November 24, 2024, 06:57:54 PM
Forum Rules: Read This Before Posting


Topic: Gas + Liquid Experiment  (Read 7619 times)

0 Members and 1 Guest are viewing this topic.

Bhavman2

  • Guest
Gas + Liquid Experiment
« on: March 25, 2006, 08:32:46 PM »
Hi,

My teacher set up this display/experiment. The set up was simple, it was basically a piece of wood, and a tube attached in a U shape.
There was liquid in the tube, both sides of the tube where open at the top.

(a drawing is attached)

Day 1- The liquid level on the right was higher than the left water level
Day 2- Both liquid level's were equal
Day 3- The liquid level on the right was lower than the left water level


The teacher admitted to have "helped" it along but the rest had to be explained in terms of PV=nRT


Any help will be welcome... Thanks
 :)
« Last Edit: March 26, 2006, 03:21:40 PM by Bhavman2 »

Offline mike

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 1245
  • Mole Snacks: +121/-35
  • Gender: Male
Re:Gas + Liquid Experiment
« Reply #1 on: March 25, 2006, 08:40:57 PM »
Are you sure if was open at both ends?
There is no science without fancy, and no art without facts.

Bhavman2

  • Guest
Re:Gas + Liquid Experiment
« Reply #2 on: March 25, 2006, 09:14:00 PM »
Yes. I thought of in terms of a manometer. It could have been somehow measuring the pressure in the room, or of some gas. Well, atleast that was my  crazy thoery until i was told that both sides were open.   :-\

Offline billnotgatez

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 4431
  • Mole Snacks: +224/-62
  • Gender: Male
Re:Gas + Liquid Experiment
« Reply #3 on: March 25, 2006, 09:58:25 PM »
Is the teacher putting in a heavy gas that takes some time to dissipate before class? Maybe alternating sides after dissipation.

Otherwise I think he is putting a clear stopper in one end or the other giving an appearance of open ends at a distance.  This is an old magic trick.


Thinking about it again, I am not sure the heavy gas thing would work.
« Last Edit: March 25, 2006, 10:02:23 PM by billnotgatez »

groovymango21

  • Guest
Re:Gas + Liquid Experiment
« Reply #4 on: March 25, 2006, 11:25:25 PM »
so ok, right:

pressure= P, volume =V, and temperature =T

PV=nRT

For the pressure and temperature, if both ends of the tube were open
you could say it was standard pressure and temperature.

http://searchsmb.techtarget.com/sDefinition/0,,sid44_gci539342,00.html

both ends of the tube could not  have been open in this experiment though, the water on sides of the tube would have been the same. Pressure on one side is diffeernt to the other for a reason here.
Are you sure you have said everything?

Bhavman2

  • Guest
Re:Gas + Liquid Experiment
« Reply #5 on: March 26, 2006, 02:59:40 PM »
It is exactly how it was set up, and also there was no clear plug involved... i have NO IDEA how it was possibleeeee  ???

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27861
  • Mole Snacks: +1813/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Gas + Liquid Experiment
« Reply #6 on: March 26, 2006, 05:22:36 PM »
1. What was the liquid - water? Are you sure it was homogenous?

2. What was the level difference?

3. What was the tube diameter?

4. What is the height of the empty part left above the liquid level?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Bhavman2

  • Guest
Re:Gas + Liquid Experiment
« Reply #7 on: March 26, 2006, 05:30:55 PM »
1. We were told not to worry about the liquid but i do believe it plays some role in this
2. The difference was about 1/2 inch
3. The tube diameter was 1/2 inch
4. 5 Inches



Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27861
  • Mole Snacks: +1813/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Gas + Liquid Experiment
« Reply #8 on: March 26, 2006, 06:17:35 PM »
1. We were told not to worry about the liquid but i do believe it plays some role in this
2. The difference was about 1/2 inch
3. The tube diameter was 1/2 inch
4. 5 Inches

So the ratio of gas hight inside the tube to water level difference was about 10. Pressure can be calculated as dh (density times column height). To replace 1/2 inch of water with a gas one need to use column of that gas that is higher then the column of water in the ratio of their densities. This means one will need gas with density 10 times lower than water, about 100 g/L. The most dense gas I am aware off is UF6 - over 6 times too light.

I was also thinking about cooling and heating tube ends to change the density of air inside, but similar thinking as presented above shows that it is not possible to obtain differences large enough with so short tube.

No idea how it was done, unless your teacher was cheating ;)
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links