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 `aT_i`, the value of a is

Explanation

To solve the problem, we will use the adiabatic process relationship for an ideal gas. The relationship states that for an adiabatic process, the product of the temperature and the volume raised to the power of (gamma - 1) is constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature \( T_i \)
- Final temperature \( T_f = a T_i \)
- Initial volume \( V_i \)
- Final volume \( V_f = \frac{1}{32} V_i \)
- For a diatomic ideal gas, \( \gamma = \frac{7}{5} \)

2. **Use the Adiabatic Relation:**
The adiabatic condition can be expressed as:
\[ T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1} \]

3. **Substituting Known Values:**
Substitute \( T_f \) and \( V_f \) into the equation:
\[ T_i V_i^{\gamma - 1} = (a T_i) \left(\frac{1}{32} V_i\right)^{\gamma - 1} \]

4. **Cancel \( T_i \) from Both Sides:**
Since \( T_i \) is common on both sides, we can cancel it:
\[ V_i^{\gamma - 1} = a \left(\frac{1}{32} V_i\right)^{\gamma - 1} \]

5. **Rearranging the Equation:**
Rearranging gives us:
\[ V_i^{\gamma - 1} = a \cdot \frac{V_i^{\gamma - 1}}{32^{\gamma - 1}} \]

6. **Dividing Both Sides by \( V_i^{\gamma - 1} \):**
Assuming \( V_i^{\gamma - 1} \neq 0 \), we can divide both sides:
\[ 1 = \frac{a}{32^{\gamma - 1}} \]

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

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

9. **Calculating \( 32^{\frac{2}{5}} \):**
We can express \( 32 \) as \( 2^5 \):
\[ a = (2^5)^{\frac{2}{5}} = 2^2 = 4 \]

10. **Final Result:**
Thus, the value of \( a \) is:
\[ a = 4 \]

### Conclusion:
The final temperature of the gas after adiabatic compression is \( 4 T_i \).