Hi, this is a very material-dependent question. But you may make a model with a capillary of your material, corresponding gas and pressure, then decide.

Use Laplace-Washburn equation as following:

pore diameter~4S.cos(theta)/Dp

S=Surface tension of water (S=75.5 mN/m at 25 oC)

thete=pore surface contact angle~0

Dp=Differential pressure at the pore opening, the pressure that pushed the water into the pore.

(Dp=Ro*g*h+Patm, Ro is the water density, g is gravitational acceleration (9.8 m/s2), h is the water height from the pore opening up to the water free surface, and Patm is the pressure at the water free surface).

This depends a lot of the wetting conditions of water in your porous medium. It should be treated hydrophobic in order to avoid spontaneous imbibition. Capillary pressure Pc = 4*S*cos(theta)/Dp has to be compared to Pg - Ph, where Ph is hydrostatic pressure Ph = rho*g*h, and Pg being the pressure of your gas pushing on the top of the tube.

S : surface tension of water with ambient atmosphere

theta : contact angle for water with the material from which the tube is made, including the effect of a surface coating

Dp : pore diameter

rho : density of water

h : depth of water in contact with the porous

g : gravity constant

Your condition for the tube to remain tight impies : Pc + Ph < Pg

Hence hoosing theta > 90° (with appropriate coating) allows to have Pc negative, and the above condition is fullfilled if Pg is large enough or if Dp is small enough.

If you cannot coat the tube surface and Pc >0, then you will have to push very strong so that Pg > Pc + Ph, and you will have to choose Dp large, which will imply that your injecting gas will bubble in your water.

Thanks Dimitar P. Mitev Ali Nematollahzadeh and Philippe Brunet

I will follow your recommendation to calculate pore size.