= Solution
A spatially uniform <pressure> is a useful leading approximation when the <sound speed> makes the acoustic crossing time much shorter than the <radiative cooling> and <thermal conduction> times. The motions must remain at low <Mach number>, and gravity and other stresses must be negligible. Sound then redistributes <pressure> before appreciable thermal evolution occurs. Small <pressure> corrections can still drive the slow motion; imposing exactly zero <pressure> gradient in the full momentum equation at every order would be a stronger assumption.
The <equation of state> gives the specific <internal energy of an ideal gas> as $e=p/[(\gamma-1)\rho]$. With $D/Dt=\partial_t+v\partial_x$, the <continuity equation> is $D\rho/Dt=-\rho v_x$. Differentiating $e$ therefore yields
$$
\rho\frac{De}{Dt}=\frac1{\gamma-1}\left(\frac{Dp}{Dt}+pv_x\right).
$$
Insert this into the <internal energy> equation, whose compressional term is $-pv_x$, and use spatially uniform <pressure>, so $Dp/Dt=p_t$. The result is
$$
\boxed{\frac{p_t}{\gamma-1}+\frac{\gamma p}{\gamma-1}v_x+\rho^2\Lambda(T)-\partial_x(\lambda T_x)=0.}
$$
For temporally constant <pressure> put $A=\mu p/\mathcal R$, so the <mass density> is $\rho=A/T$. The <continuity equation> now gives $v_x=T^{-1}DT/Dt$. Multiplication of the last equation by $(\gamma-1)T/(\gamma p)$, with $\lambda=\lambda_0T^\alpha$, gives
$$
\boxed{\frac{DT}{Dt}+C\frac{\Lambda(T)}{T}-\frac{\gamma-1}{\gamma}\frac{\lambda_0T}{p}\partial_x(T^\alpha T_x)=0,\qquad C=\frac{\gamma-1}{\gamma}\left(\frac\mu{\mathcal R}\right)^2p.}
$$
This is the <isobaric conductive cooling in mass coordinates> transformation before changing coordinates.
There is a boundary assumption implicit in the proposed <Lagrangian coordinate>. For a mass integral from a fixed origin, integration of the <continuity equation> gives
$$
m_x=\rho,\qquad m_t=-\rho v+\rho(0,t)v(0,t),\qquad \frac{Dm}{Dt}=\rho(0,t)v(0,t).
$$
Thus $m$ is material when the base <mass flux> vanishes, as appropriate for the stationary cold reservoir below. If the origin permits <mass flux>, replace $m$ by $m-\int^t\rho(0,s)v(0,s)\,ds$; this corrected label has zero <material derivative>. In a material mass coordinate, $D/Dt=\partial_t|_m$ and $\partial_x=(A/T)\partial_m$. In particular
$$
\partial_x(T^\alpha T_x)=\frac{A^2}{T}\partial_m(T^{\alpha-1}T_m).
$$
The conduction coefficient therefore becomes $C\lambda_0$, with the same $C$ as the cooling coefficient. With $\tau=Ct$, the equation is
$$
\boxed{T_\tau+\frac{\Lambda(T)}{T}-\lambda_0\partial_m(T^{\alpha-1}T_m)=0.}
$$
In the hot gas there is no volumetric <radiative cooling>. The initial <temperature> is uniform and the cold boundary introduces no finite mass length. The remaining nonlinear <diffusion equation> is invariant under $m\mapsto bm$, $\tau\mapsto b^2\tau$ with the same <temperature> scale. Its natural <similarity solution> is therefore
$$
T=T_0f(\xi),\qquad \xi=\frac{m}{\sqrt{\lambda_0T_0^{\alpha-1}\tau}},\qquad
\boxed{f(0)=0,\quad f(\infty)=1.}
$$
The far-field condition also recovers the initial <temperature> at every fixed $m>0$ as $\tau\downarrow0$. Substitution gives the <ordinary differential equation>
$$
(f^{\alpha-1}f')'+\frac\xi2 f'=0.
$$
For $\alpha=1$, this reduces to $f''+\xi f'/2=0$. Integrating first for $f'$ gives $f'=K\exp(-\xi^2/4)$. The two <boundary conditions> fix $K=1/\sqrt\pi$, hence the <error-function isobaric cooling front> is
$$
\boxed{f(\xi)=\operatorname{erf}(\xi/2),\qquad T(m,\tau)=T_0\operatorname{erf}\!\left(\frac{m}{2\sqrt{\lambda_0\tau}}\right).}
$$
Here the <error function> has normalization $2/\sqrt\pi$, as printed in the original PDF; the converted TeX's $2/\pi$ is incorrect. The following original sketch shows the <temperature> profiles. The cooling layer broadens as $\sqrt\tau$, and at fixed positive mass label the <temperature> falls as time increases, while the remote hot gas remains at $T_0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-71-cooling.png]
{title=Isobaric conductive cooling profiles at increasing scaled times}
For $\alpha=1$, the magnitude of the <heat flux> into the cold reservoir is
$$
L=\left.\lambda T_x\right|_{0^+}=\lambda_0 A T_m(0,\tau).
$$
Although the <thermal conductivity> tends to zero there, the physical <temperature gradient> diverges: their product has this finite limit. The stationary reservoir has no incoming advective energy flux and radiates the conducted energy immediately. Since $T_m(0,\tau)=T_0/\sqrt{\pi\lambda_0\tau}$, the radiated power per unit area is
$$
\boxed{L=\frac{A T_0\sqrt{\lambda_0}}{\sqrt{\pi\tau}}=T_0\sqrt{\frac{\gamma p\lambda_0}{(\gamma-1)\pi}}\,t^{-1/2}.}
$$
Thus \b[the requested exponent is $\boxed{k=1/2}$]. The initial <divergence> is the usual idealization of an instantaneous <temperature> discontinuity in a <diffusion equation>.
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