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In order to calculate the number of O2 molecules per unit volume in surface air, we can first determine the number of moles of air in a unit volume and then find the number of O2 molecules using the given percentage composition of the air and Avogadro's number. Given conditions: - Temperature (T) = \(20.0^\circ C\) = 293 K (converted to Kelvin) - Pressure (P) = 1.00 atm - Air composition (by volume): 78% N2, 21% O2, and 1% Ar Using the ideal gas law, rewrite the equation for the number of moles (n), as follows: $$ n = \frac{PV}{RT} $$ For 1 m³ of air, convert the volume into liters (V) to meet the requirements of the ideal gas law (1 m³ = 1000 L). Plug in the numbers: $$ n = \frac{1.00 \times 1000}{0.0821 \times 293} = 40.37 \mathrm{mol\ of\ air} $$ Now, find the number of moles of O2 in 1 m³ of air by using the given percentage composition, and then calculate the number of O2 molecules using Avogadro's number: $$ \mathrm{Moles\ of\ O}_{2} = 0.21\times 40.37 = 8.48 \mathrm{mol} $$ $$ \mathrm{Number\ of\ O}_{2\mathrm{molecules}} = 8.48 \times 6.022\times 10^{23} = 5.10\times10^{24} \mathrm{molecules/m^3} $$

With the given seawater density (\(\rho\)) = 1025 kg/m³ and depth (d) = 100 m, we can calculate the pressure exerted by the water column (P_water) on the diver using the formula: $$ P_{\mathrm{water}} = \rho g d $$ where g = 9.81 m/s² (acceleration due to gravity). Therefore, $$ P_{\mathrm{water}} = 1025\times 9.81\times 100 = 1.0\times10^6\,\mathrm{N/m^{2}} $$ To work with atmospheric units, we need to convert N/m² to atm, where 1 atm = 101325 N/m²: $$ P_{\mathrm{water(atm)}} = \frac{1.0\times10^6}{101325} = 9.87\,\mathrm{atm} $$ The total pressure at depth (P_total) is the sum of atmospheric pressure and the pressure by the water column: $$ P_{\mathrm{total}}=1.00+9.87=10.87\,\mathrm{atm} $$

We need to keep the same number of O2 molecules per unit volume at this depth as in surface air. Therefore, we must first calculate the number of moles of O2 needed for these conditions using the ideal gas law: $$ n_{\mathrm{O2}} = \frac{P_{\mathrm{total}}V_{\mathrm{tank}}}{RT} $$ We are considering 1 m³, so V_tank = 1000 L, and the temperature is assumed the same as the surface, T = 293 K. $$ n_{\mathrm{O2}} = \frac{10.87\times 1000}{0.0821\times 293} = 445.4\,\mathrm{mol} $$ Now, calculate the percentage composition of O2 in the tank for these conditions: $$ \mathrm{O}_{2\mathrm{Percentage}} = \frac{445.4}{445.4 + n_{\mathrm{He}}} \times 100 $$ Since we want the same number of O2 molecules per unit volume, we have: $$ 8.48\times 6.022\times 10^{23} = 445.4\times 6.022\times 10^{23}\times \frac{n_{\mathrm{O}_{2}}}{n_{\mathrm{O}_{2}}+n_{\mathrm{He}}} $$ Solve for the ratio of oxygen and helium: $$ \frac{n_{\mathrm{O}_{2}}}{n_{\mathrm{O}_{2}}+n_{\mathrm{He}}} = \frac{8.48}{445.4} = 0.019 = 1.9\% $$ Thus, for a diver going to a depth of 100 meters, the tank should contain 1.9% O2 and the remaining 98.1% should be He.