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 of finding the value of \( a \) for a diatomic ideal gas compressed adiabatically to \( \frac{1}{32} \) of its initial volume, we can follow these steps:

### Step 1: Understand the relationship for adiabatic processes
For an adiabatic process involving an ideal gas, the relationship between temperature and volume can be expressed as:
\[ T V^{\gamma - 1} = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio, which for a diatomic gas is \( \frac{7}{5} \).

### Step 2: Set up the initial and final conditions
Let:
- \( T_i \) = initial temperature
- \( V_i \) = initial volume
- \( T_f \) = final temperature
- \( V_f = \frac{1}{32} V_i \) = final volume

Using the adiabatic condition, we can write:
\[ T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1} \]

### Step 3: Substitute the known values
Substituting \( V_f = \frac{1}{32} V_i \) into the equation gives:
\[ T_i V_i^{\frac{7}{5} - 1} = T_f \left(\frac{1}{32} V_i\right)^{\frac{7}{5} - 1} \]

### Step 4: Simplify the equation
This simplifies to:
\[ T_i V_i^{\frac{2}{5}} = T_f \left(\frac{1}{32^{\frac{2}{5}}} V_i^{\frac{2}{5}}\right) \]

### Step 5: Cancel \( V_i^{\frac{2}{5}} \) from both sides
Cancelling \( V_i^{\frac{2}{5}} \) from both sides gives:
\[ T_i = T_f \cdot \frac{1}{32^{\frac{2}{5}}} \]

### Step 6: Solve for \( T_f \)
Rearranging gives:
\[ T_f = T_i \cdot 32^{\frac{2}{5}} \]

### Step 7: Express \( a \)
From the relationship \( T_f = a T_i \), we can equate:
\[ a = 32^{\frac{2}{5}} \]

### Step 8: Calculate \( a \)
Calculating \( 32^{\frac{2}{5}} \):
\[ 32 = 2^5 \implies 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4 \]

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