Question

The Van der Waals equation of state for some gases can be expressed as:
$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$
Where P is the pressure, V is the molar volume, and T is the absolute temperature of the given sample of gas and a, b and R are constants. The dimension of constant b are
(a). $\left[ M{{L}^{5}}{{T}^{-2}} \right]$
(b). $\left[ M{{L}^{-1}}{{T}^{-2}} \right]$
(c). $\left[ {{L}^{3}} \right]$
(d). $\left[ {{L}^{6}} \right]$

Answer 1

- Hint: To find the answer of this question we need to apply the concept of dimensional correctness of an equation. We can add or subtract two quantities only if the units of the two quantities are the same i.e. dimension of the quantities need to be the same. Applying this concept to the above equation we can find the dimension of the constant b.

Complete step-by-step solution -
For an equation to be correct, the equation should be dimensionally correct. i.e. the quantities on both sides of the equation should have the same dimension or the quantities related with each other by addition of subtraction should have the same dimension.
In the given equation we have in the left-hand side, we have that the constant b is related to V with subtraction. For this to be right the dimension of b should be the same as that of V.
Now, here, V is the volume of the gas. So, the dimension of constant b is equal to that of the dimension of volume.
Dimension of volume is given as $\left[ {{L}^{3}} \right]$
So, dimension of constant b will be $\left[ {{L}^{3}} \right]$

The correct option is (C).

Note: We can also find the dimension of applying this concept.
Here, the dimension of $\dfrac{a}{{{V}^{2}}}$ will be the same as the dimension of P.
Now, dimension of P is $\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
Dimension of V is $\left[ {{L}^{3}} \right]$
So, dimension of ${{V}^{2}}$ will be $\left[ {{L}^{6}} \right]$
So, dimension of a will be, $\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]\times \left[ {{L}^{6}} \right]=\left[ {{M}^{1}}{{L}^{5}}{{T}^{-2}} \right]$
We also find the dimension by simplifying the given equation and then equating the dimension of each quantity related with addition and subtraction in both sides of the equation.