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28 April, 02:08

The pressure inside a water droplet is higher than the surroundings, this is related to surface tension. If the surface tension of a spherical droplet is 73mN/m and has a diameter of 2mm, then what is the gauge pressure rise (Pg) between droplet and the surroundings (to within 10%) ?

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  1. 28 April, 04:51
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    Gauge pressure rise = 292 Pa

    Explanation:

    We are given;

    Diameter of the water droplet is d = 2mm = 0.002m

    Radius = 0.002/2 = 0.001

    Surface tension; σ = 73 mNm = 73 x 10^ (-3) m

    So we have to find the pressure rise.

    Now, pressure rise is given by the formula;

    ΔP = P_inside - P_atmosphere = 4σ/r

    Where r is radius.

    Thus, plugging in the relevant values, we have;

    ΔP = (4 x 73 x 10^ (-3)) / 0.001 = 292 pa
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