do you know science?

temp. Then open the outlet valve. Let the air

That's a new one on me, I don't keep up wioth the latest chemistry research.

But anyhow, that effect refers to the gas EXTERIOR to the container. The question here involves the gas remaining inside.

Though the Wikipedia article is poorly written, typical of that site. Which means you cited an incomprehensible reference, one which YOU uncomprehend.

The quiz is motivated by the ideal gas law. I alwways thought, one equation, multiple degrees of freedom, how the heck can you determine anything?

So with this pressurized water bottle, if we examine the formula: pv = nRT Considering the initial and final conditions, which variable(s) vary, on each side?

As expected, everyone believes a pressure drop must imply a temperature drop -

Perhaps the nozzle makes a difference, I don't know -

--
Rich
Reply to
RichD
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Hi Rich, So if you take two bottles, one full of gas and the other empty. And then you connect them and let the gas flow from one to other, then the gas filled bottle gets cold and the bottle that started empty gets hot. An d if you take both bottles as the whole system, then it looks like thermo p redicts that the temperature of the total system doesn't change. It seems like the exact temperature difference will depend on how the gas is transfe red.... after all I could always let the gas spin some generator and get en ergy out of the system.

George H.

Reply to
gghero27

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periment" is a trifle meglomaniac. Then again, our current exercise in digg ing up and burning every last scrap of fossil carbon we can find to raise t he CO2 level in the atmosphere is an even larger scale experiment.

r heads around the preliminary results, to the extent that they claim that it's not working.

ce for a Sydney audience to register scorn or derision.

I stumbled in this and immediately thought of Phil (and a few others) colorful language ;)

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-Lasse

Reply to
Lasse Langwadt Christensen

As another muddy point, the article discusses the temperature drop, as the expelled gas expands into some volume. In the example given, air escapes into the atmosphere. How does one measure the 'expansion', how do you even define it?

--
Rich
Reply to
RichD

James Prescott Joule died in 1889. William Thompson - Lord Kelvin - hung ar ound to 1907. The work was done in 1852, 160 years ago.

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The work was in thermodynamics, which is physics, rather than chemistry (th ough you will get taught about it if you study chemistry at university. I g ot taught about in my second year, in 1961. There's been quite a lot of stu ff discovered since then.

The thermodynamics covers that too.

That second year course on thermodynamics took me quite a while to comprehe nd, but I did get my head around it. Thermodynamics is notoriously difficul t to teach. The wikipedia article struck me as a well-written popularisatio n, but it's always difficult to gauge the quality of material that is teach ing people stuff you already understand.

That's only true for an ideal gas. The Joule-Thompson effect varies between gases, so it isn't covered by the ideal gas equation. In fact what's going on is that the van de Waals forces between the gas molecules (which decrea se as the 6th power of molecular separation) decrease as the gas expands, a nd this slows down the average speed of the gas molecules (which means that they are cooler) - for everything except hydrogen, helium and neon at room temperature. There's another effect which works the other way (as spelled out on the web-site) but it's smaller for your pressurised bottle of wet ai r at room temperature.

--
Bill Sloman, Sydney
Reply to
Bill Sloman

e, where it evaporates, this cools the rest of the water inside. Not sure what they are called, but I say them in Australia many moons ago. (I guess they work best in a dry climate.)

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--
Bill Sloman, SEydney
Reply to
Bill Sloman

ide, where it evaporates, this cools the rest of the water inside. Not sur e what they are called, but I say them in Australia many moons ago. (I gue ss they work best in a dry climate.)

Excellent, Thanks, me mum's from Perth in WA.

They hung 'em on the 'roo bars of their trucks. ('roo is short for kangaroo, which fill the same niche as deer in the US.)

George H.

Reply to
George Herold

That assumes no friction in the nozzles/pipe or other losses. If you include the universe as the "system", then there is no energy lost, either. ;-)

Reply to
krw

Interestingly, there's an important sense in which that isn't true.

In an expanding universe, if you have a galaxy that's moving a bit faster than it should be for its position, it will appear to have a very large kinetic energy. But as it moves with respect to other objects, it starts to (proportionally speaking) catch up to those with the same velocity.

Thus its idiosyncratic motion becomes less and less with time, so what you'd call its kinetic energy decreases without being dissipated or transferred anywhere else.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

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