10. Given that line segment $A B$ is the diameter of sphere $O$ with radius 2, points $C, D$ are on the surface of sphere $O$, $C D=2, A B \perp C D$, $45^{\circ} \leqslant \angle A O C \leqslant 135^{\circ}$, then the range of the volume of tetrahedron $A B C D$ is $\qquad$ .
Given $\angle A O C=\theta$, construct $C G \perp A G$, and connect $D G$,
then $A B \perp$ plane $C D G \Rightarrow \triangle A B C \cong \triangle A B D$.
Thus, $C G=D G=2 \sin \theta, C D=2$
$\Rightarrow \sin \angle G C D=\frac{\sqrt{4 \sin ^{2} \theta-1}}{2 \sin \theta}, D F=\frac{\sqrt{4 \sin ^{2} \theta-1}}{\sin \theta}$,
hence $V_{A B C D}=\frac{1}{3} S_{\triangle A B C} \cdot D F$
$=\frac{1}{3} \cdot 2 \cdot 2 \sin \theta \cdot \frac{\sqrt{4 \sin ^{2} \theta-1}}{\sin \theta}=\frac{4}{3} \sqrt{4 \sin ^{2} \theta-1}$
$\in\left[\frac{4}{3}, \frac{4 \sqrt{3}}{3}\right]$. Therefore, the volume range of the tetrahedron $A B C D$ is $\left[\frac{4}{3}, \frac{4 \sqrt{3}}{3}\right]$.