5. Given that the five vertices of a regular quadrilateral pyramid $P-ABCD$ lie on the same sphere. If the volume of the pyramid is $V$, then the minimum value of the surface area of the sphere is $\qquad$

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5. $\frac{9 \pi \sqrt[3]{9 V^{2}}}{4}$.

Let $h, a, R$ be the height, the side length of the base, and the radius of the circumscribed sphere of a regular quadrilateral pyramid, respectively, then
$$
V=\frac{1}{3} a^{2} h \Rightarrow a^{2}=\frac{3 V}{h} .
$$

By the perpendicular diameter theorem, we know $\left(\frac{a}{\sqrt{2}}\right)^{2}=h(2 R-h)$. Therefore,
$$
2 R=\frac{a^{2}+2 h^{2}}{2 h}=\frac{3 V}{2 h^{2}}+h=\frac{3 V}{2 h^{2}}+\frac{h}{2}+\frac{h}{2} \geqslant \frac{3 \sqrt[3]{3 V}}{2} .
$$

Equality holds if and only if $h=\sqrt[3]{3 V}$. Thus, the surface area of the sphere
$$
S=4 \pi R^{2} \geqslant \frac{9 \pi \sqrt[3]{9 V^{2}}}{4} .
$$