The pressure `P_(1)and density d_(1)` of a diatomic gas `(gamma=(7)/(5))` change to `P_(2) and d_(2)` during an adiabatic operation .If `(d_(2))/(d_(1))=32, then (P_(2))/(P_(1))`is

Options

A. a. 76
B. b. 128
C. c. 168
D. d. 298

Explanation

To solve the problem, we need to find the ratio \( \frac{P_2}{P_1} \) given that the density ratio \( \frac{d_2}{d_1} = 32 \) and the gas is diatomic with \( \gamma = \frac{7}{5} \).

### Step-by-step Solution:

1. **Understand the Adiabatic Process:**
For an adiabatic process, the relationship between pressure \( P \) and density \( d \) can be expressed as:
\[ P \cdot V^\gamma = \text{constant} \]
where \( \gamma \) is the heat capacity ratio.

2. **Relate Volume and Density:**
Since volume \( V \) is related to density \( d \) by the equation \( V = \frac{m}{d} \) (where \( m \) is the mass, which is constant), we can rewrite the adiabatic condition in terms of density:
\[ P \cdot \left(\frac{m}{d}\right)^\gamma = \text{constant} \]
This leads to:
\[ P \cdot d^{-\gamma} = \text{constant} \]
or
\[ P \propto d^\gamma \]

3. **Set Up the Ratios:**
From the proportionality, we can write:
\[ \frac{P_1}{P_2} = \left(\frac{d_1}{d_2}\right)^\gamma \]

4. **Substitute the Given Density Ratio:**
We know \( \frac{d_2}{d_1} = 32 \), which means:
\[ \frac{d_1}{d_2} = \frac{1}{32} \]
Therefore, substituting this into the equation gives:
\[ \frac{P_1}{P_2} = \left(\frac{1}{32}\right)^{\frac{7}{5}} \]

5. **Calculate the Power:**
Now, calculate \( \left(\frac{1}{32}\right)^{\frac{7}{5}} \):
\[ \frac{P_1}{P_2} = \frac{1}{32^{\frac{7}{5}}} \]
Since \( 32 = 2^5 \), we can rewrite this as:
\[ 32^{\frac{7}{5}} = (2^5)^{\frac{7}{5}} = 2^7 \]
Thus:
\[ \frac{P_1}{P_2} = \frac{1}{2^7} \]

6. **Reciprocate to Find \( \frac{P_2}{P_1} \):**
To find \( \frac{P_2}{P_1} \), we take the reciprocal:
\[ \frac{P_2}{P_1} = 2^7 = 128 \]

### Final Answer:
\[ \frac{P_2}{P_1} = 128 \]