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## 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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2021-09-03T13:45:45+00:00
2021-09-03T13:45:45+00:00 1 Answers
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## Answers ( )

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.