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Step-by-Step Solution
Step 1: Identify the key principle (Adiabatic Process)
For an adiabatic process involving an ideal gas, the relation between pressure and volume is given by
$PV^\gamma = \text{constant}$,
where $\gamma = \frac{C_p}{C_v}$ (the ratio of specific heats).
Step 2: Express pressure in terms of density
Since $V = \frac{m}{\rho}$ for a given mass $m$, we can rewrite the equation as
$P \left(\frac{m}{\rho}\right)^\gamma = \text{constant}$.
Because $m$ (the mass of the gas) is constant, we get
$P \propto \rho^\gamma.$
Hence,
$\frac{P_f}{P_i} = \left(\frac{\rho_f}{\rho_i}\right)^\gamma.$
Step 3: Substitute the given values
We are told that the density of a diatomic gas becomes 32 times its initial value, i.e.
$\rho_f = 32 \, \rho_i.$
For a diatomic gas,
$\gamma = \frac{7}{5}.$
Substituting into the proportionality relation gives
\[
\frac{P_f}{P_i} = \left(\frac{\rho_f}{\rho_i}\right)^\gamma = 32^{\frac{7}{5}}.
\]
Step 4: Simplify the expression
Note that $32 = 2^5.$ Hence,
\[
32^{\frac{7}{5}} = \left(2^5\right)^{\frac{7}{5}} = 2^{5 \times \frac{7}{5}} = 2^7 = 128.
\]
Therefore,
\[
\frac{P_f}{P_i} = 128 \quad \implies P_f = 128 \, P_i.
\]
Step 5: State the final answer
The final pressure is 128 times the initial pressure. Hence,
$n = 128.
