The relation obeyed by a perfect gas during an adiabatic process is...

For an adiabatic process, the relation between pressure and volume is given by $pV^{\gamma} = \text{constant}$.

Given the relation is $pV^{3/2} = \text{constant}$, we have $\gamma = \dfrac{3}{2}$.

Using the ideal gas equation $pV = nRT$, the relation between temperature and volume for an adiabatic process is $TV^{\gamma - 1} = \text{constant}$.

Therefore, $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.

Substituting the given values $T_1 = T$, $V_1 = V$, $V_2 = \dfrac{V}{2}$, and $\gamma = \dfrac{3}{2}$:

$T(V)^{\dfrac{3}{2} - 1} = T_2 \left(\dfrac{V}{2}\right)^{\dfrac{3}{2} - 1}$

$T(V)^{\dfrac{1}{2}} = T_2 \left(\dfrac{V}{2}\right)^{\dfrac{1}{2}}$

$T_2 = T \left(\dfrac{V}{V/2}\right)^{\dfrac{1}{2}} = T(2)^{\dfrac{1}{2}} = \sqrt{2}T$

Answer: $\sqrt{2}\,\text{T}$