| Topic: |
Science > Physics |
| User: |
"Ramesh Kumar Bhandari" |
| Date: |
05 Jul 2004 08:04:50 PM |
| Object: |
Calculating gas pressure |
Hi all,
I have been puzzled by a gas problem. I have an evacuated and sealed
quartz glass tube (1 cm in diameter and 20 cm in length). The residual
gas pressure inside the tube is 1 Pa at room temperature (300 K). Now,
one end of the horizontally positioned tube is heated to 500 K, while
the other end is still maintained at room temperature by circulating
water. Will there be same pressure everywhere inside the tube? How can
I calculate it?
I'd really appreciate your kind suggestions.
Ramesh
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| User: "John T Lowry" |
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| Title: Re: Calculating gas pressure |
06 Jul 2004 10:29:38 AM |
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"Ramesh Kumar Bhandari" <bhandaritwo@yahoo.com> wrote in message
news:ed1fc058.0407051704.8bbc30c@posting.google.com...
Hi all,
I have been puzzled by a gas problem. I have an evacuated and sealed
quartz glass tube (1 cm in diameter and 20 cm in length). The residual
gas pressure inside the tube is 1 Pa at room temperature (300 K). Now,
one end of the horizontally positioned tube is heated to 500 K, while
the other end is still maintained at room temperature by circulating
water. Will there be same pressure everywhere inside the tube? How can
I calculate it?
I'd really appreciate your kind suggestions.
Ramesh
I believe the pressure will be (remain) the same everywhere inside the
tube because if there were a pressure difference established, or even a
pressure fluctuation, the resulting force would create a flow from high
to low pressure.
If you think in terms of the ideal gas law p = R*g*rho*T, what you'll
have is rho(x) and T(x), with their product constant at p/(R*g). I.e.,
where higher T, then lower density rho at that location.
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| User: "John Popelish" |
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| Title: Re: Calculating gas pressure |
05 Jul 2004 08:21:52 PM |
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Ramesh Kumar Bhandari wrote:
Hi all,
I have been puzzled by a gas problem. I have an evacuated and sealed
quartz glass tube (1 cm in diameter and 20 cm in length). The residual
gas pressure inside the tube is 1 Pa at room temperature (300 K). Now,
one end of the horizontally positioned tube is heated to 500 K, while
the other end is still maintained at room temperature by circulating
water. Will there be same pressure everywhere inside the tube? How can
I calculate it?
I'd really appreciate your kind suggestions.
Ramesh
The only thing that can support a pressure difference in one part of
the volume of stationary fluid compared to some other part of the
volume is the effect of gravity on the density of the fluid, and the
elevation difference. Since your volume has very little elevation
differences, this effect is nil.
To the extent that the gas follows the ideal gas laws (no condensation
or other extreme deviations), I think the pressure is a function of
the volume weighted average temperature. In other words, if 1/4 of
the volume is at 500k and 3/4 of the volume is at 300k, then the
effective temperature you use to calculate the pressure is
(1/4)*300k+(3/4)*500k=450k
--
John Popelish
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| User: "" |
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| Title: Re: Calculating gas pressure |
06 Jul 2004 11:32:23 AM |
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In article <40E9FEB0.47488871@rica.net>, John Popelish <jpopelish@rica.net> writes:
Ramesh Kumar Bhandari wrote:
Hi all,
I have been puzzled by a gas problem. I have an evacuated and sealed
quartz glass tube (1 cm in diameter and 20 cm in length). The residual
gas pressure inside the tube is 1 Pa at room temperature (300 K). Now,
one end of the horizontally positioned tube is heated to 500 K, while
the other end is still maintained at room temperature by circulating
water. Will there be same pressure everywhere inside the tube? How can
I calculate it?
The only thing that can support a pressure difference in one part of
the volume of stationary fluid compared to some other part of the
volume is the effect of gravity on the density of the fluid, and the
elevation difference. Since your volume has very little elevation
differences, this effect is nil.
Agreed. I don't think there is any controversy here. The pressure will
be essentially constant throughout the tube.
To the extent that the gas follows the ideal gas laws (no condensation
or other extreme deviations), I think the pressure is a function of
the volume weighted average temperature. In other words, if 1/4 of
the volume is at 500k and 3/4 of the volume is at 300k, then the
effective temperature you use to calculate the pressure is
(1/4)*300k+(3/4)*500k=450k
I don't think that mean temperature is the relevant parameter here.
Rather, I think that we need to be looking at the harmonic mean.
Pick a molecular weight and look up Boltzman's constant.
We can figure a number of moles of the gas based on the
initial conditions. pv=nRt. Solve for n.
[It's clear that Boltzman's constant and the molecular weight of the
gas will drop out of the final result. So make things easy
and set them both to one if you like. Or leave them in and see
where they cancel].
Now assume some temperature distribution. For simplicity, let's make
it linear. T(0) = 500k. T(20) = 300k. T(x) = 500-10x
[This is where things are likely to go non-physical. I don't know
if thermal conductivity of glass and of a [convective!] ideal gas is
suitably close to constant over the range from 300 to 500k. We could
find ourselves with a nasty differential equation and wind up using
numerical methods to solve this.]
We haven't lost any gas when we heated the tube. So final moles
is equal to initial moles.
final moles = integral from 0 to 20 of incremental moles per unit length
= integral from 0 to 20 of tube-cross-sectional-area*p/R/t(x) dx
= p*tube-cross-sectional-area/R * integral from 0 to 20 of 1/t(x) dx
Evaluate the integral. Set initial moles to final moles and solve for
final pressure.
As you can see, it's the [volume] average of the inverse temperature that's
relevant. That's the harmonic mean (or the inverse thereof).
John Briggs
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