In an adiabatic process, the density of a diatomic gas becomes $$32n$$ times its initial value. The final pressure of the gas is found to be $$n$$ times the initial pressure. The value of $$n$$ is:

Let us denote the initial state of the di-atomic gas by the subscript 1 and the final state after the adiabatic change by the subscript 2.

So, the initial pressure and density are $$P_1$$ and $$\rho_1$$, while the final pressure and density are $$P_2$$ and $$\rho_2$$ respectively.

According to the statement of the problem, the density becomes $$32$$ times its initial value. Therefore

$$\rho_2 \;=\; 32\,\rho_1.$$

We are told that the final pressure is $$n$$ times the initial pressure. Hence

$$P_2 \;=\; n\,P_1.$$

Because the change is adiabatic, we must use the adiabatic relation that connects pressure and volume. First we recall the formula

$$P\,V^{\gamma} \;=\; \text{constant}$$

where $$\gamma$$ is the ratio of specific heats $$C_P/C_V$$. For a di-atomic ideal gas (with no vibrational modes excited in the temperature range considered) we have

$$\gamma \;=\;\frac{7}{5}.$$

To involve the density, we note that density is inversely proportional to volume for a fixed mass of gas. In symbols, if $$m$$ is the mass,

$$\rho \;=\;\frac{m}{V}\quad\Longrightarrow\quad V \;=\;\frac{m}{\rho} \;\propto\; \frac{1}{\rho}.$$

Substituting $$V = \dfrac{1}{\rho}$$ (ignoring the constant mass factor which cancels out when we form ratios) into the adiabatic relation, we obtain

$$P\Bigl(\frac{1}{\rho}\Bigr)^{\gamma} \;=\; \text{constant}.$$

Multiplying both sides by $$\rho^{\gamma}$$ gives

$$P \;\rho^{-\gamma}\;=\;\text{constant}\quad\Longrightarrow\quad P\;\rho^{-\gamma} = \text{constant}.$$ Therefore pressure and density are connected by

$$P \;\propto\; \rho^{\gamma}.$$

Using this proportionality for the two states, we write

$$\frac{P_2}{P_1} \;=\;\Bigl(\frac{\rho_2}{\rho_1}\Bigr)^{\gamma}.$$

Now we substitute the given multiples:

$$\frac{P_2}{P_1} \;=\; n,$$

and

$$\frac{\rho_2}{\rho_1} \;=\; 32.$$

Hence

$$n \;=\; 32^{\gamma}.$$

Putting $$\gamma = \dfrac{7}{5}$$, we have

$$n \;=\; 32^{\,\tfrac{7}{5}}.$$

We rewrite 32 as a power of 2 for convenience:

$$32 \;=\; 2^{5}.$$

Therefore

$$n \;=\; (2^{5})^{\tfrac{7}{5}}.$$

Using the law of exponents $$(a^{b})^{c} = a^{\,bc},$$ we obtain

$$n \;=\; 2^{\,5 \times \tfrac{7}{5}} \;=\; 2^{7}.$$

Now $$2^{7} = 128.$$

So we have found

$$n \;=\; 128.$$

Hence, the correct answer is Option 3.