Question

Mixtures of helium and oxygen are used in scuba diving tanks to help prevent "the bends". For a particular dive, 27.5 L of oxygen gas at 25.0°C and 1.84 atm, and 7.90 L of helium gas at 25.0°C and 2.95 atm were pumped into a tank with a volume of 6.00 L at 25.0°C. Calculate (in atm): 1. the partial pressure of oxygen 2. the partial pressure of helium 3. the total pressure in the tank at 25.0°C.

Answer 1

The ideal gas law will be used for calculation. It is as below - PV = nRT whene, P-gas pressure (atm) V- gas volume (L) na moles of gas R- Universal gas constant = 0.082057 Latm mot' k' T - absolute gas temperature (K) B .. Moles of o, pumped into the tank Mo, = RT NOV - PV RT - 1.84 atm x 27.5 x 0.082057 katm moi' k * (25+273)K = 2.069 mol page-1

And moles of He gas pumped into the tank - "He = PV RT 2.95 atm x 7,90 L 0.082057 Latm mol' k' x (25+273)K - 0.953 mol Total number of moles present in the tank 3.022 mol not nhe = (2.069 + 0.953) mol = So, total presswie in the tank - Potal = n R Tiank tank 3.022 mot x 0.082057 Łalm mo k' x (25+273) K 6.00 K = 12.316 atm Now, partial pressure ob oxygen - total pressure x mol fraction of oxygen = Ptotal no 2.069 moles = 12.316 atm X - F = 8.432 atm 3.022 mores Ano partial pressure of helium 0.953 molés = = 12.316 atm x 2022 mores 3.884 atm . page-2)

o we will repont the answers in three significant bigures as allt the data are given in three significant tigures, - 1. the partial pressure of oxygen [8.43 atm 2. the partial pressure of helium (3.88) atm 3. the total pressure in the tank at 25.0°C 12.3) atm m um (page-3