A gas centrifuge is a long hollow cylinder of radius 0.12 m that spins about its own axis, as shown…

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A gas centrifuge is a long hollow cylinder of radius $0.12\ \text{m}$ that spins about its own axis, as shown in the figure. Its wall moves at $400\ \text{m/s}$ . It holds uranium hexafluoride gas, a mixture of two kinds of molecule: $^{238}\text{UF}_6$ with molar mass $0.352\ \text{kg/mol}$ and $^{235}\text{UF}_6$ with molar mass $0.349\ \text{kg/mol}$ . The gas turns with the cylinder as a whole, at a steady $320\ \text{K}$ everywhere. Treat each kind of molecule as an ideal gas on its own, and ignore gravity. Take $R = 8.314\ \text{J/(mol K)}$ .

The end view of a hollow cylinder of radius 0.12 m spinning about its axis, which is marked by a dot at the centre. A curved arrow shows the sense of rotation, and an arrow along the rim shows the wall moving at 400 m/s. The cylinder is filled with uranium hexafluoride gas at 320 K.

For the $^{238}\text{UF}_6$ alone, what is the ratio of its pressure at the wall to its pressure on the axis?
The separation factor is the ratio of the number of $^{235}\text{UF}_6$ molecules to $^{238}\text{UF}_6$ molecules near the axis, divided by the same ratio near the wall. What is it?
What wall speed would give a separation factor of $1.20$ in one machine at the same temperature?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find how the pressure of one gas grows from the axis to the wall, then compare the two kinds of molecule.

Step 1. The pressure law. A thin ring of gas at radius $r$ moves in a circle. The net push from the pressure inside and outside it supplies the centripetal force, so $\dfrac{dp}{dr} = \rho\omega^2 r$ . With $\rho = \dfrac{pM}{RT}$ :

\displaystyle \frac{dp}{p} = \frac{M\omega^2}{RT}\,r\,dr \quad\Rightarrow\quad \frac{p(r)}{p(0)} = \exp\left(\frac{M\omega^2 r^2}{2RT}\right)

At the wall $\omega r$ is the wall speed $v_w$ .

Step 2. The heavy gas (a). The top line is $0.352\ \text{kg/mol} \times 160000\ \text{m}^2/\text{s}^2 = 56320\ \text{J/mol}$ , so $\dfrac{Mv_w^2}{2RT} = \dfrac{56320\ \text{J/mol}}{5321.0\ \text{J/mol}} = 10.585$ , so the ratio is $e^{10.585} = 3.95 \times 10^{4}$ . Nearly all the gas is pressed against the wall.

Step 3. Separation (b). For each kind the number density follows its own partial pressure. Dividing the two laws,

\displaystyle \alpha = \exp\left(\frac{(M_{238} - M_{235})v_w^2}{2RT}\right) = \exp\left(\frac{0.003 \times 160000}{5321.0}\right) = e^{0.0902} = 1.0944

The lighter molecules gather a little more towards the axis.

Step 4. A better machine (c). Set $\alpha = 1.20$ : $v_w^2 = \dfrac{2RT\ln 1.20}{\Delta M} = \dfrac{5321.0 \times 0.1823}{0.003}\ \text{m}^2/\text{s}^2$ , so $v_w = 569\ \text{m/s}$ .

Answer: (a) about $3.95 \times 10^{4}$ . (b) $1.0944$ . (c) $569\ \text{m/s}$ .

Check. The separation depends only on the difference of the molar masses, which is why a small difference needs a very fast wall. The temperature enters as $RT$ , the scale of the molecules' energy, against $\tfrac12\Delta M v_w^2$ , the extra energy the turning gives the heavier molecules. This is the same law as the density of an isothermal atmosphere, with the centrifugal potential in place of gravity.

Watch out. The two gases do not share one pressure law. Using the average molar mass for both gives no separation at all.

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