Answer

Verified

21.6k+ views

**Hint:**By considering the other terms as the constant than the $\left( {P + \dfrac{a}{{{V^2}}}} \right)$. By using this term only, the dimension of the $a$ can be determined. By keeping the $a$ in one side and the other terms in the other side, the dimension of $a$ can be determined.

**Complete step by step solution**

Given that,

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.

By considering the term,

$\left( {P + \dfrac{a}{{{V^2}}}} \right) = 0$

By rearranging the terms, then the above equation is written as,

$\left| P \right| = \left| {\dfrac{a}{{{V^2}}}} \right|$

By keeping the term $a$ in one side and the other terms in other side, then the above equation is written as,

$a = P \times {V^2}\,................\left( 1 \right)$

Now, the dimensional formula of each terms is,

The dimension of the pressure is given as,

$P = \dfrac{F}{A} = \dfrac{{ma}}{A}$

The unit of the above equation is written as,

$P = \dfrac{{kgm{s^{ - 2}}}}{{{m^2}}}$

By substituting the dimension in the above equation, then

$P = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}}$

The dimensional formula of the volume is given by,

$V = {V^2}$

The unit of the above equation is written as,

$V = {\left( {{m^3}} \right)^2}$

Then the above equation is written as,

$V = {m^6}$

By substituting the dimension in the above equation, then

$V = {L^6}$

By substituting the dimensional formula in the equation (1), then the equation is written as,

$a = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} \times {L^6}$

By rearranging the terms, then the above equation is written as,

$a = ML{T^{ - 2}} \times {L^{ - 2}} \times {L^6}$

On further simplification of the power, then

$a = M{L^5}{T^{ - 2}}$

**Hence, the option (A) is the correct answer.**

**Note:**Here the dimension of $a$ is asked, so that the term $\left( {P + \dfrac{a}{{{V^2}}}} \right)$ is taken. If the dimension of the $b$ is asked, then this term $\left( {V - b} \right)$ is taken and the solution is done like we discussed the step by step to determine the dimension formula in the above solution.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main

Find the points of intersection of the tangents at class 11 maths JEE_Main

For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

Other Pages

The electronic configuration of chloride ion is A 2 class 11 chemistry JEE_Main

When the diameter of the conductor is doubled then class 12 physics JEE_Main

Equipotential at a great distance from a collection class 12 physics JEE_Main

A household uses the following electric appliances class 12 physics JEE_Main

Define 1 Ohm resistance A student has a resistance class 12 physics JEE_Main

A metallic rod PQ of length l is rotated with an angular class 12 physics JEE_Main