Let's check it:
First, we have this formula
\(\frac{P_o*V_o}{\eta_o*T_o}=\frac{P_f*V_f}{\eta_f*T_f}\)
The pressures are the same so we can cancel, but don't leave and don't enter any mass so, let's check what stay.
\(\frac{V_o}{T_o}=\frac{V_f}{T_f}\)
We know that the final temperature is 2 times the initial temperature
so we can write
\(T_f=2T_o\)
Replacing
\(\frac{V_o}{T_o}=\frac{V_f}{2T_o}\)
\(V_o=\frac{V_f}{2}\)
So
\(\boxed{\boxed{\Huge{V_f=2V_o}}}\)
Therefore the volume is doubled.
