 Solution

For a unit mass of ideal gas, $h=u+P/\rho =u+R_{a}T$. 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 rateThis 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 $M_N,M_C$, $R_a=R/(x_N{M}_N+x_C{M}_C)$ and $c_p=[x_N{C}_{p,N}+x_C{C}_{p,C}]/(x_N{M}_N+x_C{M}_C)$. 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 $10\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$.