- 05 Sep '07 20:49An early scientist, like around 1850, decides to make a vacuum, so he figures he will start with a piston that will pull a slider out of a chamber and therefore reduce the air pressure in the chamber. So the chamber has a volume of one cubic meter, one meter on all sides and one of the sides is a slider with a perfectly good seal so if you move it out, the pressure goes down. So its a cylinder shaped container and you can see if you move the slider so the air pressure is cut in half, you have now two cubic meters of space inside the vacuum chamber.

How long does the piston have to be, that is to say, how far does the slider have to open up to get to one Milli-Torr of vacuum reading? - 05 Sep '07 21:05 / 1 edit

Assuming:*Originally posted by sonhouse***An early scientist, like around 1850, decides to make a vacuum, so he figures he will start with a piston that will pull a slider out of a chamber and therefore reduce the air pressure in the chamber. So the chamber has a volume of one cubic meter, one meter on all sides and one of the sides is a slider with a perfectly good seal so if you move it out, the ...[text shortened]... t is to say, how far does the slider have to open up to get to one Milli-Torr of vacuum reading?**

- the air follows the Ideal Gas Law

- the air starts at average ambient conditions

- the process is isothermal

I get the following:

P1*V1 = n1*R*T1

n1 = P1*V1/R*T1 = (101.3 x 10^3 Pa) * (1 m3) / (8.314 J/mol*K) * (273.15 K)

n1 = 44.61 mol

V2 = n1*R*T1/P2 = (44.61 mol) * (8.314 J/mol*K) * (273.15 K) / (0.133 Pa)

V2 = 7.60 x 10^5 m3

So if the cross section was originally 1 m2, the chamber would have to be about 760 km long. - 05 Sep '07 21:22

I do this sort of problem like this:*Originally posted by PBE6***Assuming:**

- the air follows the Ideal Gas Law

- the air starts at average ambient conditions

- the process is isothermal

I get the following:

P1*V1 = n1*R*T1

n1 = P1*V1/R*T1 = (101.3 x 10^3 Pa) * (1 m3) / (8.314 J/mol*K) * (273.15 K)

n1 = 44.61 mol

V2 = n1*R*T1/P2 = (44.61 mol) * (8.314 J/mol*K) * (273.15 K) / (0.133 Pa)

V2 = 7.60 x 10^5 ...[text shortened]...

So if the cross section was originally 1 m2, the chamber would have to be about 760 km long.

P1V1=n1RT1

P2V2=n2RT2

R= P2V2/n2T2 = P1V1/n1T1

n1=n2

P2V2/T2=P1V1/T1

Assuming STP, We know T1, P1 and P2, so

P2T1/V1P1=T2/V2

Assuming an isothermal process (I would have missed this realization)

T1=T2

P2/V1P1 = 1/V2

V2=P1V1/P2

Now just plug in the numbers. - 06 Sep '07 01:52

Oh yes, i did indeed read the post above yours. I was just making sure that you understood the dynamics of this rather easy mathematical, ah, what do ya call it. I had a vaccuum cleaner that just needed the drive belt replaced. What's with all the math formulas?*Originally posted by AThousandYoung***Ideal Gas Law. Did you read the post above mine?** - 06 Sep '07 03:45 / 3 edits

The problem in this equation is solved through high school level physics or chemistry, or at worst, college level introductory chemistry courses.*Originally posted by smw6869***Oh yes, i did indeed read the post above yours. I was just making sure that you understood the dynamics of this rather easy mathematical, ah, what do ya call it. I had a vaccuum cleaner that just needed the drive belt replaced. What's with all the math formulas?**

There's a lot of math in physics. It's not complex math (well, not in this problem) but one needs access to the formulas various scientists have figured out. This problem requires a common physics or chemistry equation to answer called the Ideal Gas Law or the simpler Law that has the name of some scientist and led to the discovery of the Ideal Gas Law. If you're familiar with that law or look it up in your science textbook you'll recognize the variables I used. If you like I can explain them to you. Other than the variables and what they mean, it's basic algebra I was taught in 8th grade.

Don't you teach gifted kids? Are you an English teacher or something?

If you look at the problem, it's full of meters and cylinders and milliTorr (I think that was the pressure unit mentioned) and cubic meters and there's a few numbers, too. How can it not be obvious this is a math problem? - 06 Sep '07 04:31

errrrr, ah, i don't teach gifted kids. I teach gifted carpenters.....framers and finish. Please don't explain the variables to me. I remember nothing of physics or chemistry, but i can build a house from foundation to completed job. I surely thought you would know that your vaccuum cleaner only needed a new belt. Do you understand now?*Originally posted by AThousandYoung***The problem in this equation is solved through high school level physics or chemistry, or at worst, college level introductory chemistry courses.**

There's a lot of math in physics. It's not complex math (well, not in this problem) but one needs access to the formulas various scientists have figured out. This problem requires a common physics or ch ...[text shortened]... eters and there's a few numbers, too. How can it not be obvious this is a math problem? - 06 Sep '07 13:58 / 1 editGood job everyone! 760,000 meters it is! I work in semi-conductor cleanrooms, have for a long time, and I didn't even have to have math to do this problem, I knew 760 torr=1 Atm. (STP) and in millitorr there is 1000 Mt per torr so one atmosphere is 760,000 millitorr so I knew right away given a chamber of one cubic meter and you move out one wall, it would end up 760,000 meters down the road to give you a vacuum reading of one millitorr. BTW, one millitorr is a very good vacuum reading to achieve from mechanical vacuum pumps but is just the start for the real vacuum you need inside most semiconductor manufacturing machines, like my specialty, Ion Implanters. I worked at Varian Ion Implanter division for 20 years, worked around the world on them, including a 3 year stint in Jerusalem with my wife and three of our kids.

In millitorr terms, the vacuums we require are a minimum of 1000 times lower than one millitorr and we consider a very good vacuum to be 100,000 times lower than one millitorr. Compared to outer space, however, even that level of vacuum is way high, if a cosmic ray rammed into that level of vacuum it would quickly be transformed into a million particles. - 06 Sep '07 20:03

Oh, I see. No, it's a physics problem, not a mechanical one. I am too much of an egghead to ever actually have taken apart a vaccuum cleaner.*Originally posted by smw6869***errrrr, ah, i don't teach gifted kids. I teach gifted carpenters.....framers and finish. Please don't explain the variables to me. I remember nothing of physics or chemistry, but i can build a house from foundation to completed job. I surely thought you would know that your vaccuum cleaner only needed a new belt. Do you understand now?**

I did replace the brake pads on my car once, but that's about it. - 06 Sep '07 20:04

Nice*Originally posted by sonhouse***Good job everyone! 760,000 meters it is! I work in semi-conductor cleanrooms, have for a long time, and I didn't even have to have math to do this problem, I knew 760 torr=1 Atm. (STP) and in millitorr there is 1000 Mt per torr so one atmosphere is 760,000 millitorr so I knew right away given a chamber of one cubic meter and you move out one wall, it would ...[text shortened]... ray rammed into that level of vacuum it would quickly be transformed into a million particles.**