An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes 32 times of its initial value. If the final pressure of gas is 128 atmospheres, the value of '$$\gamma$$' of the gas is :

We are given an ideal gas initially at atmospheric pressure (1 atm) that undergoes adiabatic compression. After compression, the density becomes 32 times its initial value, and the final pressure is 128 atm. We need to find the value of $$\gamma$$, the adiabatic index (ratio of specific heats).

For an adiabatic process in an ideal gas, the relationship between pressure and volume is $$P V^{\gamma} = \text{constant}$$. Since density $$\rho$$ is mass per unit volume and mass is constant, volume $$V$$ is inversely proportional to density, i.e., $$V \propto \frac{1}{\rho}$$. Substituting this into the adiabatic relation gives:

$$ P \left( \frac{1}{\rho} \right)^{\gamma} = \text{constant} $$

This simplifies to:

$$ P \rho^{-\gamma} = \text{constant} $$

Therefore, for the initial state (1) and final state (2):

$$ P_1 \rho_1^{-\gamma} = P_2 \rho_2^{-\gamma} $$

Rearranging this equation:

$$ \frac{P_1}{P_2} = \left( \frac{\rho_2}{\rho_1} \right)^{\gamma} $$

Given that the final density $$\rho_2$$ is 32 times the initial density $$\rho_1$$, we have $$\rho_2 = 32 \rho_1$$, so:

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

Substituting the known pressures: initial pressure $$P_1 = 1$$ atm and final pressure $$P_2 = 128$$ atm, we get:

$$ \frac{1}{128} = (32)^{\gamma} $$

Note that $$(32)^{\gamma}$$ is equivalent to $$32^{\gamma}$$. We can rewrite the equation as:

$$ 32^{\gamma} = 128 $$

Now, express both sides as powers of 2. Since $$32 = 2^5$$ and $$128 = 2^7$$, substitute:

$$ (2^5)^{\gamma} = 2^7 $$

Using the exponent rule $$(a^m)^n = a^{m n}$$, simplify:

$$ 2^{5\gamma} = 2^7 $$

Since the bases are equal, set the exponents equal:

$$ 5\gamma = 7 $$

Solving for $$\gamma$$:

$$ \gamma = \frac{7}{5} = 1.4 $$

Hence, the value of $$\gamma$$ is 1.4. Comparing with the options: A (1.5), B (1.4), C (1.3), D (1.6), the correct answer is Option B.