Answer:
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
