Question

If you have a glass capillary tube that is 0.275 mm in radius, and you push it into water then the water will rise, but if you push the capillary into mercury in will create a meniscus below the surface of the fluid. If air is the gas phase, then the surface tension of water at 140oF is 66.2 mN/m and the surface tension of mercury is 470 mN/m. If the contact angle of water on glass is 10oand mercury on glass is 130o, then calculate BOTH height changes.

Answer 1

Step-by-step explanation:

From the given information, the rise in a capillary tube h =
`(2T cos \theta )/(r \rho g)`

where:

For the height of water:

Surface Tension T = 66.2 mN/m = 66.2 N/m

θ = 10⁰

Cos θ = cos 10 = 0.985

radius r = 0,257 mm = 0.275 × 10⁻³ m

density of water
`\rho` = 1000 kg/m³

g = 9.8 m/s²

∴

replacing our values:

h =
`(2T cos \theta )/(r \rho g)`

h =
`(2* 66.2 * 10^(-3) * 0.985)/(0.275 * 10^(-3) * 1000* 9.8)`

h = 0.0484 meteres

h = 48.3 mm

Since the height h =
`(2T cos \theta )/(r \rho g)`

r = 0.275 mm = 0.275 × 10⁻³ m

the density of mecury now
`\rho_H{g}}`

= 13593 kg/m³

the surface tension of the mercury
`T_(Hg) =470 * 10^(-3) \ N/m`

θ = 130⁰

Cos θ = cos 130 = -0.6428

Using the same above equation:

h =
`(2* 470 * 10^(-3) * (-0.6428) )/(0.275 * 10^(-3) * 13593 * 9.8)`

h = -0.016494 m

h = 16.49 mm