Answer:
a) (vα΅£βββ/vα΅£βββ) = 1.00429
where vα΅£βββ represents the vα΅£ββ for 235-UF6 and vα΅£βββ represents the vα΅£ββ 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 vα΅£ββ for an atom or molecule is given by
vα΅£ββ = β(3RT/M)
where R = molar gas constant = J/mol.K
T = absolute temperature in Kelvin
M = Molar mass of the molecules.
ββ
Let the vα΅£ββ of 235-UF6 be vα΅£βββ
And its molar mass = Mβ = 349.0 g/mol
vα΅£βββ = β(3RT/Mβ)
β(3RT) = vα΅£βββ Γ βMβ
For the 238-UF6
Let its vα΅£ββ be vα΅£βββ
And its molar mass = Mβ = 352.0 g/mol
β(3RT) = vα΅£βββ Γ βMβ
Since β(3RT) = β(3RT)
vα΅£βββ Γ βMβ = vα΅£βββ Γ βMβ
(vα΅£βββ/vα΅£βββ) = (βMβ/βMβ) = [β(352)/β(349)]
(vα΅£βββ/vα΅£βββ) = 1.00429
b) Recall
vα΅£βββ = β(3RT/Mβ)
vα΅£βββ = β(3RT/Mβ)
(vα΅£βββ - vα΅£βββ) = 1 m/s
[β(3RT/Mβ)] - [β(3RT/Mβ)] = 1
R = 8.314 J/mol.K, Mβ = 349.0 g/mol = 0.349 kg/mol, Mβ = 352.0 g/mol = 0.352 kg/mol, T = ?
βT [β(3 Γ 8.314/0.349) - β(3 Γ 8.314/0.352) = 1
βT (8.4538 - 8.4177) = 1
βT = 1/0.0361
βT = 27.7
T = 27.7Β²
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.