8. In a regular tetrahedron $ABCD$ with edge length $\sqrt{2}$, it is known that $\overrightarrow{A P}=\frac{1}{2} \overrightarrow{A B}, \overrightarrow{A Q}=\frac{1}{3} \overrightarrow{A C}, \overrightarrow{A R}=\frac{1}{4} \overrightarrow{A D}$. If point $K$ is the centroid of $\triangle BCD$, then the volume of the tetrahedron $K P Q R$ is $\qquad$.
$$
\begin{array}{l}
\text { As shown in the figure, let } \overrightarrow{A F}=\lambda \overrightarrow{A E}=\frac{\lambda}{2} \overrightarrow{A C}+\frac{\lambda}{2} \overrightarrow{A D} \\
=\frac{3 \lambda}{2} \overrightarrow{A Q}+2 \lambda \overrightarrow{A R} \Rightarrow \frac{3 \lambda}{2}+2 \lambda=1 \Rightarrow \lambda=\frac{2}{7} . \\
\text { Let } \overrightarrow{A G}=\mu \overrightarrow{A K}=\frac{\mu}{3} \overrightarrow{A B}+\frac{2 \mu}{3} \overrightarrow{A E} \\
=\frac{2 \mu}{3} \overrightarrow{A P}+\frac{7 \mu}{3} \overrightarrow{A F}, \text { then } \frac{2 \mu}{3}+\frac{7 \mu}{3}=1 \\
\Rightarrow \mu=\frac{1}{3} . \text { Therefore, } G K=2 A G \\
\Rightarrow V_{K P Q R}=2 V_{A P Q R}=2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} V_{A B C D} \\
=\frac{1}{12} \cdot \frac{\sqrt{2}}{12} \cdot(\sqrt{2})^{3}=\frac{1}{36} .
\end{array}
$$