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Q. The unit $JPa^{-1}$ is equivalent to

Some Basic Concepts of Chemistry

Solution:

$JPa^{1};$ Unit of work is Joule and unit of pressure is Pascal. Dimension of Joule, i.e., work $= F \times = MLT^{-2}\times L= \left[ML^{2}T^{-2}\right]$
$\frac{1}{Pa}=\frac{1}{Pressure}=\frac{1}{F/A}=\frac{1\times A}{F}=\left[M^{-1}LT^{-2}\right]$
So, $JPa^{-1}=\left[L^{2}\times L\right] =\left[L^{3}\right]$