Question

The ratio of excluded volume (b) to molar volume of a gas molecule is:
A.1
B.2
C.3
D.4

Answer 1

Hint: The constant ‘b’ is the van der Waals constant for correction in volume. It is also known as the excluded volume or co-volume.
The numerical value of ‘b’ is found to be four times the actual volume occupied by the gas molecules.

Complete step by step answer:
The van der Waals equation of state has been formulated by incorporating two corrections in the ideal gas equation ${\text{PV = nRT}}$ by taking into account the molecular volume and the intermolecular forces. It is the modified gas equation for real gases. Mathematically, it is written as:
$\left( {{\text{P + }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{a}}}}{{{{\text{V}}^{\text{2}}}}}} \right)\left( {{\text{V - nb}}} \right) = {\text{nRT}}$
Here, P represents the pressure of the gas in atmospheres, V represents the volume of the gas in litres and n represents the number of moles of the gas. ‘a’ and ‘b’ are van der Waals constants for correction in pressure and volume respectively.
Also, the term $\dfrac{{{{\text{n}}^{\text{2}}}{\text{a}}}}{{{{\text{V}}^{\text{2}}}}}$ denotes the pressure correction and is also called the cohesive pressure. The term nb denotes the volume correction and is also called the co-volume.
From hint, it can be said that the numerical value of the excluded volume or co-volume ‘b’ can be written as:
${\text{b}} = 4 \times \dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}}$
Here, ‘r’ is the radius of each gas molecule.
Since gas molecules are spherical, therefore the molar volume of a gas molecule is equal to the volume of a sphere of radius r.
Thus, molar volume of a gas molecule $ = \dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}}$ .
So, the required ratio of excluded volume (b) to molar volume of a gas molecule is $ = \dfrac{{\text{b}}}{{\dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}}}}$ .
Substitute the value of b and we get:
$\dfrac{{4 \times \dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}}}}{{\dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}}}} = 4$

So, the required ratio is 4 and so the correct option is D.

Note:
The significance of ‘a’ is that it is related to the intermolecular attractive forces. Greater the value of ‘a’, stronger are the intermolecular attractive forces. Gases which have greater value of ‘a’ can be liquefied more easily and vice – versa.
‘b’ denotes the compressible volume per mole of gas. Hence, units of ‘b’ must be the units of volume i.e., ${{\text{m}}^{\text{3}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ or ${\text{d}}{{\text{m}}^{\text{3}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ or ${\text{litmo}}{{\text{l}}^{{\text{ - 1}}}}$ .