6. In tetrahedron $ABCD$, $\angle ABC = \angle BAD = 90^{\circ}$, $AB = 1$, $CD = 2$, and the angle between $AD$ and $BC$ is $30^{\circ}$. Then the volume of the circumscribed sphere of the tetrahedron is $\qquad$.
6. $\frac{13 \sqrt{13}}{6} \pi$.

As shown in Figure 3, draw a line $AE$ through $A$ parallel to $BC$, and make $AE = BC$, then quadrilateral $ABCE$ is a rectangle.
Figure 3

From $BA \perp AD, BA \perp AE$, we know that $AB \perp$ plane $ADE$, so $DE \perp CE$. Let $O$ be the center of the circumscribed sphere of tetrahedron $ABCD$. It is easy to know that point $E$ is also on the circumscribed sphere. Draw the projections of point $O$ on plane $ABCD$ and plane $ADE$, and denote them as $O'$ and $O''$ respectively. It is easy to know that points $O'$ and $O''$ are the centers of the circumcircles of rectangle $ABCD$ and $\triangle ADE$ respectively. Let the midpoint of line segment $AE$ be $F$, then $O'F \perp AE, O''F \perp AE$. Therefore,
$$
AO''=\frac{DE}{2 \sin 30^{\circ}}=\frac{\sqrt{2^{2}-1}}{2 \cdot \frac{1}{2}}=\sqrt{3}, \quad OO''=FO'=\frac{1}{2} AB=\frac{1}{2} .
$$

Thus, the radius of the circumscribed sphere is $OA=\sqrt{AO''^2 + O''O^2}=\frac{\sqrt{13}}{2}$, so the volume of the circumscribed sphere is $\frac{4}{3} \pi OA^3=$ $\frac{13 \sqrt{13}}{6} \pi$.