= Solution
For a unit <mass> of <ideal gas>, $h=u+P/\rho=u+R_aT$. Since $u$ depends only on <temperature>, $c_p=dh/dT=c_v+R_a$, giving $\boxed{R_a=c_p-c_v}$. For a dry parcel in adiabatic, quasistatic ascent, the <first law of thermodynamics> is $c_p\,dT=\rho^{-1}dP$. Environmental <hydrostatic equilibrium> gives $dP/dz=-\rho g$, so a convectively well-mixed layer has the <dry adiabatic lapse rate>
$$
\boxed{\Gamma_d=-\frac{dT}{dz}=\frac{g}{c_p}.}
$$
This assumes negligible condensation, approximately constant $g$, well-mixed composition and efficient convection; a radiative or stably stratified layer need not follow it. For mole fractions $x_N,x_C$ and molar masses $M_N,M_C$, $R_a=R/(x_NM_N+x_CM_C)$ and $c_p=[x_NC_{p,N}+x_CC_{p,C}]/(x_NM_N+x_CM_C)$. Molecular vibrational excitation can make $c_p$ depend on <temperature>; then $\Gamma_d=g/c_p(T)$. Nitrogen dominance near terrestrial temperatures gives the familiar order $10\,\mathrm{K\,km^{-1}}$.
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