7. In the tetrahedron $O-ABC$, it is known that
$$
OA=OB=OC=\sqrt{3}, AC=2\sqrt{2}, AB=2,
$$

and $OB \perp AC$. A sphere with radius 1 is constructed with $O$ as the center. Then the volume of the part of the tetrahedron $O-ABC$ that is not inside the sphere is $\qquad$
7. $\frac{2}{3}-\frac{\pi}{9}$.

Consider a cube with edge length 2. Place point $O$ at the center of the cube, and points $A$, $B$, and $C$ at three vertices of the cube, all on the same face.
Therefore, the part where the sphere intersects with the tetrahedron is $\frac{1}{12}$ of the sphere.
Thus, the required volume is $\frac{2^{3}}{12}-\frac{1}{12} \times \frac{4 \pi}{3}=\frac{2}{3}-\frac{\pi}{9}$.