 Solution

Denote the heated and cold vessels by subscripts $h,c$. For argon, the opposing effusive particle rates are equal in the steady state:Because $\overline{v}\propto \sqrt{T}$ and $p=n{k}_{B}T$,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 to be $V$ and the combined system sealed. Initially it contains $2p_{0}V/(k_B{T}_0)$ atoms. HenceLet $q=\sqrt{T_1/T_0}$. Combining this inventory with $p_h=q{p}_c$ gives the closed-vessel thermal transpiration resultHeating 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.