Q181 / 200 · TS EAMCET 2002 · Physics
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.
- Correct answerA$\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]$
- B$\left[\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^{-1}\right]$
- C$\left[\mathrm{ML}^2 \mathrm{~T}^2\right]$
- D$\left[\mathrm{MLT}^{-2}\right]$
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 $V$
$$
\begin{aligned}
& {[b]=\left[\mathrm{L}^3\right]} \\
& \therefore \quad \frac{b}{a}=\frac{L^3}{\mathrm{ML}^5 \mathrm{~T}^{-2}}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right] \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'A'.