Use the adiabatic relation for an ideal gas

For an adiabatic process of an ideal gas: $T V^{\gamma-1} = \text{constant}$

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

Find $\gamma$ for a diatomic ideal gas

For a diatomic gas (neglecting vibration), the degrees of freedom are $f=5$

Therefore, $C_V = \frac{f}{2}R = \frac{5}{2}R, \qquad C_P = C_V + R = \frac{7}{2}R$

So, $\gamma = \frac{C_P}{C_V} = \frac{7/2}{5/2} = \frac{7}{5}$

Thus, $\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}$

Substitute the volume change

The gas is compressed to $\dfrac{1}{32}$ of its initial volume, so $V_f = \frac{V_i}{32}$

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

we get $T_f = T_i \left(\frac{V_i}{V_i/32}\right)^{2/5} = T_i (32)^{2/5}$

Simplify

Since $32 = 2^5$ therefore, $32^{2/5} = (2^5)^{2/5} = 2^2 = 4$

So, $T_f = 4T_i$

Comparing with $T_f = aT_i$, we get $a=4$

Comparison with stored answer

Derived answer: $4$

Stored correct answer: $4$

They match.