Question

Find out the unit and dimensions of the constants $a$ and $b$ in the van der waal's equation $\left[ {P + \dfrac{a}{{{V^2}}}} \right](V - b) = RT$, where $P$ is pressure, $V$ is volume, $R$ is gas constant, and $T$ is temperature.

Answer 1

For determining dimensions of given quantities, we can, firstly, find the equivalent dimensions in both left and right sides by evaluating dimensions of every term. And then equating both sides, precisely, comparing powers of each term of equivalent dimensions on both sides.

Complete step by step answer:
We know that,
Dimension of $P$, pressure, is $[{M^1}{L^{ - 1}}{T^{ - 2}}]$
Dimension of $V$, volume, is $[{L^3}]$
Where $M$ is mass
$L$ is length
$T$ is time
Dimension and unit of $\dfrac{a}{{{V^2}}}$ will be same as $P$ and
Dimension and unit of $V$ will be same as $b$
So, we get,
$\dfrac{{{\text{Dimension of a}}}}{{{{[{L^3}]}^2}}} = [{M^1}{L^{ - 1}}{T^{ - 2}}]$
On simplifying, we get,
${\text{Dimension of a = [}}{{\text{M}}^1}{{\text{L}}^{ - 1}}{{\text{T}}^{ - 2}}{\text{][}}{{\text{L}}^3}{{\text{]}}^2}$
On further evaluating, we get,
${\text{Dimension of a = [}}{{\text{M}}^1}{{\text{L}}^5}{{\text{T}}^{ - 2}}{\text{]}}$
Now,
$\dfrac{{{\text{Unit of a}}}}{{{{\left( {{\text{Unit of volume}}} \right)}^2}}} = {\text{Unit of Pressure}}$
On solving, we get,
$\dfrac{{{\text{Unit of a}}}}{{{{\left( {{m^3}} \right)}^2}}} = N{m^{ - 2}}$
Where, $N$ is Newton and $m$ is metre
So, we get,
${\text{Unit of a = }}N{m^4}$
Now,
${\text{Dimension of b = Dimension of Volume}}$
So we get,
${\text{Dimension of b = [}}{{\text{L}}^3}{\text{]}}$
${\text{Unit of b = Unit of volume}}$
So, unit of b will be,
${\text{Unit of b = }}{m^3}$

Note: We used the property, dimension and unit of $\dfrac{a}{{{V^2}}}$ is same as $P$ in van der wall’s equation because we can add or subtract two terms if and only if they have same dimensions and same units. Same holds for $V$ and $b$.