavier molecules require greater thermal energy and higher intermolecular forces.
Therefore smaller value of ‘b’ and larger value of ‘a’ cause easy liquefaction.

Hence the correct option is D.

Note:
If temperature is high, pressure is low. The van der waals constant a and b becomes insignificant and the van der waals equation becomes the ideal gas equation.