Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S25 / 1

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 $10\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$.

The usual estimate adds a grey, infrared-absorbing radiative equilibrium upper atmosphere to the convective lower atmosphere. Consider a very optically thin upper layer: it receives upwelling infrared flux approximately $F_0$ from below, absorbs a small fraction $d\tau$, and emits $\sigma T^{4}d\tau$ into each hemisphere. Radiative balance givesfor $F_0=300\mathrm{W}{\mathrm{m}}^{-2}$. The corresponding effective emission temperature is $(F_0/\sigma {)}^{1/4}\simeq 270\mathrm{K}$. Without shortwave absorption, the model has no solar-heated stratospheric inversion.

This skin estimate requires the grey absorption/emission and angular approximations just stated. Horizontal mixing and no shortwave absorption alone do not determine an exact minimum for a spectrally selective nitrogen–carbon-dioxide atmosphere: absorption coefficients and spectral windows can change its upper-atmosphere radiative equilibrium.

Downward-from-space optical depth at wavelength $\lambda$ iswhere ${\kappa }_{\lambda }$ is the mass absorption coefficient. A nearly isothermal, hydrostatic equilibrium, compositionally well-mixed atmosphere with height-independent ${\kappa }_{\lambda }$ has $\rho \propto e^{-z/H}$ and $\tau \propto e^{-z/H_{\tau }}$ with $H_{\tau }=H=R_{a}T/g$. Thus temperature, mean molecular mass and gravity set the basic scale height. Changes in the absorber mixing ratio, pressure broadening, temperature dependence of line strengths, collision-induced absorption and departures from hydrostatic equilibrium modify it. For example if local absorption scales as ${\rho }^2$ in an isothermal atmosphere, $H_{\tau }=H/2$.

Use a convective layer $T(z)=T_s-\Gamma z$ capped at an isothermal minimum $T_t$ at height $z_t$. Let the initial optical depth be $\tau (z)={\tau }_s{e}^{-z/H_{\tau }}$ and ${\tau }_t={\tau }_s{e}^{-z_t/H_{\tau }}$. Within the lapse-rate layer, $d{B}_{\lambda }/d\tau =B^{\prime }\Gamma H_{\tau }/\tau$; above it the derivative vanishes. The supplied Schwarzschild equation therefore givesAssume the additional carbon dioxide multiplies the relevant opacity by $a=1.1$, leaving temperatures, the lapse rate and $H_{\tau }$ initially fixed. All material surfaces then acquire $a$ times their previous optical depth. HenceFor ${\tau }_t\ll 1$ and ${\tau }_s\gg 10$, the specified approximation givesFor illustration, $\Gamma =6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ and $H_{\tau }=7\mathrm{km}$ give $-4.34\mathrm{K}$; these are illustrative, not supplied data. The inequalities put the effective emitting level $\tau \sim 1$ inside the decreasing-temperature layer, well above an opaque surface and below the thin isothermal cap. Increasing opacity raises that emitting level by $H_{\tau }\ln 1.1$, so it is colder. In a spectral window or above a temperature inversion the approximation and possibly the sign change. Opacity proportional to mixing ratio is also an assumption, not exact for every absorption mechanism.