Question

In a series of experiments, the U.S. Navy developed an undersea habitat. In one experiment, the mole percent composition of the atmosphere in the undersea habitat was 79.0% He, 17.0% N₂, and 4.00% O₂. What will the partial pressure of each gas be when the habitat is at a depth below sea level where the pressure is 5.41 atm?

Answer 1

Given that the mole percent composition of He = 79.0%

So the number of moles of He is 79.0 mol

the mole percent composition of N₂ = 17.0%

So the number of moles of N₂ is 17.0 mol

the mole percent composition of O₂ = 4.00%

So the number of moles of O₂ is 4.00 mol

Total number of moles , n = number of moles of He + number of moles of N₂ + number of moles of O₂

                                                             = 79.0 + 17.0 + 4.00

                                        = 100 moles

Mole fraction of He is , X = number of moles of He / total number of moles

                                        = 79.0 / 100

                                       = 0.79

Mole fraction of He is , X' = number of moles of N₂ / total number of moles

                                        = 17.0 / 100

                                       = 0.17

Mole fraction of He is , X '' = number of moles of O₂ / total number of moles

                                        = 4.00 / 100

                                       = 0.04

Partial pressure of He ,p_He = total pressure x mole fraction of He

                                            = 5.41 atm x 0.79

                                            = 4.27 atm

Partial pressure of He ,p_N2 = total pressure x mole fraction of N₂

                                            = 5.41 atm x 0.17

                                            = 0.92 atm

Partial pressure of He ,p_O2 = total pressure x mole fraction of O₂

                                            = 5.41 atm x 0.04

                                            = 0.22 atm