Gas law help for a home project

In summary, the conversation discusses the concept of adding additional vapor to a fixed volume container holding gas under pressure, and the effects this can have on temperature and heat energy. The conversation also delves into the possibility of vapor condensation and the impact of pressure differentials on this process. There is also mention of constraints and simplifications that may affect the results, and the need for further research and analysis in order to determine the exact outcomes. Overall, the conversation highlights the complexity and variability involved in understanding and predicting the behavior of fluids under pressure and temperature changes.
  • #1
bitman
17
0
Hi

I have a fixed volume container which holds gas under pressure in a part liquid/part vapour form at normal temperatures.

If I add additional vapour to the container it should condense to increase the liquid level, however the vapour pressure should remain the same.

Would this cause any change of temperature in the vessel ? As the vapour is undergoing a change of state from gas to liquid is energy given off. I know that when vapour is removed heat energy is taken by the gas, does this mean that when vapour is added heat energy is given off to make the temperature increase?

Hope someone whose has a much better grasp of gas law physics can help me on this.

Many thanks in advance.

Bitman
 
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  • #2
Yes, when the liquid changes phase, there is a heat of transformation ( vaporization or evaporation ). The temperature in the vessel should decrease because the liquid absorbs heat in order to condense.
 
  • #3
So just to confirm. If I add vapour to the vessel some of the vapour will condense (to maintain the vapour pressure) and the temperature will fall.

This seems counter intuitive as when we have a barbeque the gas being used by the barbeque cause a drop in temperature of the cylinder. But in this case the liquid is turning to vapour.

I don't see how energy can be absorbed turning from gas to liquid and liquid to gas.

Can someone confirm which case is true.

Best Regards

Bitman
 
  • #4
Argh! I 'm going to jettison here. Someone answer
A) The temperature in the vessel should decrease because the liquid absorbs heat in order to condense.
B) The temperature in the vessel should decrease because the liquid releases heat in order to condense.
C) The temperature in the vessel should increase because the liquid absorbs heat in order to condense.
D) The temperature in the vessel should increase because the liquid releases heat in order to condense.




See Wiki: "Enthalpy of Vaporization"
 
  • #5
Hi Helios

Wiki says
"enthalpy changes of condensation are always negative (heat is released by the substance)."

but finding out by how much is not easy as the energy released varies with pressure and temperature (looking at the wiki page)

I think I'm going to have to do a great deal more research.

thanks

Bitman
 
  • #6
I know that evaporation causes a liquid to cool, so I expect condensation to cause a temperature increase.

The molecules lose energy upon condensation; vapor is a higher energy state than liquid. So it would make sense that this lost energy becomes heat.
 
  • #7
Hi
Thinking about it that makes sense. OK Now for the tricky question.

If the vapour that was to be added to the vessel was at a higher pressure, using an expansion nozzle the vapour would expand and absorb heat energy.

Is this true ?

If so how can I derive (Anyone derive ?) a formula which will tell me how much higher pressure is needed to offset the heating effect caused by condensation. I'm very happy to accept an approximate solution if one exists.

I know this is a lot to ask and whilst I did well in my O Level physics GCE, it was a long time ago (close to half a century). So a virtual beer to anyone who can assist me - on me.

Bitman

(AKA the Mad Mechanic)
 
  • #8
Hi bitman,
bitman said:
Hi

I have a fixed volume container which holds gas under pressure in a part liquid/part vapour form at normal temperatures.

If I add additional vapour to the container it should condense to increase the liquid level,
First problem is that you haven't specified some additional constraints on the system. Specifically, do you want to consider the fluid to be adiabatically compressed or not (ie: do you want to consider the fluid in question to be undergoing this process with no heat transfer). A second constraint is the thermal gradient throughout the fluid. For simplicity, we generally would like to consider the fluid to be at the same temperature throughout. However, this isn't generally realistic, and I’ll ignore this constraint in the response below. Another simplification is the fluid pressure gradient from the top on down. Fluid pressure increases as we go down through the fluid, so this can also cause thermal gradients and/or allow a gradual variation in the fluid's state. All of these are very real phenomena we see when dealing with cryogenic liquids in tanks for example.

I think at least for a real system, and for small variations in pressure, you are incorrect in assuming that some of the vapor will condense. On the other hand, we can consider an ‘ideal’ system in which the temperature of the fluid throughout is a constant. In this case, it may be that some of the fluid condenses.

I was surprised to find for example, that adding heat to an enclosed container might actually condense the fluid in some (rather extreme) circumstances. So I wouldn’t make any sweeping generalizations about what will happen in your case either. That said, this situation pops up quite a bit in the kind of systems I work on. In those cases, pressure rises, no vapor condenses, the liquid gets subcooled slightly (because of the increase in pressure), and the vapor warms to some degree. The liquid doesn’t warm up hardly at all (the amount of increase in temperature is negligable) however, the amount of increase in temperature of the gas can be significant in comparison. This may remain true for small variations in pressure, but for large variations, dT grows, and heat transfer between the gas and liquid will become a factor, possibly condensing some vapor over time. In such cases, it’s best to examine the system in question in more detail.

bitman said:
… OK Now for the tricky question.

If the vapour that was to be added to the vessel was at a higher pressure, using an expansion nozzle the vapour would expand and absorb heat energy.

Is this true ?
Don’t know yet. Might help to explain your device in detail, preferably with sketches to show what you’re trying to accomplish.
 
  • #9
I think that gas expansion of the injected vapour will cause the overall temperature to fall.
I also think that heat of condensation will cause the temperature to rise.
So that if the heat lost in expasion of the added vapour is greater than the heat gained by condensation the temperature will fall. If its less it will rise.

I also think there's a very good chance that I'm wrong.

I shall try some tests with butane (because its vapour pressures are quite low and I've got some :-)) to see If I can work out what's actually happening. I can control the pressure of the injected gas by putting the cylinder in a water bath and changing the temperature.

If anyones got any better ideas they'd be more than welcome.

Many thanks for the help.

Bitman
 

1. What is the ideal gas law?

The ideal gas law is a mathematical relationship that describes the behavior of an ideal gas. It states that the pressure of a gas is directly proportional to its temperature and number of moles, and inversely proportional to its volume. This law can be written as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

2. How can I use the ideal gas law for my home project?

The ideal gas law can be used to calculate the pressure, volume, temperature, or number of moles of a gas in a closed system, as long as the other variables are known. This can be useful for projects such as building a homemade air compressor or understanding the behavior of gases in a confined space.

3. What is the difference between an ideal gas and a real gas?

An ideal gas is a theoretical gas that follows the ideal gas law perfectly under all conditions. A real gas, on the other hand, may deviate from the ideal gas law at high pressures or low temperatures due to intermolecular forces. However, at moderate temperatures and pressures, most gases behave similarly to an ideal gas.

4. How do I convert between units in gas law calculations?

To convert units in gas law calculations, you can use the following conversion factors: 1 atm = 101.3 kPa = 760 mmHg = 760 torr and 1 mole = 22.4 L at STP (standard temperature and pressure). Make sure to use the appropriate units for pressure (atm, kPa, mmHg, or torr) and volume (L or mL) in your calculations.

5. Can the ideal gas law be applied to all gases?

No, the ideal gas law is only applicable to ideal gases, which are gases that have no intermolecular forces and occupy negligible volume compared to the container they are in. Real gases may deviate from the ideal gas law due to intermolecular forces, so it is important to consider the behavior of the specific gas in your project when using the ideal gas law.

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