7. Given that the center $O$ of a sphere with radius $\frac{\sqrt{22}}{2}$ is the center of a regular tetrahedron $\Gamma$, and the total length of the curves formed by the intersection of the sphere's surface with the four faces of the tetrahedron $\Gamma$ is $8 \pi$. Then the volume of the tetrahedron $\Gamma$ is $\qquad$ .
$7.18 \sqrt{2}$ or $54 \sqrt{6}$.
Let the edge length of the tetrahedron $\Gamma$ be $a$, and the radius of the small circle be $r$.
If a face of the tetrahedron $\Gamma$ intersects the sphere as shown in Figure 3, then the circumference of the small circle is $2 \pi$, and the radius is 1.
Therefore, the distance from the center of the sphere to the face of the tetrahedron is
$$
\begin{array}{l}
\frac{\sqrt{6} a}{12}=\sqrt{R^{2}-r^{2}}=\frac{3 \sqrt{2}}{2} \\
\Rightarrow a=6 \sqrt{3} \Rightarrow V=\frac{\sqrt{2}}{12} a^{3}=54 \sqrt{6} .
\end{array}
$$

If a face of the tetrahedron $\Gamma$ intersects the sphere as shown in Figure 4, let $D$ be the midpoint of $A C$. By the problem,
$$
\begin{array}{l}
\overparen{A B}=\frac{2 \pi}{3}, \angle A O_{1} B=\frac{2 \pi}{3 r} . \\
\text { Also } \angle B O_{1} C=\frac{\pi}{3}, \angle A O_{1} D=\frac{2 \pi}{3}-\frac{\pi}{3 r}, \\
O_{1} D=\frac{\sqrt{3}}{6} a=r \cos \angle A O_{1} D . \\
\text { By } \frac{\sqrt{6} a}{12}=\sqrt{R^{2}-r^{2}}=\sqrt{\frac{11}{2}-r^{2}} \\
\Rightarrow O_{1} D=\frac{\sqrt{3}}{6} a=\sqrt{11-2 r^{2}} \\
\Rightarrow \cos \left(\frac{2 \pi}{3}-\frac{\pi}{3 r}\right)=\sqrt{\frac{11}{r^{2}}-2} \\
\Rightarrow r=2, a=6, V=\frac{\sqrt{2}}{12} a^{3}=18 \sqrt{2} .
\end{array}
$$

In summary, the volume of the tetrahedron $\Gamma$ is $18 \sqrt{2}$ or $54 \sqrt{6}$.