# Physics question?

There is water of 20 degrees Celsius in a pot. There is a sphere-shaped air bubble of volume 0,5mm^3 at the bottom of the pot. The pressure inside the air bubble is 10kPa greater than it is over the water where the atmospheric pressure is 100kPa.

1) Find the depth the air bubble is in.

2) Find the number of air and water vapor molecules inside the air bubble and what percentage do these molecules make in relation to all molecules.

Use water density as 1t/m^3, surface tension coefficient is 73 mN/m, partial pressure of the saturated water vapor – 2,3kPa, strength of the gravitational field – g=9,8N/kg

Relevance

The question is badly worded and the surface tension is not needed for the required degree of accuracy.  But see if this helps:

If P1 is the pressure inside the bubble and P2 is the pressure *just* outside the bubble then

P1 -  P2 = 2T/R (standard formula for bubble in liquid)

V = ⁴/₃πR³

R = ∛[3V/(4π)] = ∛[3*0.5/(4π)] = 0.492mm = 4.92*10^-4 m

P1 -  P2 = 2T/R = 2*73*10^-3/(4.92*10^-4) = 297Pa (=0.3kPa approx.)

The pressure in the bubble is 100+10=110kPa.  So the contribution from surface tension (0.3kPa) is negligible.

_____________________

1) Depth (h) is found using ΔP = ρgh where ΔP is pressure difference between surface and depth h.  Ignoring surface tension, take ΔP=10kPa.  Note 1t/m³ = 1000kg/m³.

h = P/(ρg) = 10*10^3/(1000*9.8) = 1.02m (round to 1.0m)

(If you wanted to take the surface tension into account you would use ΔP=10 – 0.3 = 9.7kPa which gives h = P/(ρg) = 9.7*10^3/(1000*9.8) = 0.99m (round to 1.0m))

2)  The pressure due to water molecules is 2.3kPa.

The pressure due to air molecules is 110 - 2.3 = 107.7kPa

(If you wanted to take surface tension into account you would use 107.7-0.3 = 107.4kPa.)

For air:

P = 107.7*10^3 Pa

V = 0.5^10^-9 m³

R = 8.31 J/(mol.K)

T = 20+273 = 293K

Plug these values into PV=nRT to find n (number of moles).  Then multiply n by Avogadro’s number to find number of air molecules. You can do the arithmetic!

Same calculation for water molecules but use P =2.3kPa

The question about %s is not clear.  Note n∝P for a given volume and temperature.  So you don’t need to use number of molecules:

% of air molecules compared to total = 107.7/110 * 100 = 98%

% of water molecules compared to total = 2.3/110 * 100 = 2%

But maybe something else is being asked