Answer
$\bar{c}_p = \bar{c}_v+R_u$, see proof in the step-by-step solution.
Work Step by Step
From the definition of enthalpy:
$dH=dU+pdV$
Dividing by the number of mols in the system:
$d\bar{h} = d\bar{u} + pd\bar{v}$
For an ideal gas:
$PV=nR_uT \rightarrow \bar{v}=\frac{R_uT}{P}$
$d\bar{h} = d\bar{u} + p.d(\frac{R_uT}{P})$
$\bar{c}_pdT = \bar{c}_vdT+R_udT$
$\bar{c}_p = \bar{c}_v+R_u$
