Thecatch83
I smoke Opus like candy!
So if you place your digital hygrometer on the top tray, or down on the bottom near the humidifcation source.....wheat reading would be more correct, higher or lower in RH?
Assuming the the hygrometers are correctly calibrated they are both correct. They are correctly reading the humidity at their location. See Ron's post (top of the page) up a few about temp. differences and the resulting humidity differences. That is why is is helpful to have a fan to stir or mix the air.So if you place your digital hygrometer on the top tray, or down on the bottom near the humidification source.....wheat reading would be more correct, higher or lower in RH?
I can't argue with that other than to say that it would be just about impossible to actually test this. First of all, Avogadro's law deals with "ideal gases". Second, it would be just about impossible to introduce beads into completely dry air and seal the system before any moisture is released. I think we are at an impasse. Avogadro's law is certainly accepted as true, as is the one you reference. I'll have to see if I can find some more info that might clear things up.Ok, just to give you something to ponder with your next cigar. :thumbsup: Assume that the cigars are slightly under-humidified and the box starts with completely dry air, and some properly hydrated beads. We know that the beads will release moisture to the air which in turn is absorbed by the sticks. With this scenario, the USA Today quote predicts total failure. As no water would be able to go into the air, and not then be available for the sticks.
We can argue over my use of the word "light" and "heavy" as those are really not the best terms when talking about density, but I can't find anything that refutes my statement that humid air is lighter (less dense) than dry air.The density of a mixture of dry air molecules and water vapor molecules is:
D =\frac{P_d}{R_dT} + \frac{P_v}{R_vT}
D = density, kg/m3
Pd = pressure of dry air, Pascals
Pv = pressure of water vapor, Pascals
Rd = gas constant for dry air, J/(kg*degK) = 287.05
Rv = gas constant for water vapor, J/(kg*degK) = 461.495
T = temperature, degK = deg C + 273.15
As you can see, the largest value for D will result when Pv is zero.
:rofl:You getting all this, JNT? The test will count for 30% of your grade...
Jason,Randy,
We can argue over my use of the word "light" and "heavy" as those are really not the best terms when talking about density, but I can't find anything that refutes my statement that humid air is lighter (less dense) than dry air.
Thanks stroke !:rofl:
All I know is OB couldn't have chosen a better avatar.
Isn't this exactly the opposite of what you've said earlier, or am I misunderstanding?One of the important predictions made by Avogadro is that the identity of a gas is unimportant in determining the P-V-T properties of the gas. This behavior means that a gas mixture behaves in exactly the same fashion as a pure gas.
[ame="http://en.wikipedia.org/wiki/Humidity"]Humidity - Wikipedia, the free encyclopedia@@AMEPARAM@@/wiki/File:Cloud_forest_mount_kinabalu.jpg" class="image"><img alt="" src="http://upload.wikimedia.org/wikipedia/commons/thumb/8/83/Cloud_forest_mount_kinabalu.jpg/220px-Cloud_forest_mount_kinabalu.jpg"@@AMEPARAM@@commons/thumb/8/83/Cloud_forest_mount_kinabalu.jpg/220px-Cloud_forest_mount_kinabalu.jpg[/ame]Humid air is less dense than dry air because a molecule of water (M ≈ 18 u ) is less massive than either a molecule of nitrogen (M ≈ 28) or a molecule of oxygen (M ≈ 32). About 78% of the molecules in dry air are nitrogen (N2). Another 21% of the molecules in dry air are oxygen (O2). The final 1% of dry air is a mixture of other gases. For any gas, at a given temperature and pressure, the number of molecules present in a particular volume is constant - see ideal gas law. So when water molecules (vapor) are introduced into that volume of dry air, the number of air molecules in the volume must decrease by the same number, if the temperature and pressure remain constant. (The addition of water molecules, or any other molecules, to a gas, without removal of an equal number of other molecules, will necessarily require a change in temperature, pressure, or total volume; that is, a change in at least one of these three parameters. If temperature and pressure remain constant, the volume increases, and the dry air molecules that were displaced will initially move out into the additional volume, after which the mixture will eventually become uniform through diffusion.) Hence the mass per unit volume of the gas—its density—decreases. Isaac Newton discovered this phenomenon and wrote about it in his book Opticks.[2]
For everyone else we work inside and gases are usually contained in containers, like our traveldor or humidors. With these containers the volume is fixed and cant change and we hope that they dont leak too badly. So the increase in moles from the addition of water must change something else. That becomes the pressure. Pressure is not hard to understand, just think of it as the mass of all the air above your head pushing down on you. At sea level this pressure supports 760 mmHg (millimeters of mercury). Your body and everything else must push back with an equal force, or collapse. From the Daltons law link you provided:So when water molecules (vapor) are introduced into that volume of dry air, the number of air molecules in the volume must decrease by the same number, if the temperature and pressure remain constant. (The addition of water molecules, or any other molecules, to a gas, without removal of an equal number of other molecules, will necessarily require a change in temperature, pressure, or total volume; that is, a change in at least one of these three parameters. If temperature and pressure remain constant, the volume increases, and the dry air molecules that were displaced will initially move out into the additional volume, after which the mixture will eventually become uniform through diffusion.)
So for us pressure changes. Of the 760 mmHg of pressure 78% is due to nitrogen and 21% oxygen, so essentially 593 mmHg for N and 160 mmHg for O. The equilibrium water vapor pressure at a temp of 70 deg F is 47mmHg. This would be the reading for water at 100% RH. So we are shooting for 30.5 mmHg of water (65% RH) for our stogies. Not much of an increase in the overall pressure of the system inside our container.All molecules in the gas have access to the entire volume of the system, thus V is the same for both nitrogen and oxygen. Similarly, both compounds experience the same temperature.
P = P nitrogen + P oxygen
The above equation is called Dalton's Law of Partial Pressure, and it states that the pressure of a gas mixture is the sum of the partial pressures of the individual components of the gas mixture.