Question

What is the significance of Van Der Waals constants? Out of ${{\text{N}}_{\text{2}}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ which one will have:
i)larger value of a
ii)larger value of b

Answer 1

Hint: Van der Waals constants a and b are positive values and are characteristics of the individual gas. If a gas behaves ideally, both a and b are zero. The constant a provides correction for the intermolecular forces. Constant b adjusts for the volume occupied by the gas particles.

Complete step by step answer:
The ideal gas law treats the molecules of a gas as point particles with perfectly elastic collisions. This works for dilute gases. But as the concentration of gases increases, they begin to deviate from ideality. To account for the deviation of gas, the Vander Waal constants a and b are used in the ideal gas equation.
For option i, a depends on the intermolecular force of attraction between gaseous molecules. As we know that, ${\text{N}}{{\text{H}}_{\text{3}}}$ is a polar molecule and can form hydrogen bonds with each other. Thus, the intermolecular forces between ${\text{N}}{{\text{H}}_{\text{3}}}$ molecule will be greater than ${{\text{N}}_{\text{2}}}$ which has only weak Vander Waal forces. Thus, ${\text{N}}{{\text{H}}_{\text{3}}}$ has greater value of a.
For option ii, b depends on the volume occupied by the gas molecules. In the given gases, ${\text{N}}{{\text{H}}_{\text{3}}}$ has a molar mass of ${\text{17 g mo}}{{\text{l}}^{{\text{ - 1}}}}$ while ${{\text{N}}_{\text{2}}}$ has a molar mass of ${\text{28 g mo}}{{\text{l}}^{{\text{ - 1}}}}$ . So, clearly, ${{\text{N}}_{\text{2}}}$ is a bigger molecule than ${\text{N}}{{\text{H}}_{\text{3}}}$ despite having less atoms. Thus, the value of b is larger for ${{\text{N}}_{\text{2}}}$.
$\therefore $ ${\text{N}}{{\text{H}}_{\text{3}}}$ has a higher value of a because of hydrogen bonding while ${{\text{N}}_{\text{2}}}$ has a higher value of b due to its large size.

Note:
In case of real gas, the ideal gas equation gets converted to:
${\text{[P + a(}}\dfrac{{\text{n}}}{{\text{V}}}{{\text{)}}^{\text{2}}}{\text{](}}\dfrac{{\text{V}}}{{\text{n}}}{\text{ - b) = RT}}$
where P is the pressure, V is the volume, n is the moles, R is a constant ${\text{(8}}{\text{.314 J }}{{\text{K}}^{{\text{ - 1}}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{)}}$ and T is the temperature. For ideal gases, the values of a and b are zero.