= Solution
Denote the heated and cold vessels by subscripts $h,c$. For <argon>, the opposing effusive <particle> rates are equal in the steady state:
$$
n_h\overline v(T_1)=n_c\overline v(T_0).
$$
Because $\overline v\propto\sqrt T$ and $p=nk_BT$,
$$
\boxed{\frac{p_h}{\sqrt{T_1}}=\frac{p_c}{\sqrt{T_0}},
\qquad \frac{p_h}{p_c}=\sqrt{\frac{T_1}{T_0}}.}
$$
This is <thermal transpiration> in the free-molecular limit. It fixes a ratio; the initial conditions determine the absolute pressures.
For the stated two identical vessels, take their fixed <volume (thermodynamics)> to be $V$ and the combined system sealed. Initially it contains $2p_0V/(k_BT_0)$ atoms. Hence
$$
\frac{p_hV}{k_BT_1}+\frac{p_cV}{k_BT_0}=\frac{2p_0V}{k_BT_0}.
$$
Let $q=\sqrt{T_1/T_0}$. Combining this inventory with $p_h=qp_c$ gives the <closed-vessel thermal transpiration> result
$$
\boxed{p_c=\frac{2p_0q}{q+1},\qquad
p_h=\frac{2p_0q^2}{q+1},\qquad
\Delta p=p_h-p_c=\frac{2p_0q(q-1)}{q+1}.}
$$
Heating makes the hot side less dense but at higher <pressure>. Both expressions reduce to $p_0$ when $T_1=T_0$. For a small imposed <temperature> rise $\Delta T$, the <pressure> difference starts as $p_0\Delta T/(2T_0)$.
If instead the cold vessel is interpreted as an externally maintained <pressure> reservoir with $p_c=p_0$, the answer becomes $\Delta p=p_0(q-1)$. That alternative requires the fixed-pressure reservoir assumption; it does not conserve the initial <particle> inventory of two sealed equal-volume vessels. Merely describing the vessels as large does not keep their pressures fixed through the redistribution. The pressures in the boxed answer assume the natural closed-vessel interpretation.
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