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

Options

A. 1.5
B. 1.4
C. 1.3
D. 1.6

Explanation

To solve the problem, we will use the relationship between pressure and density for an adiabatic process. The key relationships we will use are:

1. The relationship between pressure and density in an adiabatic process:
\[ P \propto \rho^\gamma \]
where \( P \) is pressure, \( \rho \) is density, and \( \gamma \) is the adiabatic index.

2. The initial and final states of the gas can be expressed as:
- Initial pressure \( P_1 = 1 \) atm (atmospheric pressure)
- Final pressure \( P_2 = 128 \) atm
- Initial density \( \rho_1 \)
- Final density \( \rho_2 = 32 \rho_1 \)

### Step-by-Step Solution:

1. **Write the relationship for the initial and final states:**
\[ \frac{P_1}{P_2} = \left( \frac{\rho_1}{\rho_2} \right)^\gamma \]

2. **Substitute the known values:**
- \( P_1 = 1 \) atm
- \( P_2 = 128 \) atm
- \( \rho_2 = 32 \rho_1 \)

Thus, we have:
\[ \frac{1}{128} = \left( \frac{\rho_1}{32 \rho_1} \right)^\gamma \]

3. **Simplify the density ratio:**
\[ \frac{1}{128} = \left( \frac{1}{32} \right)^\gamma \]

4. **Express \( \frac{1}{32} \) in terms of powers of 2:**
\[ 32 = 2^5 \quad \Rightarrow \quad \frac{1}{32} = 2^{-5} \]

5. **Substituting this back into the equation:**
\[ \frac{1}{128} = (2^{-5})^\gamma \]

6. **Express \( \frac{1}{128} \) in terms of powers of 2:**
\[ 128 = 2^7 \quad \Rightarrow \quad \frac{1}{128} = 2^{-7} \]

7. **Equate the powers of 2:**
\[ 2^{-7} = 2^{-5\gamma} \]

8. **Since the bases are the same, set the exponents equal:**
\[ -7 = -5\gamma \]

9. **Solve for \( \gamma \):**
\[ 5\gamma = 7 \quad \Rightarrow \quad \gamma = \frac{7}{5} = 1.4 \]

### Final Answer:
The value of \( \gamma \) for the gas is \( \frac{7}{5} \) or \( 1.4 \).