4. As shown in Figure 1, it is known that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with a side length of 3, and $P, Q, R$ are points on the edges $A B, A D, A A_{1}$ respectively, with $A P=A Q=A R=1$. Then the volume of the tetrahedron $C_{1} P Q R$ is
4. $\frac{4}{3}$.

Obviously, $A C_{1} \perp$ plane $P Q R$.
Since $A P=A Q=A R=1$, we have,
$$
\begin{array}{l}
P Q=Q R=R P=\sqrt{2} . \\
\text { Also, } A C_{1}=3 \sqrt{3}, \\
V_{A-P Q R}=\frac{1}{3} \times \frac{1}{2} \times 1^{2} \times 1=\frac{1}{6}, \\
V_{C_{1}-P Q R} \\
=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times(\sqrt{2})^{2} \times 3 \sqrt{3}-V_{A-P Q R} \\
=\frac{4}{3} .
\end{array}
$$