In the equation $\left(\mathrm{P}+\frac{a}{\mathrm{~V}^2}\right)(\mathrm{V}-b)=\mathrm{RT}$, where P is…

In the equation $\left(\mathrm{P}+\frac{a}{\mathrm{~V}^2}\right)(\mathrm{V}-b)=\mathrm{RT}$, where P is pressure, V is volume, T is temperature, R is universal gas constant, $a$ and $b$ are constants. The dimensions of $a$ are
  1. $\mathrm{ML}^{-1} \mathrm{~T}^{-2}$
  2. $\mathrm{ML}^5 \mathrm{~T}^{-2}$
  3. $\mathrm{M}^0 \mathrm{~L}^3 \mathrm{~T}^0$
  4. $\mathrm{ML}^3 \mathrm{~T}^{-2}$

Solution

$\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$ By principle of homogeneity, $\begin{aligned} & {[\mathrm{p}]=\left[\frac{\mathrm{a}}{\mathrm{v}^2}\right]} \\ & \Rightarrow[\mathrm{a}]=[\mathrm{pV}][\mathrm{V}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\left[\mathrm{L}^3\right]=\left[\mathrm{M} \mathrm{~L}^5 \mathrm{~T}^{-2}\right] \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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