Answer:
a) (vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429
where vᵣₘₛ₁ represents the vᵣₘₛ for 235-UF6 and vᵣₘₛ₂ represents the vᵣₘₛ for 238-UF6.
b) T = 767.34 K
c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.
Explanation:
The vᵣₘₛ for an atom or molecule is given by
vᵣₘₛ = √(3RT/M)
where R = molar gas constant = J/mol.K
T = absolute temperature in Kelvin
M = Molar mass of the molecules.
₁₂
Let the vᵣₘₛ of 235-UF6 be vᵣₘₛ₁
And its molar mass = M₁ = 349.0 g/mol
vᵣₘₛ₁ = √(3RT/M₁)
√(3RT) = vᵣₘₛ₁ × √M₁
For the 238-UF6
Let its vᵣₘₛ be vᵣₘₛ₂
And its molar mass = M₂ = 352.0 g/mol
√(3RT) = vᵣₘₛ₂ × √M₂
Since √(3RT) = √(3RT)
vᵣₘₛ₁ × √M₁ = vᵣₘₛ₂ × √M₂
(vᵣₘₛ₁/vᵣₘₛ₂) = (√M₂/√M₁) = [√(352)/√(349)]
(vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429
b) Recall
vᵣₘₛ₁ = √(3RT/M₁)
vᵣₘₛ₂ = √(3RT/M₂)
(vᵣₘₛ₁ - vᵣₘₛ₂) = 1 m/s
[√(3RT/M₁)] - [√(3RT/M₂)] = 1
R = 8.314 J/mol.K, M₁ = 349.0 g/mol = 0.349 kg/mol, M₂ = 352.0 g/mol = 0.352 kg/mol, T = ?
√T [√(3 × 8.314/0.349) - √(3 × 8.314/0.352) = 1
√T (8.4538 - 8.4177) = 1
√T = 1/0.0361
√T = 27.7
T = 27.7²
T = 767.34 K
c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.
