A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is `T_i` (in Kelvin) and the final temperature is a `T_i`, the value of a is

Explanation

To solve the problem, we will use the adiabatic condition for an ideal gas, which relates temperature and volume. The steps are as follows:

### Step-by-Step Solution:

1. **Understand the Adiabatic Process**: For an adiabatic process involving an ideal gas, the relationship between temperature (T) and volume (V) is given by:
\[ TV^{\gamma - 1} = \text{constant} \]
where \(\gamma\) is the heat capacity ratio.

2. **Identify the Values**: Given that the gas is diatomic, we know:
\[ \gamma = \frac{C_p}{C_v} = \frac{7}{5} = 1.4 \]
The initial temperature is \(T_i\) and the initial volume is \(V_1\). The final volume \(V_2\) is given as:
\[ V_2 = \frac{1}{32} V_1 \]
The final temperature is given as \(T_f = a T_i\).

3. **Apply the Adiabatic Condition**: Using the initial and final states in the adiabatic equation:
\[ T_i V_1^{\gamma - 1} = T_f V_2^{\gamma - 1} \]
Substituting the known values:
\[ T_i V_1^{\gamma - 1} = (a T_i) \left(\frac{1}{32} V_1\right)^{\gamma - 1} \]

4. **Simplify the Equation**: Cancel \(T_i\) from both sides (assuming \(T_i \neq 0\)):
\[ V_1^{\gamma - 1} = a \left(\frac{1}{32} V_1\right)^{\gamma - 1} \]

5. **Rearranging the Equation**: This can be rewritten as:
\[ V_1^{\gamma - 1} = a \cdot \frac{V_1^{\gamma - 1}}{32^{\gamma - 1}} \]
Dividing both sides by \(V_1^{\gamma - 1}\) (assuming \(V_1 \neq 0\)):
\[ 1 = \frac{a}{32^{\gamma - 1}} \]

6. **Solve for \(a\)**: Rearranging gives:
\[ a = 32^{\gamma - 1} \]

7. **Substituting \(\gamma\)**: Substitute \(\gamma = \frac{7}{5}\):
\[ a = 32^{\frac{7}{5} - 1} = 32^{\frac{2}{5}} \]

8. **Calculate \(32^{\frac{2}{5}}\)**:
\[ 32 = 2^5 \implies 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4 \]

### Final Answer:
Thus, the value of \(a\) is:
\[ \boxed{4} \]