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Ideal Gas Law help please

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Posted by Morris Dovey on December 24, 2010, 5:38 pm
 

Given the Ideal Gas Law: PV = nRT

If some volume of an ideal gas is captured in a closed container at STP
(293.15K, 1 atm), it appears that V, n, and R should be constant.

Rearranging gives P = nRT/V = (n/V)RT

I recall that at STP, 1 mole of a gas occupies a 22.4 L volume - which
should allow me to write n/V = 1/22.4 and use R = 0.08205746 Latm/Kmol
to get

   P = (0.08205746/22.4)*T = 0.00366328*T atm

To arrive at a (P,T) relationship that is independent of the actual
number of moles of the gas and/or the actual volume of the container.

This plots as a straight line passing through (P=0,T=0) with a slope of
(approximately) 0.00366328 atm/K (or atm/C)

This seems too easy - and strikes me as too small. Have I made a wrong
assumption or miscalculated somewhere here?

--
Morris Dovey
http://www.iedu.com/DeSoto/
PGP Key ID EBB1E70E


Posted by Ahem A Rivet's Shot on December 24, 2010, 6:24 pm
 
On Fri, 24 Dec 2010 11:38:49 -0600


    As indeed it should be given that n and V are held constant while P
and T vary.


    A quick sanity check (273.15*0.00366328) gives 1 atm (1.00062493)
at 0 C so yes you have it right - it's a linear relationship which goes
through (0K, 0atm) and (273.15K, 1atm).

--
Steve O'Hara-Smith                          |   Directable Mirror Arrays
C:>WIN                                      | A better way to focus the sun
The computer obeys and wins.                |    licences available see
You lose and Bill collects.                 |    http://www.sohara.org/

Posted by Morris Dovey on December 24, 2010, 8:21 pm
 On 12/24/2010 12:24 PM, Ahem A Rivet's Shot wrote:


Thanks! I wasn't sure of my own sanity - not long ago I plotted out
Pressure vs Temperature for water between 0°C and 374°C in that same
container:

   http://www.iedu.com/DeSoto/Misc/H2O_VP.png  and
   http://www.iedu.com/DeSoto/Misc/H2O_VP_PSI.png

which, at at the upper end of that range, had a slope (rate of change)
of approximately 2.3 atm/°C - which is more than 600 times the rate for
an ideal gas. I expected that there would be a difference, but wasn't
prepared to so large a difference.

I had been working with fluidynes (a Stirling-cycle engine with fluid
pistons), heating and cooling air to produce a high/low pressure cycle,
but I'm now thinking that I should be thinking about a water-only engine
- heating and cooling water to be producing much larger pressure swings.

I read somewhere that if water is heated beyond its critical
temperature, it acts like an ideal gas. Do you suppose that means that
when the temperature exceeds the critical temperature, the pressure will
suddenly drop?

Or would the pressure continue to rise, but at the much lower rate of an
ideal gas?

It seems interesting either way. :)

--
Morris Dovey
http://www.iedu.com/DeSoto/
PGP Key ID EBB1E70E


Posted by Ahem A Rivet's Shot on December 24, 2010, 9:18 pm
 On Fri, 24 Dec 2010 14:21:49 -0600


    Wow that's a huge difference, I certainly wouldn't have predicted
one that large.


    Yeah I've been watching with interest your posts on them.


    Massive pressure swings but only very small displacements possible I
think you'll find.


    The latter.

--
Steve O'Hara-Smith                          |   Directable Mirror Arrays
C:>WIN                                      | A better way to focus the sun
The computer obeys and wins.                |    licences available see
You lose and Bill collects.                 |    http://www.sohara.org/

Posted by Morris Dovey on December 30, 2010, 10:14 pm
 On 12/24/2010 3:18 PM, Ahem A Rivet's Shot wrote:


<scratches head>

I'm not so sure. Curbie pointed me at an article on manometers at

   http://www.efunda.com/formulae/fluids/manometer.cfm

and the formula shown there tells me that the displacement (/h/ in the
drawing) can be calculated by dividing the difference in pressures by
the product of the fluid density (1000 kg/m^3 for water) and
acceleration due to gravity (9.81 m/s^-2).

That would seem to say that if several atmospheres of pressure could be
applied to the left side of the manometer, then the displacement /h/
might be substantial.

If I'm understanding things correctly, raising the pressure on the left
side from 1 to 4 atm should produce a displacement of approx 30 m - and
if the pressure source can be made to oscillate in that range, then we
should be able to use it as the basis for a pump capable of pushing
water up ~30m...

--
Morris Dovey
http://www.iedu.com/DeSoto/
PGP Key ID EBB1E70E


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