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Attempt to combine Charles and Boyles laws

  • Thread starter johns123
  • Start date
34
0
1. Homework Statement

At altitude of 160 km, the density of the air is 1.5 x 10^-9 kg/m^3

also the temp at that altitude is 500 K ( that's what it says! )

What is the pressure at that altitude?


2. Homework Equations

pV = nRT and mass density of air = "rho"

the 1-values are at STP, the 2-values are at altitude

p1 = 1ATM ; rho1 = 29 kg/m^3 ; T1 = 273 K

And I don't see that either vol or temp is being
held constant, so just Boyle's or Charle's law
cannot apply alone ( in my opinion ).


3. The Attempt at a Solution


I think I can establish a ratio between these values at altitude
and the same values at STP and calc the unknown p2. Do you agree?

p1 * V1 n1 * R * T1
_______ = ___________ Where the Rs cancel

p2 * V2 n2 * R * T2


so p2 = p1 [ ( n2 V1 T2 ) / ( n1 V2 T1 )


but ( n2/V2 ) / (n1/V1) should equal ( rho2 / rho1 )

giving p2 = p1 * ( rho2/rho1 ) * T2/T1
 

Andrew Mason

Science Advisor
Homework Helper
7,536
317
1. Homework Statement

At altitude of 160 km, the density of the air is 1.5 x 10^-9 kg/m^3

also the temp at that altitude is 500 K ( that's what it says! )

What is the pressure at that altitude?
I think you are supposed to assume the air has the same molecular mass as it has near the surface (ie. M = 29 g/mol = .029 kg/mol)

At 160 km altitude you are into the thermosphere. The molecules are moving around rather quickly but not a lot of total energy due to low density.

Work out the number of moles / m3. (n/V) and then use PV=nRT to find the pressure.

AM
 
34
0
OK. Thanks. I'm struggling with concepts here .. a battle of definitions. I'll solve it your way and post that.
 
34
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Finally, I think I understand it ( I hope )

If air density at altitude is 1.5 x 10^-9 kg/m^3 at 500K and ( your hint ) M = .29 kg/mole that gives

1.5 x 10^-9 / .29 = 5.172 x 10^-9 moles/m^3 .. which is ( your hint ) = n/V

and now use pV = nRT or p = (n/V)RT = 5.172 x 10^-9 ( 8.81 joules/molesK ) 500k ( and the units work )

so finally ( after 2 weeks ) p = 42.98 x 10^-9 pascals ( I sure hope so .. thanks )
 

Andrew Mason

Science Advisor
Homework Helper
7,536
317
Finally, I think I understand it ( I hope )

If air density at altitude is 1.5 x 10^-9 kg/m^3 at 500K and ( your hint ) M = .29 kg/mole that gives

1.5 x 10^-9 / .29 = 5.172 x 10^-9 moles/m^3 .. which is ( your hint ) = n/V
My hint was to use M = .029 Kg/mol. Where did you get .29 Kg/mol?

and now use pV = nRT or p = (n/V)RT = 5.172 x 10^-9 ( 8.81 joules/molesK ) 500k ( and the units work )

so finally ( after 2 weeks ) p = 42.98 x 10^-9 pascals ( I sure hope so .. thanks )
The approach is correct. But where do you get R=8.81 J/mol K? And you appear to have forgotten to multiply by T. So your answer is about 4 orders of magnitude too low. Other than that.....

AM
 
34
0
I think I need to learn to type :-) it's 8.31 J/mk. I was just copying out of my notebook. Try again.

n/V = 1.5 x 10^-9 kg/m^3 / .029 kg/mole = 51.724 x 10^-9 moles/m^3

and using p = (n/V)RT = 51.724 x 10^-9 ( 8.31 ) 500K = 214913.22 x 10^-9 pa ( I really hate this calculator )

thanks
 

Andrew Mason

Science Advisor
Homework Helper
7,536
317
I think I need to learn to type :-) it's 8.31 J/mk. I was just copying out of my notebook. Try again.

n/V = 1.5 x 10^-9 kg/m^3 / .029 kg/mole = 51.724 x 10^-9 moles/m^3

and using p = (n/V)RT = 51.724 x 10^-9 ( 8.31 ) 500K = 214913.22 x 10^-9 pa ( I really hate this calculator )

thanks
You should round to 2 or 3 significant figures and express the answer in normalized scientific notation. But the answer is correct.

AM
 
34
0
Thanks Andrew. I think this one problem holds the key to a world of conversions that are left out of the books. I am using 3 books trying to learn one chapter.
 

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