Question

The unit of van der Waals constant ‘a’ in SI units is:
A. $Pa{m^5}mo{l^{ - 2}}$
B.$Pa{m^6}mo{l^{ - 1}}$
C. $Pa{m^6}mo{l^{ - 3}}$
D. $Pa{m^6}mo{l^{ - 2}}$

Answer 1

The van der Waals equation is a better fit for behaviour of real gas than ideal gas equation.

Formula used: The expression for van der Waals equation for n mole of gas is given as:
$(P + \dfrac{{a{n^2}}}{{{V^2}}})(V - nb) = nRT$
Where P is pressure of gas, V is volume of gas, R is gas constant, T is temperature and n is the number of moles of gases.

Complete step by step answer:
According to n mole of gas van der Waals equation is given as:
$(P + \dfrac{{a{n^2}}}{{{V^2}}})(V - nb) = nRT$
Where a and b are the van der Waals constant.
Unit of a can be calculated as:
Pressure defect for n moles of a gas is given as:
$p = \dfrac{{a{n^2}}}{{{V^2}}} \\\ \Rightarrow a = \dfrac{{p \times {V^2}}}{{{n^2}}} \\\ \Rightarrow a = \dfrac{{pressure \times {{(volume)}^2}}}{{{{(mole)}^2}}} \\\$
So the unit of a is related to the dimensions of pressure and volume.
If pressure is expressed in pascals and volume in ${m^3}$then the unit of a will be:
$a = \dfrac{{Pa \times {{({m^3})}^2}}}{{{{(mol)}^2}}} \\\ \Rightarrow a = Pa{\text{ }}{m^6}{\text{ }}mo{l^{ - 2}} \\\$
Hence the correct option is answer D.
And if the pressure is in atmosphere and volume is in $d{m^3}$then the unit will be:
$a = \dfrac{{atm \times {{(d{m^3})}^2}}}{{{{(mol)}^2}}} \\\ \Rightarrow a = atm{\text{ d}}{m^6}{\text{ }}mo{l^{ - 2}} \\\$

So, the correct answer is “Option D”.

Note: The van der Waals constant ‘a’ and ‘b’ are characteristic properties of a particular gas. The constant ‘a’ corrects the force of attraction between gas particles and larger the value of ‘a’ stronger is the force of attraction between particles. The value of ‘a’ also relates to the boiling point of a gas and the gasses such as ${H_2}$ and He needs to be cooled to absolute zero to be condensed to liquid form. The constant ‘b’ is also known as co-volume.