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## Homework Statement

Consider an experiment on the International Space Station, which is illustrated below. A cylindrical capillary of length L= 10. [cm] and inner diameter of dc=500. [µm] (sealed at one end), is positioned to contact a droplet of water, D = 1.0 [cm], which is floating in the ISS laboratory. Water is completely hydrophilic with the material of that capillary (contact angle is θ ≈ 0o). The temperature in the ISS laboratories is 20 [oC].

Question: How far the water will enter the capillary; i.e. determine ‘h’ as shown in the illustration

below

## Homework Equations

I assume this is the most important equation for looking at it (but nothing is really given):

h=2Tcos(theta)/(prg)

All of these are also given:

c=500 [µm]; R =250 [µm]; L= 0.1 [m]; µwater=0.001 [Pa s]; ρwater =1000 [kg/m3] ;

σwater-air =0.072 [N/m]; h = ??? [m]

## The Attempt at a Solution

Since we do not know gravity I tried to solve initially for Pint of the capillary

Pint=2T/R

Pint=576pa = 0.576kpa --> 0.00568 atm

(My assumption that pressure in ISS is that of earth at sea level)

Po= 1 atm + 0.00568 atm = 1.00568 atm

Vcap= (L-dc/2)(pi)(dc/2)^2 + (2/3)pi(dc/2)^3

Vcap= 1.96 x 10^-5 L

nair=(1atm)(1.96 x 10^-5 L)/((0.08206 Latm/molK)(293K))

nair= 8.160 x 10^-7 mol

If that's the air in the capillary before capillary action, then the air will be compressed to the new pressure:

Po=1.00568 atm

Vair2= (8.160 x 10^-7 mol)(0.08206 Latm/mol K)(293 K)/1.00568 atm

Vair2= 1.951 x 10^-5 L = 1.951 x 10^-8 m^3

Vair2= (L-dc/2-h)pi(dc/2)^2 + (2/3)pi(dc/2)^3

height comes out to be:

h= 5.53 x 10^-4 m

That height seems way to small to be right

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Then I thought maybe I could assume (using the internet) that gISS=gearth(0.9)

and use the equation:

h=2Tcos(theta)/(prg)

to get:

h=6.5 cm

^This answer makes a lot more sense to me but it doesn't feel like the right way to do the problem, seeing as we are not given the statement that the gravity on the ISS is 90% of that on earth. Am I missing something?

Thanks!