Answer
79 percent
Work Step by Step
This is an adiabatic process, so we know that $PV^{\gamma}$ is constant. Thus, it follows:
$PV^{\gamma}=\frac{1}{3}P(2V)^{\gamma}$
$3 = 2^{\gamma}$
$\gamma = log_2(3)=1.585$
We know that $O_2$ has 5 degrees of freedom, so $C_v=\frac{5}{2}$. We know that $Ar$ has 3 degrees of freedom, so $C_v=\frac{3}{2}$.
Thus, we find:
$1.585=\frac{(-x+3.5)}{(-x+2.5)}$
$x=.79$
Thus, the percentage of the mixture that is Argon is $\fbox{79 percent}$.
