When will it flow out?

For convenience, I will use $L_0$ and $h$ to refer to the the bottom air column height and to the mercury height, respectively. It is important to note that these quantities vary and are not fixed at the given values (per my convention). It is obvious that $L_0$ varies in the range $62.5 \leq L_0 \leq 100$ . We want to determine the required air column temperature to maintain stasis in each case.

Start with the ideal gas equation (let $A$ be the cross-sectional area of the cylinder):

$P V = n R T \\ P A \, L_0 = n R T \\ T = \alpha \, P \, L_0$

In the initial state:

$T = 250 \\ P = 75 + 25 \\ L_0 = 62.5$

This allows us to calculate $\alpha$ . Next, we sweep $L_0$ over its range and calculate the required stasis temperature. It is important to keep track of the mercury height $h$ :

$h = 25 \,\,\,\,\,\, (\text{for} \,\, 62.5 \leq L_0 < 75) \\ h = 100 - L_0 \,\,\,\,\,\, (\text{for} \,\, L_0 \geq 75)$

The pressure and temperature calculations for the air column are thus:

$P = P_0 + h \\ T = \alpha \, P \, L_0$

Below is the graph of $L_0$ vs the air column temperature. In order to push all of the mercury out of the tube, we have to raise the air column to a temperature which exceeds the maximum temperature on the graph, which happens to be $306.25$ .