Question

A very long pipe is capped at one end with a semipermeable membrane. How deep (in meters) must the pipe be immersed into the sea for fresh water to begin to pass through the membrane? Assume the water to be 20 oC and is a 0.70 M NaCl solution. The density of the seawater is 1.03 g/cm3 and the acceleration due to gravity is 9.81 m/s2.

Answer 1

Fresh water will begin to pass through the membrane when pressure of the water under the sea will be equal to the osmotic pressure.

Osmotic pressure = CRT                 [where C = concentration of the sea water in molarity]

= 0.7 moles/lit * 0.082 lit.atm.K^-1.mol^-1 * 293K

= 16.818 atm

= 101325 * 16.818 N.m^-2

Let the pipe have to be immerses y meter into the sea, then pressure at y meter

= 101325 * 16.818 N.m^-2

or, ρgh = 101325 * 16.818 N.m^-2

or, (1.03 g/cm3) * (9.81 m/s²) * h = 101325 * 16.818 N.m^-2

or, (1.03 g/cm3) * (9.81 m/s²) * h = 101325 * 16.818 N.m^-2

or, (1030 kg.m^-3) (9.81 m/s²) * h = 101325 * 16.818 N.m^-2

or, h = 101325 * 16.818 N.m^-2/(1030 kg.m^-3) (9.81 m/s²)

= (101325 * 16.818 m kg. s^-2.m^-2 )/(1030 kg.m^-3) (9.81 m/s²)

= 168.65 m