13. (2000 National High School Competition Question) A sphere is tangent to all six edges of a regular tetrahedron. If the edge length of the tetrahedron is $a$, then the volume of the sphere is $\qquad$ .
13. Fill $\frac{\sqrt{2}}{24} \pi a^{3}$. Reason: Let the regular tetrahedron be $A B C D$, the center of the sphere be $O$, the radius be $r$, the volume be $V$, the center of face $B C D$ be $O_{1}$, and the midpoint of edge $B C$ be $E$. Then, $A O_{1}=\sqrt{a^{2}-O_{1} B^{2}}=\sqrt{a^{2}-\frac{1}{3} a^{2}}=\frac{\sqrt{6}}{3} a$. From $O B^{2}=O_{1} O^{2}+O_{1} B^{2}=\left(O_{1} A-O B\right)^{2}+$ $O_{1} B^{2}$, we get $\frac{2}{3} a^{2}-\frac{2 \sqrt{6}}{3} a \cdot O B+\frac{1}{3} a^{2}=0$, hence $O B=\frac{3 a}{2 \sqrt{6}}=\frac{\sqrt{6}}{4} a$, thus $r=O E=\sqrt{O B^{2}-B E^{2}}=$ $\sqrt{\frac{3}{8} a^{2}-\frac{1}{4} a^{2}}=\frac{1}{2 \sqrt{2}} a$. Therefore, $V=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \cdot \frac{1}{16 \sqrt{2}} a^{3}=\frac{\sqrt{2}}{24} \pi a^{3}$.