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Initial condition: Scuba diver is naturally buoyant at constant depth, so initial net force is 0. The upward buoyant force, which only depends on the volume (and thus the weight) of the water displaced by the diver remains the same. Conclusion: The new net force will thus only be the change of the diver's total weight, which is the weight lost from the 10L $N_2$ gas. Solution: Using the Ideal Gas Law $PV = nRT$ , and the pressure $P=P_{atm}+\rho_{water} g d$ , the mol of $N_2$ gas is 1.24 mol. Using the molecular mass 28g/mol, the mass of the $N_2$ gas is = 34.72g = 0.03472 kg. Finally, the weight of the $N_2$ gas is (0.03472 kg)(9.8 m/s^2) = 0.34N

The diver and gas is originally neutrally bouyant, which means that the total initial net force on the diver is zero. In order to find the net force after the gas is lost, we need to find the mass of the gas. We could use the ideal gas law to find this, but first we need to know the pressure 20 m below the surface of the water. This is $p_{20}=p_{surface}+\rho g d=302225$ Pa, where d is 20 m. The ideal gas law then yields $N=p_{20} V/(RT)=1.24$ mol of gas (make sure to convert units properly).

The mass of the gas lost is therefore $m=(0.028)(1.24)=0.0348$ kg. The left over diver/gas system is lighter by $(0.0348)(9.8)=0.341$ N but the displaced volume and bouyant force is still the same as before. Therefore the net force upwards is 0.341 N.