Question

A scuba tank has a volume of 2800 cm3. For very deep dives the tank is filled with 50 percent (by volume) pure oxygen and and 50 percent pure helium.

(a) How many molecules are there of each type in the tank if it is filled at 20°C to a gauge pressure of 10 atm?

(b) What is the ratio of the average kinetic energies of the two types of molecule?

(c) What is the ratio of the rms speeds of the two types of molecule?

Answer 1

(a) Given the total volume = 2800 cm3 = 2800 mL = 2800 mL x (1L / 1000 mL) = 2.8 L

Since He and O2 are present 50 % by volume,

Volume of O2 = 2.8 L x (50/100) = 1.4 L

Volume of He = 2.8 L - 1.4 L = 1.4 L

Temperature, T = 20 DegC = 273 + 20 = 293 K

Pressure, P = 10 atm

Now number of moles can be calculated from ideal gas equation.

PV = nRT

=> n = PV/RT = (10 atm x 1.4 L) / (0.0821 L.atm.mol-1K-1 x 293K) = 0.582 mol He

Since both have same volume, moles of O2 = 0.582 mol O2

Hence molecules of He = 0.582 mol He x (6.023 x 10²³ molecules / 1 mol He) = 3.51x10²³ molecules He (answer)

molecules of O2 = 0.582 mol O2 x (6.023 x 10²³ molecules / 1 mol He) = 3.51x10²³ molecules O2 (answer)

(b) Kinetic energy of a molecule can be alcualted form the following frmulae

KE = (3/2) x nRT

Since both have same number of moles, the ratio of average kinetic energies of the two types of molecule is 1 (answer)

(c) Molar mass of O2, M_O2 = 32 g/mol

Hence rms speeed of O2 molecules, V(O2) = underroot(3RT / M_O2) = underroot(3RT / 32)

Molar mass of He, M_He = 4 g/mol

Hence rms speeed of He molecules, V(He) = underroot(3RT / M_He) = underroot(3RT / 4)

V(O2) / V(He) = [underroot(3RT / 32)] / [underroot(3RT / 4)] = 1/22 = 1/2.82 (answer)