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

*> *

*> 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.*

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

and T vary.

*> This plots as a straight line passing through (P=0,T=0) with a slope of*

*> (approximately) 0.00366328 atm/K (or atm/Â°C)*

*> *

*> This seems too easy - and strikes me as too small. Have I made a wrong*

*> assumption or miscalculated somewhere here?*

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:

*> 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).*

Thanks! I wasn't sure of my own sanity - not long ago I plotted out

Pressure vs Temperature for water between 0Â°C and 374Â°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/Â°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

*> On 12/24/2010 12:24 PM, Ahem A Rivet's Shot wrote:*

*> *

*> > 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).*

*> *

*> Thanks! I wasn't sure of my own sanity - not long ago I plotted out*

*> Pressure vs Temperature for water between 0Â°C and 374Â°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/Â°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.*

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

one that large.

*> I had been working with fluidynes (a Stirling-cycle engine with fluid*

*> pistons), heating and cooling air to produce a high/low pressure cycle,*

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

*> 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.*

Massive pressure swings but only very small displacements possible I

think you'll find.

*> 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?*

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:

*> Massive pressure swings but only very small displacements possible I*

*> think you'll find.*

<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

>> 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.