Question

A submarine obtains its fresh (pure) water supply by reverse osmosis. A sem-permeable membrane that allows only water to pass through is placed on the hull of the submarine. As the submarine descends into the ocean, the pressure inside is maintained at 1 atm while the pressure outside increases due to hydrostatic pressure. At a certain depth, this pressure difference across the membrane will become sufficient to cause water to diffuse into the submarine, against the water concentration gradient. The following information is available:

Answer 1

Mole fraction of the salt = 0.05

Basis : 100 mole of Ocean water

Moles of salt = 0.05*100 = 5 mol

Moles of water = (1-0.05)*100 = 95 mol

Mass of 95 mol of water = Moles * Molar mass = 95 mol * 18 g/mol = 1710 g

Volume of water = Mass of water / density = 1710 g/ 1 g/cm³ = 1710 cm³

Concentration of Salt in ocean = Moles of salt / Volume of water = 5/1710 mol/cm³ = 0.002924  mol/cm³ = 2924 mol/m³

Osmotic pressure ($\Delta \pi$) = $\pi_D$ - $\pi_F$

$\pi_D$ =  CRT = 2924 * 8.314 * T = 24310.136*T Pa

T = 290 - h/100 K

$\pi_D$ = 24310.136* ( 290 - h/100) Pa

$\pi_F$ = CRT = 0 . Assuming no salt or negligible amount of salt present in fresh water.

$\Delta \pi$ = 24310.136* ( 290 - h/100) Pa

$\Delta P$ = Pout - Pin

Pout = Patm + $\rho _{sw}$ gh = 101325 + 1145 *9.81*h = 101325 + 11232.45*h Pa

Pin = 1 atm = 101325 Pa

$\Delta P$ = Pout - Pin = 101325 + 11232.45*h - 101325 = 11232.45*h Pa

The water flux is given as,

Q = Lp*($\Delta P - \Delta \pi$)

Just before water starts flowing in to low concentration area,

$\Delta P - \Delta \pi$ = 0

On substituting the values of the above terms,

24310.136* ( 290 - h/100) = 11232.45*h

2.164*(290 - h/100) = h

h = 614.29 m

Minimum depth = 614.29 m