Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: paul.fr on 10/07/2007 12:56:18
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Ever used a gas bottle for heat or a barbque? sometimes the bottom of the bottle will freeze up and ice form around it. why is that?
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Simple answer is Gay-Lussac's law - which simply states that pressure is inversely proportional to temperature (back to the bicycle pump for generating heat).
The gas in the bottles is under pressure, but the heat generated when the gas was placed under pressure has long since been dissipated. When you use up the gas, you reduce the pressure in the bottle, which then lowers the temperature, and hence the freezing.
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Simple answer is Gay-Lussac's law - which simply states that pressure is inversely proportional to temperature (back to the bicycle pump for generating heat).
The gas in the bottles is under pressure, but the heat generated when the gas was placed under pressure has long since been dissipated. When you use up the gas, you reduce the pressure in the bottle, which then lowers the temperature, and hence the freezing.
If we're talking about liquified gas, e.g. buthane, then the reason of the cooling is vaporisation heat: any substance in a condensed state require heat to transform into gaseous state. If the process is very fast, the heat that the vapour/gas has taken from the liquid (or solid, in the case of sublimation) cannot be taken immediately, in turn, from the environment, so the liquid has a fast reduction in temperature and it can breeze. This is the way to produce artificial snow, for example.
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I'm a long way from sure that it's anything to do with Gay Lussac's law. Ideal gases don't cool or heat when they expand. Real gases do but the effect is small and, in some cases (eg H2) the effect goes the other way. Hydrogen heats up when you let it out of a high pressure cylinder.
As lightarrow says the effect here is due to latent heat required to boil th eliquid butane.
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I am not doubting that the heat of evaporation is greater than the heat of expansion, but I don't see how you can argue that there is only minimal heat caused by compression - how else would diesel engines work?
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Heats of evaporation are much greater than heats of expansion or general warming it is the energy required to evaporate the butane that causes the bottom of the cylinder to get cold the you can prove this by noting that it is only the area covered by the liquid butane that gets cold. Thats why as the cylinder runs out this happens more often because there is less area for the cylinder to absorb heat
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"I am not doubting that the heat of evaporation is greater than the heat of expansion, but I don't see how you can argue that there is only minimal heat caused by compression - how else would diesel engines work?"
Good question. Nevertheless hydrogen warms up on expansion. The effect is due to non ideal behaviour and G-L's law is to do with ideal gases.
http://en.wikipedia.org/wiki/Joule-Thomson_effect
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Either way, the reason that Paul's gas cylinder gets frosty, and why his fridge keeps the burgers cold beforehand, is that gas under pressure is being expanded from the cylinder, so fresh liquid evaporates to replace the gas that left the cylinder.
As it does so it "expands" to form a gas and this requires energy, which is supplied by the environment. Hence the frost.
In the fridge, a compressed refrigerant gas is squirted under pressure into a high-volume condenser. As the gas expands it gets cold.
Chris
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Good question. Nevertheless hydrogen warms up on expansion. The effect is due to non ideal behaviour and G-L's law is to do with ideal gases.
http://en.wikipedia.org/wiki/Joule-Thomson_effect
Very interesting. It does not actually contradict what I believed, but merely places limiting parameters around it.
What is particularly interesting is that, if understand correctly, gasses (and/or combinations of gasses) can be caused to expand (in the right circumstances), or be compressed, with neither loss nor any gain in energy.
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Good question. Nevertheless hydrogen warms up on expansion. The effect is due to non ideal behaviour and G-L's law is to do with ideal gases.
http://en.wikipedia.org/wiki/Joule-Thomson_effect
Very interesting. It does not actually contradict what I believed, but merely places limiting parameters around it.
What is particularly interesting is that, if understand correctly, gasses (and/or combinations of gasses) can be caused to expand (in the right circumstances), or be compressed, with neither loss nor any gain in energy.
A gas cools when it expands making work on the external environment; if it expands without making work, then it can cool or heat, as Bored Chemist said, because of another effect (if I remember correctly it should be Joule-Thomson effect) due to the deviation from ideal of a real gas. An ideal gas does not vary its temperature on expansion without making work (free expansion).
In the case of the initial question, it's not actually a free expansion, because there is the external air, but there is a difference in pressure, so it's half free-expansion and half not.
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In a butane cylinder where there is a mix of liquid and gas the pressure is approximatelyconstant so there is very little cooling to the cylinder by the gas expanding in the cylinder the main place where the gas expands is in the nozzle of the cooker where it is burnt. As I said before the main cooling to a cylinder of butane while it is being used is the heat flow into the liquid butane to turn it into gas you can see that clearly because it is the part of the cylinder that is covered by liquid not the part that is in contact with the gas that gets cold.
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In a butane cylinder where there is a mix of liquid and gas the pressure is approximatelyconstant so there is very little cooling to the cylinder by the gas expanding in the cylinder the main place where the gas expands is in the nozzle of the cooker where it is burnt. As I said before the main cooling to a cylinder of butane while it is being used is the heat flow into the liquid butane to turn it into gas you can see that clearly because it is the part of the cylinder that is covered by liquid not the part that is in contact with the gas that gets cold.
I agree totally with you. I was just answering to George about why a gas cools when expands.
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Latent heat of vaporisation? Classical Physics rules. KE of the molecules in the liquid is lost in the process of liberating molecules from the surface into the vapor form.
Also, a gas 'cools' as it expands because there is work done (a pressure X volume change ) this energy comes from the KE of the molecules in the gas.
The average KE IS the temperature, so the temperature drops.
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Except, as has been pointed out, the pressure change inside the cylinder is small- the pressure really changes at the regulator and / or the jet. Talking about the temprerature change of the gas as it expands rather ignores the fact that it's the liquid that gets cold- that's why (as, again, has been pointed out) the tank ices up to roughly the level of the liquid.
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OK then, it's latent heat of vaporisation - the heat is lost at the surface. Heat enters the liquid butane from below - slowly because it's a liquid with poor conductivity(?) and through the steel case - a good conductor with a low specific heat capacity, compared with the butane - so it gets cold.
I have noticed that you can 'stimulate' a tired gaz cannister to release its butane by holding it in warm hands and shaking - adding heat / energy to the system.
On the other hand - there is a volume change; the volume that goes out into the burner. My equation(delta p X delta v) holds; work is done, one way or another - so energy is lost. The actual pressure won't change much because, as gas is lost through the regulator, it is released from the surface, to maintain equilibrium at the surface.
It's probably a bit of both - it's not an ideal gas but, either way, it doesn't contravene any of our beloved laws.
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OK then, it's latent heat of vaporisation - the heat is lost at the surface. Heat enters the liquid butane from below - slowly because it's a liquid with poor conductivity(?) and through the steel case - a good conductor with a low specific heat capacity, compared with the butane - so it gets cold.
I have noticed that you can 'stimulate' a tired gaz cannister to release its butane by holding it in warm hands and shaking - adding heat / energy to the system.
On the other hand - there is a volume change; the volume that goes out into the burner. My equation(delta p X delta v) holds; work is done, one way or another - so energy is lost. The actual pressure won't change much because, as gas is lost through the regulator, it is released from the surface, to maintain equilibrium at the surface.
It's probably a bit of both - it's not an ideal gas but, either way, it doesn't contravene any of our beloved laws.
Let's consider an ideal case: there is no air outside, you are in the outer space. Do you think there would still be cooling of the gas inside the cylinder, in that case?
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"Heat enters the liquid butane from below "
How does it know to do this? who tells the heat not to go in through the side?
Anyway, there's no question the PV work is done when the gas expands, but the pressure inside the cylinder is nearly constant (actually it will fall as the liguid gets cold) and the volume of the cylinder is also constant so, from the cylinder's point of view there isn't much of a PV change. Since it's the cylinder that gets cold and the PV change happens somewhere else (the regulator or whatever) you need something other than the PV change to explain the cooling of the cylinder. That makes it look to me like it's the latent heat change that cools the cylinder rather than the PV change (which, with a long tube, could be made to happen in the next county).
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How does it know to do this? who tells the heat not to go in through the side?
That's in the second half of my sentence!
I have thought about the thing a bit more.
There is energy transferred from the gas as it expands through the nozzle but, as molecules leave the surface and go into the gas, this internal energy is, effectively replaced. As you say, there is no net volume or pressure change; lost gas is replaced from the liquid surface. The energy comes from the vaporisation process so the liquid gets cooler.
If this were done in outer space (no sunlight either) , the temperature would drop and drop until the liquid temperature reached the boiling point. The rate of gas loss would then drop considerably.
I am not sure of the significance of lightarrow's outer space model - there would still be work done whether or not there were any air to push against. A rocket engine in space still does work on the expelled gases. The escaping gas molecules would be taking their KE with them - there would be a pressure gradient along the nozzle, so would there not be cooling on the way through?
During evaporation, it's the molecules with the highest KE that are lost - so the average KE (i.e. temperature) will drop.
I remember the old primus stoves (paraffin) used the heat from the burner flame to vaporize the paraffin on its way to the jet. In that case the vapour pressure at room temp is very low; with lpg it is marginal. Propane is better for external use in winter, I believe.
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An open container of butane in space would cool well below it's normal boiling point. I suspect that it would cool until it froze then it would continue to cool by sublimation. The pressure's mighty low out in space.