Naturally found uranium consists of 99.274%, 238U, 0.720%235U, and 0.006% 233U (

Question

Naturally found uranium consists of 99.274%, 238U, 0.720%235U, and 0.006% 233U (by mass). As we have seen, 235U is the isotope that can undergo a nuclear chain reaction. Most of the 235U used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, UF6(g).

A.) What is the mass of UF6 in a 30.0-L vessel of UF6 at a pressure of 690 torr at 360 K?

B.) What is the mass of 235U in the sample described in part A? The atomic mass of uranium-235 is 235.044 u.

C.) Now suppose that the UF6 is diffused through a porous barrier and that the change in the ratio of 238U and 235U in the diffused gas can be described by the equation r1r2=urms1urms2=3RT/M13RT/M2=M2M1. What is the mass of 235U in a sample of the diffused gas analogous to that in part A?

D.) After one more cycle of gaseous diffusion, what is the percentage of 235UF6 in the sample? (In your calculations unrounded values from previous parts should be used.)

Explanation / Answer

Tol solve this we can use the PV=nRT.

We can use this equation because the UF6 is in gas phase.

As we know n (number of mol) is equal to n=m/M

m is the mass and M is the molar weight.

With this we can substitute them and have the next equation.

PV=(m/M)RT.

If we need to determine the mass, we can clear m

m=PV/MRT

And solve the equation.

UF6 molar weight is 352.02 g/mol

m= [(690 torr)(30.0L)]/[(352.02 g/mol)( 62.3636 TorrL/molK)(360K)]

m=2.61x10-3 g of UF6

With this mass we can determine the mass of isotope U235

2.61x10-3 g (0.720/100)= 1.87x10-5 g of U235

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