Answer
\[\begin{align}
& 1\text{ J}=1\text{ kg }{{\text{m}}^{\text{2}}}\text{/}{{\text{s}}^{\text{2}}} \\
& \text{m}\,\,\text{=}\,\,\text{kg,}\,\,{{\text{v}}^{\text{2}}}\,=\,{{\left( \text{m/s} \right)}^{2}},\text{thus,mv}\,\,\text{=}\,\,\text{kg}\,{{\text{m}}^{\text{2}}}\,{{\text{s}}^{\text{2}}} \\
& \text{P}\,\,\text{=}\,\,\text{N/}{{\text{m}}^{2}}\,=\,\,\text{kg}\,\,\text{m/}{{\text{s}}^{\text{2}}}\text{/}{{\text{m}}^{\text{2}}}\,=\,\,\text{kg/m}\ {{\text{s}}^{{}}} \\
& \text{V}\,\,\text{=}\,\,{{\text{m}}^{3}}\,,\,\text{PV}\,\,=\,\text{kg}\,\,{{\text{m}}^{3}}/\text{m}{{\text{s}}^{2}}\,\,=\,\,\text{kg}\,{{\text{m}}^{\text{2}}}\text{/}{{\text{s}}^{\text{2}}} \\
\end{align}\]
Work Step by Step
The SI unit of energy is Joule (J).
\[1\text{ J}=1\text{ kg }{{\text{m}}^{\text{2}}}\text{/}{{\text{s}}^{\text{2}}}\]
Now, \[E=\frac{1}{2}m{{v}^{2}}\].
The SI unit of mass is kilogram and velocity is \[\text{m/s}\]. Thus,
\[\begin{align}
& \frac{1}{2}m{{v}^{2}}=\text{kg }{{\left( \text{m/s} \right)}^{2}} \\
& =\text{kg }{{\text{m}}^{\text{2}}}\text{/}{{\text{s}}^{\text{2}}}
\end{align}\]
The SI unit of pressure is calculated as follows:
\[\begin{align}
& P=\frac{F}{A} \\
& =\frac{ma}{A}
\end{align}\]
The SI unit of mass is kg, acceleration is \[\text{m/}{{\text{s}}^{\text{2}}}\], and area is \[{{\text{m}}^{2}}\]. Thus,
\[\begin{align}
& \frac{ma}{A}=\frac{\text{kg}\left( \text{m/}{{\text{s}}^{2}} \right)}{{{\text{m}}^{2}}} \\
& =\text{kg/(m }{{\text{s}}^{2}})
\end{align}\]
Similarly, \[E=\frac{3}{2}PV\].
The SI unit of pressure is \[\text{kg/(m}\ {{\text{s}}^{\text{2}}})\] and volume is \[{{\text{m}}^{3}}\]. Thus,
\[\begin{align}
& \frac{3}{2}PV=\left( \text{kg/(m}\ {{\text{s}}^{2}}) \right)\left( {{\text{m}}^{\text{3}}} \right) \\
& =\text{kg }{{\text{m}}^{\text{2}}}\text{/}{{\text{s}}^{\text{2}}}
\end{align}\]
The derived SI units of both these terms are those of energy.
