= Solution
\b[Balances and meanings.] The first equation is <conservation of energy> per unit area. Here $c$ is effective areal <heat capacity>, $Q$ is the <solar constant>, and $Q/4$ accounts for the ratio of the Earth's intercepting disc area to its total surface area. The <albedo> $a$ reflects a fraction of that input. The outgoing <thermal radiation> is the <Stefan–Boltzmann law> modified by $\gamma(p)$, an effective transmission factor for the <greenhouse effect>; it normally decreases as atmospheric <carbon dioxide> increases.
The <albedo> relaxes towards its <temperature>-dependent equilibrium on the <ice>-adjustment time $t_i$. The atmospheric mass inventory per unit <pressure> is $M_{\mathrm{CO}_2}A/(M_ag)$, using the <molar mass> ratio of <carbon dioxide> to air and hydrostatic atmospheric mass. Its sources and sinks are volcanic input $v$, removal by <chemical weathering> over land area $A_L$ at rate $W$, and the atmosphere-to-ocean exchange $h(p-p_s)$. The ocean equation is a molar <mass conservation> balance: $\rho_{\mathrm{H_2O}}V_{\mathrm{oc}}$ is ocean <water> mass, division by $M_{\mathrm{CO}_2}$ converts the incoming carbon-dioxide mass to moles, and $bC$ removes dissolved carbon by burial/export. With $C$ in moles per kilogram, $b$ has units of kilograms of <water>-equivalent per year.
\b[Times and reduction.] Linearize the <temperature> equation at fixed $a,p$ about <radiative equilibrium>. Its <exponential relaxation time> is
$$
\boxed{t_T=\frac{c}{4\sigma\gamma(p)T^3}=\frac{cT}{Q(1-a)}=3.00\times10^6\ \mathrm{s}=0.0952\ \mathrm y.}
$$
The equality uses $\sigma\gamma T^4=Q(1-a)/4$. A stock-over-flux thermal time, sometimes used as a scale, is four times this, $0.381\ \mathrm y$; both describe a rapid thermal response relative to the other processes.
For <pressure> in atmospheres, write $I=M_{\mathrm{CO}_2}A\,10^5/(M_ag)$ for the atmospheric carbon-dioxide mass per atmosphere. The factor $10^5$ is essential because $h$ is given per atmosphere. The <pressure> and slow ocean <exponential relaxation times> are
$$
\boxed{t_p=\frac{I}{h}=108.8\ \mathrm y,\qquad t_C=\frac{\rho_{\mathrm{H_2O}}V_{\mathrm{oc}}}{b}=1.667\times10^5\ \mathrm y.}
$$
Thus $t_T\ll t_p\ll t_i=10^4\ \mathrm y<t_C$. The leading <quasi-steady approximation> gives
$$
T=\Theta(a,p)=\left[\frac{Q(1-a)}{4\sigma\gamma(p)}\right]^{1/4},\qquad h(p-p_s)=v-A_LW.
$$
Substitution in the two remaining equations yields
$$
\boxed{t_i\dot a=B(a,p)-a,\quad B(a,p)=a_{\mathrm{eq}}(\Theta(a,p)),\qquad t_C\dot C=C_0-C,\quad C_0=\frac{v}{M_{\mathrm{CO}_2}b}.}
$$
\b[The coefficient of $\dot a$ remains $t_i$; the printed $t_T$ in the reduced equation is a typographical error.] Eliminating a fast <temperature> variable does not replace the <ice>-response time. The ocean time here is the slow time after atmospheric balance cancels the exchange flux; holding <pressure> fixed after imposing a carbon-dependent $p_s$ would describe a different conditional relaxation.
The supplied volcanic input gives
$$
\boxed{C_0=8.21\times10^{-4}\ \mathrm{mol\,kg^{-1}},}
$$
about $41.1\%$ of the stated present value $2.0\times10^{-3}$; that value is about $2.43C_0$.
\b[Carbon-dioxide levels.] Since $A_LW=3.0\times10^{11}\ \mathrm{kg\,y^{-1}}=v$, the atmospheric balance with $p_s=C/K$ gives
$$
\boxed{p_0=\frac{2.0\times10^{-3}}{7.1}=2.817\times10^{-4}\ \mathrm{atm}\approx282\ \mathrm{ppm}.}
$$
This uses the specified present ocean <concentration>, as requested, rather than replacing it by the model's smaller volcanic-only equilibrium.
After sustained additional input $v_i$, $C$ barely changes during the atmospheric adjustment. At this stage
$$
p(t)\simeq\frac{C(0)}K+\frac{v_i}{h}(1-e^{-t/t_p}),\qquad\boxed{p_{\mathrm{fast}}\simeq4.187\times10^{-4}\ \mathrm{atm}\approx419\ \mathrm{ppm}.}
$$
This is an intermediate balance, approached over several centuries. On the slower ocean time,
$$
C(t)\simeq C_\infty+[C(0)-C_\infty]e^{-t/t_C},\qquad C_\infty=\frac{v+v_i}{M_{\mathrm{CO}_2}b}=2.820\times10^{-2}\ \mathrm{mol\,kg^{-1}},
$$
with <pressure> following $p\simeq C/K+v_i/h$. Since $C(0)<C_\infty$, <pressure> continues to rise slowly. Its eventual level in this stipulated constant-parameter model is
$$
\boxed{p_\infty=\frac{C_\infty}{K}+\frac{v_i}{h}=4.109\times10^{-3}\ \mathrm{atm}\approx4109\ \mathrm{ppm}.}
$$
The <two-reservoir carbon adjustment> therefore has a rapid rise towards the frozen-ocean balance followed by a much longer increase. The figure solves the coupled linear <pressure>–ocean equations, retaining their small inventory correction to the reduced <time scales>.
\Image[/past-exam-of-the-mathematics-and-the-environment-course-of-the-university-of-oxford/2011-carbon-pressure.png]
{title=Atmospheric carbon dioxide adjustment over centuries and over the slow ocean-carbon time}
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