$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$ represents the equation of state of some gases. Where $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $a, b, R$ are the constants. The physical quantity, which has dimensional formula as that of $\frac{b^{2}}{a}$, will be:
Solution
$[a]=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]$
<br/><br/>$$
\begin{aligned}
& {[b]=\left[\mathrm{L}^{3}\right] } \\\\
& {\left[\frac{b^{2}}{a}\right]=\left[\frac{\mathrm{L}^{6}}{\mathrm{ML}^{5} \mathrm{~T}^{2}}\right] }=\left[\mathrm{M}^{-1} \mathrm{LT}^{-2}\right] \\\\
&=[\text { Compressibility] }
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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