91Ó°ÊÓ

We are given the teh salinity of Seawater is $3.5\%$.

We have to find the osmotic pressure difference between seawater and fresh water and the minimum work required to desalinate one liter of seawater.

We know that,

$V=\frac{m}{\mathrm{ÒÏ}}$

Here, m is the mass and $\mathrm{ÒÏ}$is the density.

Substitute $m=1kg$and $1000kg/m^3$for$\mathrm{ÒÏ}$

$$\begin{gathered} V=\frac{1kg}{1000kg/m^3} \\ ={10}^{-3}{m}^3 \end{gathered}$$

State the molecular weight of NaCl.

$$\begin{gathered} A=23g/mol+35.5g/mol \\ =58.5g/mole \end{gathered}$$

Writing the expression for the number of moles of NaCl,

$$\begin{gathered} n_B=\frac{numberofgrams}{formulaweightofNaCl} \\ =\frac{m}{4} \end{gathered}$$

Substitute$m=35g$and $58.5g/mol$for A,we get

$$\begin{gathered} n_B=\frac{35g}{58.5g/mole} \\ =0.5983moles \end{gathered}$$

Now,

$P_2-P_1=\frac{N_{B}kT}{V}$

Substitute nR for NK in the above expression,

$P_1-P_2=\frac{n_{B}RT}{V}$

Substituting the values, we get

$$\begin{gathered} P_2-P_1=\frac{(0.5983mole)(8.31J/mole.K)(300K)}{{10}^{-3}{m}^3} \\ =\frac{1.49×{10}^3}{{10}^{-3}}N/m^2 \\ =(1.49×{10}^{6}Pa)\left(\frac{1atm}{1.01×{10}^{5}Pa}\right) \\ =14.9atm \end{gathered}$$

Therefore, the osmatic pressure difference between seawater and fresh water is $14.9atm$.

$W=(P_2-P_1)V$

Put role="math" localid="1647756016152" $P_2-P_1=1.49×{10}^{6}Pa$and ${10}^{-3}{m}^3$ for V,we get.

$$\begin{gathered} W=(14.9×{10}^{5}Pa)({10}^{-3}{m}^{-3}) \\ =1490J \end{gathered}$$

Therefore, the work needed to desalinate one liter of sea water is $1490J$.