There are two important isotopes of uranium, 235U and 238U; these isotopes are nearly identical chemically but have different

Question

There are two important isotopes of uranium, 235U
and 238U; these isotopes are nearly identical chemically
but have different atomic masses. Only 235U is very
useful in nuclear reactors. Separating the isotopes is called
uranium enrichment (and is often in the news as of this
writing, because of concerns that some countries are
enriching uranium with the goal of making nuclear
weapons.) One of the techniques for enrichment, gas
diffusion, is based on the different molecular speeds of
uranium hexafluoride gas, UF6 . (a) The molar masses of
235U and 238UF6 are 349.0 g/mol and 352.0 g/mol,
respectively. What is the ratio of their typical speeds vrms ?
(b) At what temperature would their typical speeds differ by
1.00 m/s? (c) Do your answers in this problem imply that
this technique may be difficult?

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Thiên Di 2 weeks 2021-09-03T13:45:45+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-03T13:46:59+00:00

    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.

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