Units & Dimensions

JEE Main level, single correct, medium

57 of 149

Question 57 of 149

What is the dimensional formula of ab−1a b^{-1} in the equation (P+aV2)(V−b)=RT\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}, where letters have their usual meaning.

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Answer: (C) [ML2 T−2]\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

∵[V]=[b]\because[\mathrm{V}]=[\mathrm{b}]
∴\therefore Dimension of b=[L3]b=\left[\mathrm{L}^3\right]
$\begin{aligned}
& \&[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \\
& {[\mathrm{a}]=\left[\mathrm{PV}^2\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^6\right]}
\end{aligned}$
Dimension of a=[ML5 T−2]\mathrm{a}=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]
∴ab−1=[ML5 T−2][L3]=[ML2 T−2]\therefore \mathrm{ab}^{-1}=\frac{\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

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