7. Given that the surface area of tetrahedron $ABCD$ is $1, AB=AD, BD=\frac{\sqrt{2}}{2}$, and $\angle ABD + \angle DBC + \angle CBA = \angle ADB + \angle BDC + \angle CDA = 180^{\circ}, \angle ACB + \angle ACD + \angle BCD = 90^{\circ}$. Then the volume of tetrahedron $ABCD$ is . $\qquad$
7. $\frac{1}{24}$.

As shown in Figure 4, unfold the tetrahedron $ABCD$ on the plane $BCD$. From $\angle ABD + \angle DBC + \angle CBA = 180^{\circ}$, we know that points $A_1, B, A_2$ are collinear. Similarly, points $A_1, D, A_3$ are collinear, and $\angle A_2CA_3 = 90^{\circ}$.

Since $AB = AD$, we have $A_1A_2 = A_1A_3$, and $B, D$ are the midpoints of $A_1A_2$ and $A_1A_3$ respectively, and $A_2C = A_3C$, $2BD = A_2A_3 = \sqrt{2}$. Let the distance from point $A_1$ to $A_2A_3$ be $2a$. Then the surface area of the tetrahedron $S = S_{\triangle A_1A_2A_3} + S_{\triangle CA_2A_3} = \sqrt{2}a + \frac{1}{2} = 1$. Solving for $a$, we get $a = \frac{\sqrt{2}}{4}$.

Thus, $AC = A_2C = 1$, $AB = AD = \frac{1}{2}$. It is easy to see that $\angle BA_1D = \angle BA_2C = \angle DA_3C = 90^{\circ}$, so $AC \perp$ plane $ABD$. Therefore, $V_{C-ABD} = \frac{1}{3} AC \cdot S_{\triangle ABD} = \frac{1}{24}$.