## Essential University Physics: Volume 1 (4th Edition)

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 $\fbox{79 percent}$.