JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 17)
$$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$$ 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:
Energy density
Bulk modulus
Modulus of rigidity
Compressibility
Explanation
$[a]=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]$
$$ \begin{aligned} & {[b]=\left[\mathrm{L}^{3}\right] } \\\\ & {\left[\frac{b^{2}}{a}\right]=\left[\frac{\mathrm{L}^{6}}{\mathrm{ML}^{5} \mathrm{~T}^{2}}\right] }=\left[\mathrm{M}^{-1} \mathrm{LT}^{-2}\right] \\\\ &=[\text { Compressibility] } \end{aligned} $$
$$ \begin{aligned} & {[b]=\left[\mathrm{L}^{3}\right] } \\\\ & {\left[\frac{b^{2}}{a}\right]=\left[\frac{\mathrm{L}^{6}}{\mathrm{ML}^{5} \mathrm{~T}^{2}}\right] }=\left[\mathrm{M}^{-1} \mathrm{LT}^{-2}\right] \\\\ &=[\text { Compressibility] } \end{aligned} $$
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