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[deleted] t1_j37xy5g wrote

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Movpasd t1_j3bx7ym wrote

> How the temperature relates to other state variables is different (and difficult!) in liquids but that doesn't apply to kinetic energy

Is there a simple explanation for why this is the case? Given the presence of intermolecular potentials (which are not quadratic terms), I wouldn't expect equipartition to hold. Is the argument that this effect is negligible, and if so, how does one argue that it is?

Furthermore, does your calculation account for vibrational and rotational modes?

If you could point me to sources that cover these questions, I'd be very grateful. Thanks.

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ForeverInQuicksand t1_j386kr9 wrote

What if you took a 2ft pipe, that can be capped at both ends, and placed a valve you could open and close in the middle. Then you could place a small valve that allows a drop at a time to fall on both ends.

If you filled the left side with pure deuterated water, and the right side with pure water made with only normal hydrogen-1, and then opened the valve in the middle, while simultaneously collecting and isolating single drop samples of the water at each end of the tube over time.

By testing the samples in a mass spectrometer, wouldn’t it be possible to measure the deuterated water composition of each drop to see how long it would take both sides of the tube to release drops of the same d2O/H2O composition.

If the water molecules are distributing at a rate of 500m/s, there would be near instantaneous mixing of the two water types, as soon as the two samples touched.

I don’t think that would be the case.

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LiveNeverIdle t1_j388gxw wrote

The individual molecules travel that fast, but soon bump into other molecules. Those other molecules get bumped and speed off into still other molecules. So the molecules move very fast but don't travel very far. Eventually all of the water would mix though, which we call diffusion.

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EPIKGUTS24 t1_j39wqup wrote

It'd also mix via convection as deuterated (heavy) water is, well, heavier than regular water.

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ELDOR-King t1_j395hq5 wrote

the individual molecule speed is not necessarily the same as diffusion, as the movement is disordered. you can very easily measure this diffusion with NMR. (provided you have an NMR with pulsed field gradients. No need for valves or deuterium labelling etc.)

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