Answer
Please see the work below.
Work Step by Step
We know that
$P_1V_1^{\gamma}=P_2V_2^{\gamma}$
This can be rearranged as:
$(\frac{V_2}{V_1})^{\gamma}=\frac{P_1}{P_2}$
We plug in the known values to obtain:
$(\frac{1}{2})^{\gamma}=\frac{1}{2.55}$
$2^{\gamma}=2.55$
$\implies \gamma=\frac{Ln(2.55)}{Ln(2)}$
$\gamma=1.4$
