Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    05 Sep '07 20:49
    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 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?
  2. Standard member PBE6
    Bananarama
    05 Sep '07 21:05 / 1 edit
    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?
    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 m3

    So if the cross section was originally 1 m2, the chamber would have to be about 760 km long.
  3. Subscriber AThousandYoung
    It's about respect
    05 Sep '07 21:22
    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.
    I do this sort of problem like this:

    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.
  4. Subscriber AThousandYoung
    It's about respect
    05 Sep '07 21:23 / 1 edit
    The easier way is to simply know the formula

    P1V1 = P2V2

    I prefer to derive it from the IGL though.
  5. Standard member smw6869
    Granny
    05 Sep '07 22:41
    Originally posted by AThousandYoung
    The easier way is to simply know the formula

    P1V1 = P2V2

    I prefer to derive it from the IGL though.
    I, ah, prefere the IGL, also.

    Wha?
  6. Subscriber AThousandYoung
    It's about respect
    06 Sep '07 00:24
    Originally posted by smw6869
    I, ah, prefere the IGL, also.

    Wha?
    Ideal Gas Law. Did you read the post above mine?
  7. Standard member PBE6
    Bananarama
    06 Sep '07 00:33
    Originally posted by AThousandYoung
    The easier way is to simply know the formula

    P1V1 = P2V2

    I prefer to derive it from the IGL though.
    True, this is much easier.
  8. Subscriber AThousandYoung
    It's about respect
    06 Sep '07 00:35 / 1 edit
    Originally posted by PBE6
    True, this is much easier.
    It's an easier formula to use but I prefer to know fewer formulas from which one can derive the others. Memorization sucks.
  9. Standard member smw6869
    Granny
    06 Sep '07 01:52
    Originally posted by AThousandYoung
    Ideal Gas Law. Did you read the post above mine?
    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?
  10. Subscriber AThousandYoung
    It's about respect
    06 Sep '07 03:45 / 3 edits
    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?
    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 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?
  11. Standard member smw6869
    Granny
    06 Sep '07 04:31
    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?
    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?
  12. Subscriber coquette
    Already mated
    06 Sep '07 05:43 / 1 edit
    Forget it. It's all covered.
  13. Subscriber sonhouse
    Fast and Curious
    06 Sep '07 13:58 / 1 edit
    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 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.
  14. Subscriber AThousandYoung
    It's about respect
    06 Sep '07 20:03
    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?
    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.

    I did replace the brake pads on my car once, but that's about it.
  15. Subscriber AThousandYoung
    It's about respect
    06 Sep '07 20:04
    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.
    Nice