The pressure and density of a diatomic gas `(gamma=7//5)` change adiabatically from (p,d) to `(p^('),d^('))`. If `(d^('))/(d)=32`, then `(P^('))/(P)` should be

Answer

To solve the problem, we will use the relationship between pressure and density for an adiabatic process involving a diatomic gas.

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process**:
In an adiabatic process, the relationship between pressure (P) and density (d) is given by:
\[ P \cdot V^\gamma = \text{constant} \]
where \( \gamma = \frac{C_p}{C_v} \) is the heat capacity ratio. For a diatomic gas, \( \gamma = \frac{7}{5} \).

2. **Relating Pressure and Density**:
We can express the volume (V) in terms of density (d):
\[ V = \frac{m}{d} \]
where \( m \) is the mass of the gas. Substituting this into the adiabatic condition gives:
\[ P \left(\frac{m}{d}\right)^\gamma = \text{constant} \]
This leads to the relationship:
\[ P \cdot m^\gamma \cdot d^{-\gamma} = \text{constant} \]

3. **Setting Up the Ratios**:
For two states (initial and final), we can write:
\[ P_1 \cdot d_1^{-\gamma} = P_2 \cdot d_2^{-\gamma} \]
Rearranging gives:
\[ \frac{P_2}{P_1} = \left(\frac{d_2}{d_1}\right)^\gamma \]

4. **Substituting Given Values**:
We are given that:
\[ \frac{d_2}{d_1} = \frac{d'}{d} = 32 \]
and \( \gamma = \frac{7}{5} \). Therefore:
\[ \frac{P'}{P} = \left(32\right)^{\frac{7}{5}} \]

5. **Calculating the Ratio**:
Now we compute \( 32^{\frac{7}{5}} \):
\[ 32 = 2^5 \implies 32^{\frac{7}{5}} = (2^5)^{\frac{7}{5}} = 2^7 = 128 \]

6. **Final Result**:
Thus, the ratio of the final pressure to the initial pressure is:
\[ \frac{P'}{P} = 128 \]

### Conclusion:
The ratio \( \frac{P'}{P} \) is 128.