Adiabatic Compression of Diatomic Ideal Gas Temperature Ratio

For an adiabatic process involving an ideal gas, the relationship between temperature $T$ and volume $V$ is given by:
$T V^{\gamma - 1} = \text{constant}$

where $\gamma = \frac{C_p}{C_v}$ is the adiabatic index of the gas.

For a rigid diatomic ideal gas, the number of degrees of freedom is $f = 5$. Therefore, the adiabatic index $\gamma$ is:
$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}$

Thus, the power exponent becomes:
$\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}$

Let $T_i$ and $V_i$ be the initial temperature and initial volume of the gas, respectively. The final volume is $V_f = \frac{1}{32}V_i$ and the final temperature is $T_f = a T_i$.

Applying the adiabatic relation between the initial and final states:
$T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1}$

Substitute $T_f = a T_i$, $V_f = \frac{V_i}{32}$, and $\gamma - 1 = \frac{2}{5}$:
$T_i V_i^{2/5} = (a T_i) \left(\frac{V_i}{32}\right)^{2/5}$

Dividing both sides by $T_i V_i^{2/5}$:
$1 = a \left(\frac{1}{32}\right)^{2/5}$

Solving for $a$:
$a = 32^{2/5}$

Since $32 = 2^5$:
$a = (2^5)^{2/5} = 2^{5 \times \frac{2}{5}} = 2^2 = 4$

Hence, the value of $a$ is $4$.