
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

According to the first law of thermodynamics:

\[ Q=\Delta u + A \]

where \( Q \) is the amount of heat, \( \Delta u \) is the change in internal energy, and \( A \) is the work done by the gas. In our case,

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

where \( x \) is the displacement of the piston, \( k \) is the spring constant, and \( T \) is the temperature of the gas after expansion. Let \( P \) be the pressure of the gas after expansion and \( S \) be the cross-sectional area. Then \( k x = P S \), and the change in the volume of gas is \( \Delta V = S x \). Since the volume increased \( n \) times, after expansion the volume is \( V = \frac{n}{n-1} S x \). Using the equation of state for 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 expressions for the change in internal energy and the work into the first law of thermodynamics, we obtain:

\[ c_{v}(T - T_{0}) = -\frac{n-1}{n} \frac{R T}{2} \Rightarrow T = \frac{T_{0}}{1 + \frac{(n-1)R}{2 n c_{v}}} \]

In the initial and expanded states, the equation of state for the gas can be written as:

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

Dividing the second equation by the first, we get:

\[ \frac{P}{P_{0}} = \frac{T}{n T_{0}} = \frac{1}{n\left(1 + \frac{(n-1) R}{2 n c_{v}}\right)} \]

Present the answer in LaTex format: \boxed{Your answer}
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