The van der Waals' equation for a gas is $$ \left(P+\frac{a}{V^2}\right)(V-b)=n R T $$ where $P, V, R, T$…

The van der Waals' equation for a gas is $$ \left(P+\frac{a}{V^2}\right)(V-b)=n R T $$ where $P, V, R, T$ and $n$ represent the pressure, volume, universal gas constant, absolute temperature and number of moles of a gas, respectively. $a$ and $b$ are constants. The ratio $\frac{b}{a}$ will have the following dimensional formula.
  1. $\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]$
  2. $\left[\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^{-1}\right]$
  3. $\left[\mathrm{ML}^2 \mathrm{~T}^2\right]$
  4. $\left[\mathrm{MLT}^{-2}\right]$

Solution

van der Waals gas equation is $ \left(p+\frac{a}{V^2}\right)(V-b)=n R T $ From principle of homogeneity Dimension of $\frac{a}{V^2}=$ dimension of $P$ $\therefore$ Dimension of $a=\left[V^2\right] \times[P]$ $ \begin{aligned} {[a] } & =\left[\mathrm{L}^3\right]^2\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\ & =\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right] \end{aligned} $ Dimension of $b=$ dimension of $V$ $ \begin{aligned} & {[b]=\left[\mathrm{L}^3\right]} \\ & \therefore \quad \frac{b}{a}=\frac{L^3}{\mathrm{ML}^5 \mathrm{~T}^{-2}}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right] \\ & \end{aligned} $

Asked in: AP EAMCET 2002

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