Question

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose. The temperature of the air is constant (body temperature). The air acts as an ideal gas. Salt water has an average density of around 1.03 g/cm3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm. What is the ratio of the molar concentration of gases in Gabor's lungs at the depth of 15 meters to that at the surface

Answer 1

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

b

The number of moles of gas that must be released is  $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

 This means that the pressure at the depth of the surface would be

                $P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

                      $= 2.5 atm$

The ideal gas equation is mathematically represented as

                $PV = nRT$

Where P is pressure at the surface

           V is the volume

            R is the gas constant  = 8.314 J/mol. K

making n the subject we have

        $n = \frac{PV}{RT}$

 Considering at the surface of the water the number of moles at the surface would be

               $n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ ,$6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

              $n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

                   $= 0.2359 mol$  

To obtain the number of moles at the depth of the water we use

                $n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d$ are pressure and no of moles at the depth of the water

        Substituting values we have

              $n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

                  $= 0.5897 mol$

Now to obtain the number of moles released we have

             $n = n_d - n_s$

               $= 0.5897mol - 0.2359mol$

              $=0.3538 \ mol$

     The molar concentration at the surface  of water is

                $[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

                                $=39.31mol/m^3$

    The molar concentration at the depth  of water is

           $[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

                      $= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

         $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$