Specific volume of an air parcel So I have an unsaturated air parcel in the atmosphere that ascends adiabatically from the $1000\:\mathrm{mbar}$ level to the $700\:\mathrm{mbar}$ level with its initial temperature being $10\:°\mathrm{C}$.

I have to find the initial specific volume in $\mathrm{m^3}\:\mathrm{kg^{-1}}$.

I know that the specific volume formula is $v=1/\rho$ and that the equation of the state of an ideal gas can be written as $p=\rho RT$ where $R$ is $8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}}$

So I have $1000\:\mathrm{mbar}=\rho(8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}})(283.15\:\mathrm{K})$ and solve then for $\rho$ and then get $0.42477$ and plug that for $v=1/0.42477$ and get $2.354\:\mathrm{m^3}\:\mathrm{kg^{-1}}$ as the initial specific volume. Am I doing this right?
The ideal gas law states
$$ pV = nRT \Leftrightarrow \frac{n}{V} = \frac{p}{RT}\ .$$

For the density $\rho = m/V$ and the mass $m = M\cdot n$ with the molar mass $M$, we get a new formula for the density:
$$ \rho = M\frac{n}{V} = M\frac{p}{RT}$$

With the definition of the specific volume we get
$$ v = \frac{RT}{Mp}\ . $$

So unless you know the molar mass of your parcel of air, you will not get out a numerical value for the specific volume.