9. According to the first law of thermodynamics

$$
Q=\Delta u+A
$$

where $Q$ - the amount of heat, $\Delta u$ - the change in internal energy, $A$ - the work done by the gas. In our case

$$
Q=0, \quad \Delta u=c_{v}\left(T-T_{0}\right), \quad A=\frac{k x^{2}}{2}
$$

where $x$ - the displacement of the piston, $k$ - the stiffness of the spring, $T$ - the temperature of the gas after expansion. Let $P$ - the pressure of the gas after expansion, $S$ - the cross-sectional area. Then $k x=P S$, and the change in volume of the gas is $\Delta V=S x$. Since the volume increased by $n$ times, after expansion it became $V=\frac{n}{n-1} S x$. Using the equation of state of one mole of an ideal gas $P V=R T$, we get

$$
P \frac{n}{n-1} S x=R T, \quad k x^{2}=\frac{n-1}{n} R T, \Rightarrow A=\frac{k x^{2}}{2}=\frac{n-1}{n} \frac{R T}{2}
$$

Substituting the change in internal energy and work into the expression of the first law of thermodynamics, we get

$$
c_{v}\left(T-T_{0}\right)=-\frac{n-1}{n} \frac{R T}{2} \Rightarrow T=\frac{T_{0}}{1+\frac{(n-1) R}{2 n c_{v}}}
$$

In the initial and expanded positions, the equation of state of the gas can be represented in the form

$$
P_{0} V_{0}=R T_{0}, \quad P n V_{0}=R T
$$

Dividing the second by the first, we get:

$$
\frac{P}{P_{0}}=\frac{T}{n T_{0}}=\frac{1}{n\left(1+\frac{(n-1) R}{2 n c_{v}}\right)}
$$
Answer: $P_{0} \frac{1}{n\left(1+\frac{(n-1) R}{2 n c_{v}}\right)}$.