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If you halve the pressure of a gas… You would help me with this problem

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posted on Aug, 6 2008 @ 06:31 PM
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Hello I have this great invention on the drawing board. I’ve even given a few inventions away on here before (some good and not so good). However this invention is for me only, but I still need your help.

My Question: If you halve the pressure of a (e.g. ideal) gas with a temperature x how much (preferably in percent) does its temperature fall?

The Context…
I double the temperature of some trapped C02 at 1atm. Therefore I double its pressure to 2atm (it’s a fact) Google: link: www.google.co.uk...
But I then decide to halve its pressure (by doubling its volume).
What’s happens to the temperature?

Surely it hasn’t halved? If that were the case then wouldn’t I have created a weird C02 that does not cool down (because halving would bring it to the outside temperature), yet it has half the density of the original C02 (since it occupies twice the space).

My Research…

Google: half-the-pressure temperature
Book: Elements of Physics; Or, Natural Philosophy, General and Medical published 1856
books.google.co.uk... mz3PonY&hl=en&sa=X&oi=book_result&resnum=9&ct=result#PPA278,M1
If you halve the pressure of air of 1atm you get a drop of about 50 degrees. This explains why mountains where the air is half are not minus 140.
So air at beach is 20 cenitigrade; migrates up, expands, gets cold
See snow line or line of perpetual congelation
But if you double its pressure you always double its temperature. Weird!!!

However with this book being from 1856 I would really appreciate if someone could tell me a bit more about what happens to a temperature when it’s pressure is halved; and if possible the process.

P.S My maths can make a monkey look educated, so it would be great if you would keep it to layman’s terms.

Sorry to have to ask about this; but my idea does work (using waste heat) (from a specific industry) to make electricity. It’s just that I would love to know a bit more about the efficiency of what I indeed have.
Thanks; Lib



posted on Aug, 6 2008 @ 07:19 PM
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Boyle's Law: At a constant absolute temperature and number of gas molecules, the pressure and volume of a gas are reciprocally related to one another. Thus a doubling in volume will halve the pressure.

Charles' Law: At a constant pressure and number of gas molecules, the volume occupied by the gas is directly proportional to the absolute temperature. Thus, doubling the absolute temperature will double the volume occupied by the gas.

You can only halve the pressure by doubling the volume if the temperature is absolute constant.

So if you do not maintain a constant temperature while doubling the volume you will not halve the pressure.

So to answer your question: if you double the volume to halve the pressure, the temperature MOST be constant.

Hope this helps - good luck



posted on Aug, 6 2008 @ 07:33 PM
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reply to post by Liberal1984
 



In a closed system temperature and pressure are directly proportional. Double or halve the pressure and you will double or halve the temperature (and vice versa). Remember that when applying the gas laws you have to use the Kelvin (absolute) scale. 20 c = 293 k. To double the temperature you would have to raise the temperature of the gas to 586k. That is equivalent to 313c, not 40.

There is a problem with your atmospheric example. It does not apply to your problem because it is not a contained system. In your invention ( a contained system) the amount of gas (number of molecules) is constant. In the open atmosphere it becomes another variable. Not only does pressure change with altitude, but molar volume does as well. The gas laws still apply but it is not simply a matter of temperature and/or pressure changing, there is more going on.

I may have messed up some of the details but the basic ideas are there. Hope it helps.




[edit on 6-8-2008 by Phage]



posted on Aug, 6 2008 @ 07:52 PM
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I myself, not being a scientist, it sounds like your experiment my share some parallels with the principles of refrigeration.

A compressor creating high pressure with a gas, then rapidly expanding the gas (or reducing the pressure) causes a latent drop in temperatures. The heat is then expelled outside, or in the case of a food refrigerator is radiated out the back of the appliance.

I would research these principles because they can be applied to something tangible, rather than theoretic laws.


edit: apparently using an extremely dense gas works best, most refrigerants, eg freon have a high relative density. Co2 is fairly close to air in density.
[edit on 6-8-2008 by wrathchild]
edit#2

Also research the technology used in "heat pumps" or geo thermal heating systems.

Being a plumber in Canada these are becoming quite popular. Using the earth's heat to heat houses.

The basic principle is pumping 45 degree water underground and raising it's temperature to 50 degrees. You take that large amount of 50 degree water and turn it into a small amount of 140 degree water and heat your home with it.
[edit on 6-8-2008 by wrathchild]

[edit on 6-8-2008 by wrathchild]



posted on Aug, 6 2008 @ 07:53 PM
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Originally posted by Liberal1984
I double the temperature of some trapped C02 at 1atm. Therefore I double its pressure to 2atm (it’s a fact) Google: link: www.google.co.uk...
But I then decide to halve its pressure (by doubling its volume).


So what you did is:

First, you have CO2 at 1atm.

1) You double its pressure to 2atm. Somehow. How did you do it? Did you half the volume?

2) Then, you decrease the pressure to 1atm by doubling the volume.


Temperature doesn't come into play here, unless you doubled the temperature in step 1 instead of halving the volume.


Basically... a gas is a bunch of particles. "Pressure" is basically the force exerted against the container its in. You do this by increasing the frequency of collisions by either raising the temperature (the molecules move more and therefore impact the container more) or decrease the volume (more molecules in a confined space).



posted on Aug, 6 2008 @ 08:10 PM
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here's a few numbers...

to change 1 pound of CO2 liquid to 1 pound CO2 gas takes 247 BTU's and no temperature change.

This also works in reverse.

BTU= amount of energy or heat, to raise 1 lb of water 1 degree fahrenheit



posted on Aug, 6 2008 @ 09:22 PM
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Firstly guys thanks for the most excellent response!!!
I just posted some ideas on forum for you to look at (or have).

But I'm little closer to solving the problem; it’s my fault...

Here’s what I do…
I take 1 cubic metre of air; at 1 atmosphere at 300 Kelvin (26.85 degrees).
I then heat it (in the same space) to 600 Kelvin (326.85 degrees)
Therefore its pressure doubles to 2atm
I then allow this hot air to push back a mobile barrier until the hot airs pressure returns to 1atm. I visualise this would require the airs volume to increase to 2M3.
But I add no more heat during the expansion process.
So I’ve been told that this act of decompression will cause the airs temperature to drop?

But by how much?

Originally posted by Bluess

You can only halve the pressure by doubling the volume if the temperature is absolute constant.


Does this mean (in the process above) the temperature will be the same as it was before it was decompressed?
Or does it mean (since I add no extra heat during the expansion stage) that the hot cubic meter at 2atm will refuse to expand (under it’s own pressure) to 2 M3?
If so at what size would its volume stop expanding (and reach equilibrium with the outside (1atm) pressure?) And what would its temperature be?

Really sorry for my own thickness; I'm quite intelligent (but a bad learner).
If only you knew (actually its great you don’t know) how much mental agony this question has caused me!!!
So I cannot describe how happy I’ll be towards the first person who puts this problem at rest.

Hope this answers your questions Johnmike
And a very big thank you also to Phage and wrathchild

[edit on 090705 by Liberal1984]



posted on Aug, 6 2008 @ 10:12 PM
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the exact amount of energy applied to the air in order to create the expansion will also be lost when the area heated is allowed to decompress.

This assuming it's theoretical and there is no losses from radiation, convection and conduction.

The act of decompression will consume the energy that was originally applied. when the increased pressure moves back the mobile barrier that would require a consumption of energy(heat). that would intern lower the amount of energy(heat) that was originally applied to the experiment.



posted on Aug, 6 2008 @ 10:32 PM
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It does make one's head spin a bit doesn't it?

Allowing the pressure to return to 1atm by increasing the volume will cause the temperature to return to 26.85. But increasing the volume is not the same as "allowing" the hot air to push a barrier (piston?). In order to move the piston and anything attached to it energy will be expended (First Law). The result being that the volume will not expand to exactly 2m3 but the temperature will return to 26.85. The extent to which the expansion will occur depends on the amount of work being done. The harder the piston is to push, the lower the ultimate expansion, even if it is only pushing against atmospheric pressure.

If, on the other hand, the piston is pulled by an outside agent, the temperature will return to 26.85. It will go below that if the piston is pulled so that volume is more than doubled.

For the purposes of this discussion we assuming the vessel is "perfectly" insulated.



posted on Aug, 6 2008 @ 11:51 PM
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Phage

For the purposes of this discussion we assuming the vessel is "perfectly" insulated.

Yes. It’s easy to deduct loses from thermal conductivity ect afterwards.

Originally posted by wrathchild

the exact amount of energy applied to the air in order to create the expansion will also be lost when the area heated is allowed to decompress.


That’s what I thought but then you encounter my first problem.
If the temperature in the new 2M3 block of air has fallen back to 26.85 degrees (300Kelvin) we have a very weird product. Weird because whatever gas is inside the 2M3 block cannot cool down (well not today anyway as its temperature is the same as the outside temperature).
Yet this gas has half the density of ordinary outside air (since it came from 1M3 heated 2M3). Therefore the stuff would be like helium.
Surely not right?
I mean say we were to brake open the 2M3 container and release this gas; would we have just added 2M3 of gas to the earths atmosphere (having only taken 1M3 from it)?
If we have added 2M3 then how long will it increase the earths atmosphere? Say it cools from 26.85 to 25.85 how much would the gas shrink?
(One would have thought it would reach 1M3 at least 150Kelvin)
But perhaps it’s not worth asking as the gas never cools down?
But if the stuff never cools down then we have expanded a gas without a temperature drop.

Glad you seem to see the dilemma!!! It’s been driving me nuts


In order to move the piston and anything attached to it energy will be expended (First Law). The result being that the volume will not expand to exactly 2M3 but the temperature will return to 26.85.


Sure.
For simplicity let’s just assume all the kinetic energy is absorbed outside the 2M3 box. (Happens in nature very quickly anyway).

Extra Maths (But only if it’s useful…)
The specific heat capacity of air is near 1KJ per Kg.
www.engineeringtoolbox.com...
1 Meter cube of air weighs near 1.2 Kg
So to raise the temperature by 300 degrees-Kelvin is 1.2KJ times 300 equals 360KJ

Work Done By Expansion…
The atmosphere weighs about 10tonnes. 1 tonne is 1000KG. To lift 1Kg 1 Meter takes about 10Joules. So 10,000 Joules per tonne meter, which gives 100,000 Joules of work done in total. Don’t ask where the other 260KJ in heat goes.

Finally…
Before posting I was going to suppose that (being an enclosed system) the hot gas gets to expand (under its own pressure) without losing its temperature. I reasoned fridges work by losing heat through compression first, and transferring this “cold” by expansion 2nd.
All seems cool.
And it does (to me) seem one of the best answers. Is this really wrong?
Except: The reason why the tops of mountains (in deserts) are snowy (despite being closer to the sun lights strength than things on the ground) is supposed to be because the air pressure up there is lower. Therefore hot air blown from the ground expands on its way up, and loses temperature (causing snow even in deserts).

Therefore one would think my 1M3 of air becoming 2M3 also loses temperature?
But that entails air becoming lighter than air; without being hot, and without nuclear (reactor) physics.

As I said…
Welcome to my dilemma.
Please help!
Please help!
Please help!



[edit on 090705 by Liberal1984]



posted on Aug, 7 2008 @ 12:04 AM
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Originally posted by Liberal1984
But I add no more heat during the expansion process.
So I’ve been told that this act of decompression will cause the airs temperature to drop?

But by how much?

No, it will not. You're not really "decompressing." You're just kind of picking it apart - all that's happening is that the gas is expanding due to a temperature increase, and allowing the pressure to remain constant.

You're seeing it in a way that's too unified. There's the unified gas law (our beloved PV=nRT), but in reality you should see it as only two things happening at a time. Keep one constant. You increase temperature, volume increases. Pressure is constant.'

So, one step at a time, for fun and understanding. Science!

Though I'm going to remove units to make it easier. Breaking some sacred rule of chemistry, I know.


Originally posted by Liberal1984
I take 1 cubic metre of air; at 1 atmosphere at 300 Kelvin (26.85 degrees).

V = 1
P = 1
T = 300




Originally posted by Liberal1984
I then heat it (in the same space) to 600 Kelvin (326.85 degrees)
Therefore its pressure doubles to 2atm

V = 1
P = 2
T = 600


Temperature is doubled (x2) by outside influence. Volume is forced to remain constant due to rigid container. Therefore, pressure doubles (x2) to compensate.


Originally posted by Liberal1984
I then allow this hot air to push back a mobile barrier until the hot airs pressure returns to 1atm. I visualise this would require the airs volume to increase to 2M3.

V = 2
P = 1
T = 600


Container's size is doubled (x2) by outside influence. Gas, being a gas, expands to fill the container. Therefore, exerted pressure is halved (x0.5) to compensate for the change in volume. Temperature remains constant.



posted on Aug, 7 2008 @ 12:43 AM
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Originally posted by Johnmike
Container's size is doubled (x2) by outside influence. Gas, being a gas, expands to fill the container. Therefore, exerted pressure is halved (x0.5) to compensate for the change in volume. Temperature remains constant.


Doh!

Keep it simple stupid!

I stand sheepishly corrected.



posted on Aug, 7 2008 @ 12:47 AM
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I will come back from science class, and I will let you know all the details.



posted on Aug, 7 2008 @ 01:48 AM
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reply to post by Liberal1984
 

I take 1 cubic metre of air; at 1 atmosphere at 300 Kelvin (26.85 degrees).
I then heat it (in the same space) to 600 Kelvin (326.85 degrees)
Therefore its pressure doubles to 2atm
I then allow this hot air to push back a mobile barrier until the hot airs pressure returns to 1atm.

Congratulations!

You've invented the steam engine!



posted on Aug, 7 2008 @ 02:47 PM
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i'd like to point you to a website:

www.grc.nasa.gov...

you can search the terms given there and find better, ie. more understandable explanations or just tkae the end results and plug in the numbers, whatever floats your boat.


MBF

posted on Aug, 7 2008 @ 10:37 PM
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Originally posted by Johnmike



Originally posted by Liberal1984
I then heat it (in the same space) to 600 Kelvin (326.85 degrees)
Therefore its pressure doubles to 2atm

V = 1
P = 2
T = 600


Temperature is doubled (x2) by outside influence. Volume is forced to remain constant due to rigid container. Therefore, pressure doubles (x2) to compensate.


Originally posted by Liberal1984
I then allow this hot air to push back a mobile barrier until the hot airs pressure returns to 1atm. I visualise this would require the airs volume to increase to 2M3.

V = 2
P = 1
T = 600


Container's size is doubled (x2) by outside influence. Gas, being a gas, expands to fill the container. Therefore, exerted pressure is halved (x0.5) to compensate for the change in volume. Temperature remains constant.


Wouldn't the temperature go down when you increase the volume and decrease the pressure unless you continue to add energy to the system?



posted on Aug, 7 2008 @ 10:54 PM
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Originally posted by MBF
Wouldn't the temperature go down when you increase the volume and decrease the pressure unless you continue to add energy to the system?

No. The increase in volume is enough to fully account for the decrease in pressure, and vice versa.



posted on Aug, 8 2008 @ 10:11 PM
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Volume = 1
Pressure = 2
Temperature = 600K

Volume = 2
Pressure = 1
Temperature = 600K


Thank you so much Johnmike!!!
(As well as everybody else)

But doesn’t what you say raise a problem with nature; it goes: “Why are Mountains in hot locations cold?”
I was told it’s because if a say cubic metre of air at 300K (and ground level) is blown up by wind to say half the earth atmospheres height; then it expands by half; and causes a temperature reduction by half.

But according to what you say this should cause no change at all.

But perhaps this is just simplified classroom **** a teacher can give to shut kids up?

For starters I guess the surface to volume ratio with freezing cold space, and the air up there, is much higher than on the ground. (Not to mention the atmosphere having a far greater circumference than the earths surface).

So is it something like this instead that makes mountains in deserts cold places?

In which case thanks very, very much for you’re time and effort in putting my mind to rest. It’s because mountains are cold places, and not being able to do much maths (correctly) that I couldn’t get heads or tails of this problem.

I reckon I should add another idea away when I next log in; but I might offer it to you first. If it's any good?

Anyway thanks again.

Originally posted by Astyanax

Congratulations!
You've invented the steam engine!


Get real!!! I ain’t no crank!! You might of missed it but I did say I wouldn’t share what I’ve got; but yes the example might apply to “a hot air steam engine” type thing.

[edit on 090705 by Liberal1984]



posted on Nov, 22 2010 @ 07:33 PM
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I was wrong.


Well, sort of. I assumed isothermal conditions -- that the environment was sharing heat with the container. I ignored the case where the container is adiabatic, as Phage wanted to consider (and I stupidly ignored).

Basically, you were right. Expansion takes energy, which is expended in the form of work. During isothermal expansion, it is as I said above. The gas will expand an external pressure (which is obviously less than the pressure of the gas in the container/piston/whatever). This does work, so you lose energy. In this case, the energy goes down by the amount of work done, which in this (irreversible) case, is the external pressure times the change in volume. Assuming constant external pressure, it's just P*dV.

As a do-over:


Originally posted by Liberal1984
I take 1 cubic metre of air; at 1 atmosphere at 300 Kelvin (26.85 degrees).

V = 1
P = 1
T = 300


That's 1L = 1dm^3 = 0.001m^3, 1atm = 101.3KPa = 101300Pa, and 300K.


Originally posted by Liberal1984
I then heat it (in the same space) to 600 Kelvin (326.85 degrees)
Therefore its pressure doubles to 2atm

V = 1
P = 2
T = 600


1L = 1dm^3 = 0.001m^3, 2atm = 202.6KPa = 202600Pa, and 600K.

What happened here? At constant volume, heat (denoted here by q) was added to the system. As volume did not change, no work was done. So the total energy of the system increased by q, or the temperature that was used to increase the enthalpy and therefore temperature of the system. We can calculate the amount of energy that was needed to increase the system's temperature by 600K - 300K = 300K with the equation q = Cv * dT, where q is again heat added, dT is the change in temperature (300K), and Cv is a constant which depends on the gas being heated. So q = 300K*Cv. If the gas is H2, Cv = 20.18J/(mol*k), so q = 300K*20.18J/(mol*k), = 6054J = 6.054kJ.


Now, the gas is going to do work by expanding against an external pressure until its pressure is equal to the external pressure. External pressure is denoted Pex.


Originally posted by Liberal1984
I then allow this hot air to push back a mobile barrier until the hot airs pressure returns to 1atm. I visualise this would require the airs volume to increase to 2M3.

V = 2
P = 1
T = 600


And here is where I made my mistake. One of two things may happen, depending on whether or not the gas is insulated -- 1) The gas is not insulated, giving us isothermal expansion or 2) The gas is insulated, as Phage wished to consider, giving us adiabatic expansion.

For both cases:
Work is done by the gas. W = -Pex*dV. Since it expands until it reaches 1atm = 101.3kPa, W = -101.3kPa*(2L-1L) = -101.3kPa*1L = -101300Pa*0.001m^3 = -101.3J. 101.3J is lost by the gas to do work.


Case 1)
Isothermal Expansion: Since the material is not insulated, the loss of energy due to work will be compensated by the environment, maintaining the temperature of the gas. The total energy of the system will not change: q = -w = +101.3J. Therefore, there is no temperature difference.




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