2. In $\triangle A B C$, it is known that $A B=2, B C=4$, $\angle A B C=120^{\circ}$, and a point $P$ outside the plane satisfies $P A=P B$ $=P C=4$. Then the volume of the tetrahedron $P-A B C$ is equal to
2. $\frac{4 \sqrt{5}}{3}$.

In $\triangle A B C$, by the cosine rule, we get $A C=2 \sqrt{7}$.
Let the circumradius of $\triangle A B C$ be $R$. Then, by the sine rule, we have
$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{2 \sqrt{21}}{3} .
$$

Draw $P O \perp$ plane $A B C$, with the foot of the perpendicular being $O$.
Since $P A=P B=P C$, we know $O A=O B=O C$, which means $O$ is the circumcenter of $\triangle A B C$, thus
$$
\begin{array}{l}
P O=\sqrt{P A^{2}-R^{2}}=\frac{2 \sqrt{15}}{3} . \\
\text { Therefore, } V_{\text {pyr } P-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot P O=\frac{4 \sqrt{5}}{3} .
\end{array}
$$