What is the dimensional formula of $\mathrm{ab}^{-1}$ in the equation $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^{2}}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$, where letters have their usual meaning.

(B) $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$

(A) $\left[\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]$

(C) $\left[\mathrm{M}^{-1} \mathrm{~L}^{5} \mathrm{~T}^{3}\right]$

(D) $\left[\mathrm{M}^{6} \mathrm{~L}^{7} \mathrm{~T}^{4}\right]$

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Units & Measurement

What is the dimensional formula of

ab^{- 1}

in the equation

(P + \frac{a}{V ^{2}}) (V - b) = RT

, where letters have their usual meaning.

Answer: (B)

(P + \frac{a}{V ^{2}}) (V - b) = R T

, the term

(V - b)

must be dimensionally consistent with

V

, so

[b] = [V] = L^{3}

.
The term

a / V^{2}

must carry the dimension of pressure:

[a] = [P] [V^{2}] = (M L^{- 1} T^{- 2}) (L^{6}) = M L^{5} T^{- 2}

Hence

[a b^{- 1}] = \frac{[ a ]}{[ b ]} = \frac{M L ^{5} T ^{- 2}}{L ^{3}} = M L^{2} T^{- 2}

which is the dimension of energy.

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option