A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ 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

Step 1: Identify the given parameters.

The volume and temperature relations for the adiabatic compression are given.

$$V_f = \frac{1}{32}V_i$$ $$T_f = aT_i$$

Step 2: Determine the adiabatic index for a diatomic gas.

A diatomic ideal gas has $5$ degrees of freedom.

$$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$$

Step 3: Apply the adiabatic relation between temperature and volume.

For an adiabatic process, the temperature and volume are related by a constant.

$$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$$

Step 4: Express the ratio of temperatures.

Rearrange the relation to solve for the temperature ratio.

$$\frac{T_f}{T_i} = \left(\frac{V_i}{V_f}\right)^{\gamma-1}$$

Step 5: Calculate the value of $a$.

Substitute the given values into the ratio.

$$a = 32^{\frac{7}{5}-1}$$ $$a = 32^{2/5}$$ $$a = (2^5)^{2/5}$$ $$a = 4$$

Answer: The value of $a$ is $4$.