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# The equation of state of a real gas is given by $\left( {{\bf{P}} - \dfrac{{\bf{a}}}{{{{\bf{V}}^2}}}} \right)({\bf{V}} - {\bf{b}}) = {\bf{RT}}$, where $P, V$ and $T$ are pressure, volume and temperature respectively and R is the universal gas constant. The dimensions of the constant a in the given equation is _______.A. ${\rm{M}}{{\rm{L}}^6}\;{{\rm{T}}^{ - 2}}$B. ${\rm{M}}{{\rm{L}}^5}\;{{\rm{T}}^{ - 2}}$C. ${\rm{M}}{{\rm{L}}^7}\;{{\rm{T}}^{ - 2}}$D. ${\rm{M}}{{\rm{L}}^8}\;{{\rm{T}}^{ - 2}}$ Verified
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Hint: The dimensional formula is an equation that expresses the relationship between fundamental and derived units in terms of dimensions (equation). In mechanics, the three basic dimensions of length, mass, and time are denoted by the letters $L, M$ and $T$ respectively.

The van der Waals equation is a state equation that accounts for two properties of real gases: gas particle excluded volume and gas molecule attractive forces. The Van der Waals equation is a state equation in chemistry and thermodynamics that generalises the ideal gas law based on possible explanations why particular gases do not behave ideally. The van der Waals constants are the constants a and b. The intermolecular forces are compensated for by the constant a.

The space filled by the gas particles is compensated for by constant b. Its magnitude is the amount of one mole of atoms or molecules, and it is a corrective for finite molecular size. It is thought that there is no force between molecules in the kinetic principle of gases, but there is actually intermolecular attraction, which lowers the strain. As a result of intermolecular attraction, the correction for P occurs.

Since the molecules have a finite size, the whole volume V of the container is unavailable to them. The kinetic theory statement that molecules are point masses with no noticeable volume is false. As a result of the size of molecules, a correction for V is needed.
Dimension of $P$ = Dimension of $\left( {\dfrac{{\rm{a}}}{{{{\rm{V}}^2}}}} \right)$
Dimension of $P$ = ${\rm{M}}{{\rm{L}}^{ - 1}}\;{{\rm{T}}^{ - 2}}$
Dimension of $V$ = ${{\bf{L}}^6}$
$\Rightarrow {\rm{M}}{{\rm{L}}^{ - 1}}\;{{\rm{T}}^{ - 2}} = \dfrac{{{\rm{ Dimension of a }}}}{{{{\bf{L}}^6}}}$
$\therefore {\rm{ Dimension of }}{\bf{a}} = {\bf{M}}{{\bf{L}}^5}{{\bf{T}}^{ - 2}}$

Hence, option B is correct.

Note: The Van der Waals equation is mathematically basic, but it predicts critical behaviour and the experimentally demonstrated transformation between vapour and liquid.The Joule–Thomson effect (temperature change during adiabatic expansion), which is impossible in an ideal gas, is also sufficiently predicted and explained.