Question

The Van der waals equation of ‘n’ moles of a real gas is $\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = nRT$
Where P is the pressure, V is the volume, T is absolute temperature, R is molar gas constant and a, b, c are Van der waal constants. The dimensional formula for ab is:

\[
  (a){\text{ }}\left[ {M{L^8}{T^{ - 2}}} \right] \\
  (b){\text{ }}\left[ {M{L^6}{T^{ - 2}}} \right] \\
  (c){\text{ }}\left[ {M{L^4}{T^{ - 2}}} \right] \\
  (d){\text{ }}\left[ {M{L^2}{T^{ - 2}}} \right] \\
 \]

Answer 1

Hint: In this question use the concept that the dimension of $\left( {P + \dfrac{a}{{{V^2}}}} \right)$ is same as the dimension of

$\left[ P \right] = \left[ {\dfrac{a}{{{V^2}}}} \right]$ that is dimension of P and dimension of $\dfrac{a}{{{V^2}}}$. Use the dimension of pressure that is $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$ and dimension of volume that is$\left[ {{L^3}} \right]$. This will help approach the solution.

Complete step-by-step answer:

Now given equation is

$\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = nRT$

Where, P = pressure, V = volume, T = temperature, R = molar gas constant, n = number of moles in a gas, and a, b and c are the van der wall constants.

So according to above explanation the dimension of $\left( {P + \dfrac{a}{{{V^2}}}} \right)$ is same as the dimension of

$\left[ P \right] = \left[ {\dfrac{a}{{{V^2}}}} \right] = \left( {P + \dfrac{a}{{{V^2}}}} \right)$........... (1)

Now as we know that the dimension of pressure is $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
And the dimension of volume is $\left[ {{L^3}} \right]$

Now substitute this values in equation (1) we have,

$ \Rightarrow \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] = \dfrac{{\left[ a \right]}}{{\left[ {{{\left( {{L^3}} \right)}^2}} \right]}} = \dfrac{{\left[ a \right]}}{{\left[ {{L^6}} \right]}}$

Now simplify this we have,

$ \Rightarrow \left[ a \right] = \left[ {M{L^{ - 1 + 6}}{T^{ - 2}}} \right] = \left[ {M{L^5}{T^{ - 2}}} \right]$................... (2)

Similarly the dimension of $\left( {V - b} \right)$ is same as the dimension of

$\left[ V \right] = \left[ b \right] = \left( {V - b} \right)$........... (3)

As we know the dimension of volume is $\left[ {{L^3}} \right]$ so from equation (3) we have,
$ \Rightarrow \left[ b \right] = \left[ {{L^3}} \right]$.............................. (4)

Now multiply equation (2) and (4) we have,

$ \Rightarrow \left[ a \right]\left[ b \right] = \left[ {M{L^5}{T^{ - 2}}} \right]\left[ {{L^3}} \right]$

So the dimension of ab is

$ \Rightarrow \left[ {ab} \right] = \left[ {M{L^{5 + 3}}{T^{ - 2}}} \right] = \left[ {M{L^8}{T^{ - 2}}} \right]$

So this is the required dimension of the ab.

Hence option (A) is the correct answer.

Additional Information:
As we know that all physical quantities can be expressed in terms of seven fundamental base quantities such as mass, length, time , temperature, electric current, luminous intensity and amount of substance.

These seven quantities are called seven dimensions of the physical world.

We can use symbols instead of the names of the base quantities and they are represented in square brackets.

[M] for mass, [L] for length, [T] for time, [K] for temperature, [I] for current, [cd] for luminous intensity and [mol] for the amount of substance.

These will specify the nature of the unit and not its magnitude.

Uses of dimensional analysis:

$\left( 1 \right)$ It will be used to check the consistency of a dimensional equation.
$\left( 2 \right)$ It will be used to derive the relation between physical quantities in physical phenomena.
$\left( 3 \right)$ It will be used to change units from one system to another.

Limitations of dimensional analysis:

The very most important limitation of dimensional analysis is if the given formula carries the sum of two quantities and if only one of those quantities is provided dimensional analysis cannot be used to verify if the formula is correct or not.

For example:

$v = u$ m/sec and $v = at$ m/sec,

Both are dimensionally correct but the formula is not correct.

As the correct formula is $v = u + at$ m/sec (first law of motion).

So in dimension analysis if the equation has two variables which are added or subtracted then the dimension of both the variables must be the same otherwise the equation is not correct.

Note: Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M,L and T.