Question

The unit of van der Waals constant ‘a’ is:
A.KP a
B. KP a ${\rm{d}}{{\rm{m}}^{\rm{3}}}$
C. KP a ${\rm{d}}{{\rm{m}}^{\rm{3}}}$${\rm{mo}}{{\rm{l}}^{ - 1}}$
D. KP a ${\rm{d}}{{\rm{m}}^6}$${\rm{mo}}{{\rm{l}}^{ - 2}}$

Answer 1

Hint: We know that van der Waals equation is a modification of ideal gas law that is applied to the real gases. This equation corrects the two defects, that is, molecular volume and molecular attraction.

Complete step by step answer:
Let’s understand the van der Waals equation in detail. First, we discuss the correction of molecular volume.
We know that volume of the gas is the sum of volumes of the molecules and also of the empty spaces or voids among them. We also know that the molecules are incompressible and it is only the volume of the empty spaces which increases or decreases during the expansion or contraction respectively. van der Waal was of the opinion that as the molecular volumes remain unchanged during expansion or contraction of gases, it must be excluded from the total volume of the gas, so the corrected volume of the gas is $V - nb$ (for n mole of gas). And b is the constant for molecular volume.
Now, we discuss the correction of molecular attraction. We know that attractive forces are always present among the gas molecules and they can be neglected only under certain conditions. Previously this attraction was ignored which resulted in a defect. So, the corrected pressure is
$\left( {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right)$ ,where a is constant for molecular attraction.
Therefore, the van der Waals equation is,
$\left[ {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right]\left[ {V - b} \right] = RT$
Where, P is pressure, V is volume, R is gas constant and T is temperature, b is the constant for molecular volume and a is constant for molecular attraction and n is number of moles of gas.
Now, come to the question. We have to choose the correct unit of a.
$P = \dfrac{{{n^2}a}}{{{V^2}}}$
$ \Rightarrow a = \dfrac{{P{V^2}}}{{{n^2}}} = \dfrac{{atm\,li{t^2}}}{{mo{l^2}}}$
$ \Rightarrow a = {\rm{atm}}\,{\rm{li}}{{\rm{t}}^2}\,{\rm{mo}}{{\rm{l}}^{ - 2}} = {\rm{atm}}\,{\rm{d}}{{\rm{m}}^6}\,{\rm{mo}}{{\rm{l}}^{ - 2}}$ (1 lit=${\rm{d}}{{\rm{m}}^{\rm{3}}}$)

Therefore, the correct answer is D.

Note: Always remember that van der Waals gas equation is followed by real gas and ideal gas follows ideal gas equation (PV=nRT). At low pressure and ordinary temperature, the volume of the gas is very large and constant ‘b’ can be neglected.