An ideal diatomic gas is taken through an adiabatic process in which density increases to 32 times. If pressure increases to 'n' times. Find n

Options

A. 4
B. 8
C. 64
D. 128

Answer

To solve the problem, we need to analyze the adiabatic process of an ideal diatomic gas, where the density increases to 32 times its initial value. We will find the factor by which the pressure increases, denoted as 'n'.

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process**:
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure (P), volume (V), and density (ρ) can be described by the equation:
\[ P V^\gamma = \text{constant} \]
where \(\gamma\) (gamma) is the adiabatic index.

2. **Relating Density and Volume**:
Density is defined as:
\[ \rho = \frac{m}{V} \]
Therefore, we can express volume in terms of density:
\[ V = \frac{m}{\rho} \]

3. **Substituting Volume in the Adiabatic Equation**:
Substitute \(V\) in the adiabatic equation:
\[ P \left(\frac{m}{\rho}\right)^\gamma = \text{constant} \]
This can be rearranged to show the relationship between pressure and density:
\[ P \propto \rho^\gamma \]
Thus, we can write:
\[ \frac{P_f}{P_i} = \left(\frac{\rho_f}{\rho_i}\right)^\gamma \]

4. **Given Information**:
The problem states that the final density (\(\rho_f\)) is 32 times the initial density (\(\rho_i\)):
\[ \frac{\rho_f}{\rho_i} = 32 \]

5. **Finding the Value of Gamma for Diatomic Gas**:
For a diatomic gas, the value of \(\gamma\) is:
\[ \gamma = \frac{C_p}{C_v} = \frac{7}{5} \]

6. **Substituting Values into the Pressure Equation**:
Now substitute the values into the pressure equation:
\[ \frac{P_f}{P_i} = \left(32\right)^{\frac{7}{5}} \]
We can express 32 as \(2^5\):
\[ \frac{P_f}{P_i} = \left(2^5\right)^{\frac{7}{5}} = 2^{7} = 128 \]

7. **Conclusion**:
Therefore, the pressure increases to \(128\) times the initial pressure:
\[ n = 128 \]

### Final Answer:
The value of \(n\) is \(128\).