7. In the tetrahedron $A-B C D$, it is known that the lateral edges $A B$, $A C$, and $A D$ are pairwise perpendicular, and the areas of $\triangle A B C$, $\triangle A C D$, and $\triangle A D B$ are $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$, and $\frac{\sqrt{6}}{2}$, respectively. Then the volume of the circumscribed sphere of the tetrahedron $A-B C D$ is $\qquad$
7. $\sqrt{6} \pi$.

Let $A B=a, A C=b, A D=c$. Then
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} a b=\frac{\sqrt{2}}{2}, \\
S_{\triangle A C D}=\frac{1}{2} b c=\frac{\sqrt{3}}{2}, \\
S_{\triangle D B}=\frac{1}{2} c a=\frac{\sqrt{6}}{2} .
\end{array}
$$

Multiplying and simplifying the above three equations yields $a b c=\sqrt{6}$.
Thus, $a=\sqrt{2}, b=1, c=\sqrt{3}$.
Therefore, the radius of the circumscribed sphere of the tetrahedron $A-B C D$ is
$$
r=\frac{1}{2} \sqrt{a^{2}+b^{2}+c^{2}}=\frac{\sqrt{6}}{2} \text {, }
$$

and its volume is
$$
\frac{4}{3} \pi r^{3}=\sqrt{6} \pi \text {. }
$$