7. A 2.0 L container had 0.40 mol of He (g) and 0.60 mol of Ar (g) at 25°C.

Which gas has a greater average kinetic energy?

Which gas is traveling at a higher average speed?

If Ar has an average speed of 431 m/s, what is the average speed of He?

What is the total pressure in the container?

What are the partial pressures of He and Ar?

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U):

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N (1/2mv²) = 3/2 NKT

∴ N = nNo ... number of moleculas

∴ K = 1.380 E-23 J/K ... Boltzmann's constant

∴ No = 6.022 E23 molec/mol ... Avogadro's number

for He:

⇒ N = (0.4) (6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2) (2.4088 E23) (1.380 E-23 J/K) (298) = 1485.892 J

for Ar:

⇒ N = (0.6) (6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2) (3.6132 E23) (1.380 E-23 J/K) (298) = 2228.838 J

* * Ar gas has a greater average kinetic energy

b) He:

∴ N (1/2) mv² = (3/2) NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol) (4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √ (3 (1.380 E-23) (298) / (1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6) (39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

* * v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s) / (2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol) (0.082 atm. L/K. mol) (298 K) / (2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4) (0.082) (298) / 2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm