Is there an actual specific heat for aluminum in its gaseous state?
By &quot;bring aluminum from its liquid state to a gaseous state&quot;, I assume you mean to a gaseous state at a temperature above aluminum's boiling point—since to merely convert it from a liquid to a gas, you only need the enthalpy of vaporization.
According to https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node18.html,
most metallic vapors have molar heat capacities very close to those of ideal gases, i.e., $C_v = 3/2 R = 12.47$ J/(mol K) and $C_p = 5/2 R = 20.79$ J/(mol K).  Thus you can obtain aluminum's specific heat capacity by dividing by the atomic mass, giving $C_v = 0.46$ J/(g K)  and $C_p = 0.77$ J/(g K)
The reason for this is that metals don't form metallic bonds in the gaseous state (metallic bonding is a collective phenomenon that requires bulk metal), and are thus monatomic*.  And the heat capacities of monatomic gases are simple, since they have only translational degrees of freedom.
[*As real gases, they may not be exactly monatomic, since they can still transiently associate via van der Waals forces.  But treating them as monatomic is a good approximation.]
The exception to the above would occur when the temperature is sufficiently high to cause the metal to ionize, in which case you would have a higher heat capacity because you would now have two (or more) particles for every ionized aluminum atom. But since aluminum's first ionization energy is 5.985 eV/atom = $577$ kJ/mol, and $RT$ at $3000$ K is $25$ kJ/mol, I don't expect there to be significant ionization close to its $2792$ K boiling point. [Also, some might argue that, once it starts ionizing, it is no longer a gas, but a gas-plasma mixture.]