7. For the tetrahedron $ABCD$, $AB \perp BC$, $CD \perp BC$, $BC=2$, and the angle between the skew lines $AB$ and $CD$ is $60^{\circ}$. If the radius of the circumscribed sphere of the tetrahedron $ABCD$ is $\sqrt{5}$, then the maximum volume of the tetrahedron $ABCD$ is $\qquad$ .
$7.2 \sqrt{3}$.
Consider the right triangular prism $A B D_{1}-A_{1} C D, B C=2$, $\angle A B D_{1}=60^{\circ}$. Then the tetrahedron $A B C D$ is a tetrahedron that satisfies the given conditions, and its circumscribed sphere is the same as that of the triangular prism $A B D_{1}-A_{1} C D$.

Let the circumradius of the base $\triangle A B D_{1}$ of the triangular prism be $r$. Then
$$
r^{2}+\left(\frac{B C}{2}\right)^{2}=5 \Rightarrow r=2 .
$$

In $\triangle A B D_{1}$, by the Law of Sines we have
$$
\frac{A D_{1}}{\sin \angle A B D_{1}}=2 r \Rightarrow A D_{1}=2 \sqrt{3} \text {; }
$$

By the Law of Cosines we get
$$
\begin{array}{l}
A D_{1}^{2}=A B^{2}+B D_{1}^{2}-2 A B \cdot B D_{1} \cos \angle A B D_{1} \\
\Rightarrow A B^{2}+B D_{1}^{2}-A B \cdot B D_{1}=12 .
\end{array}
$$

By the AM-GM inequality we have
$A B \cdot B D_{1} \leqslant 12$.
$$
\begin{array}{l}
\text { Hence } V_{\text {tetrahedron } A B C D}=\frac{1}{3} V_{\text {prism } A B D_{1}-A_{1} C D} \\
=\frac{1}{3} \times \frac{1}{2} A B \cdot B D_{1} \sin \angle A B D_{1} \cdot B C \\
\leqslant 2 \sqrt{3},
\end{array}
$$

The equality holds when the triangular prism $A B D_{1}-A_{1} C D$ is a regular triangular prism.
Therefore, the maximum volume of the tetrahedron $A B C D$ is $2 \sqrt{3}$.