Chemistry: Molecular Approach (4th Edition)

\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}
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.