Number Density in the Atmosphere

In summary, the conversation discusses the Boltzmann distribution law and the calculation of the number density of molecules at different heights in the Earth's atmosphere. The discussion also includes the use of potential energy and molecular mass in the equation. The correct units for the arguments of exponential and logarithmic functions are also mentioned. The ratio of atmospheric density at 11.0 km to that at sea level is found to be very small, indicating a decrease in density with height.
  • #1
amcca064
32
0
Assume that the Earth's atmosphere has a uniform temperature of 20ºC and uniform composition, with an effective molar mass of 28.9 g/mol. a) show that the number density of molecules depends on height according to:

nv (y) = (n0) e^ -(mgy)/kBT

b) commercial jetliners typically cruise at an altitude of 11.0 km. Find the ratio of the atmospheric density there to the density at sea level.


Ok, so I really am kind of stuck here, for the normal Boltzmann distribution equation, where nv(E) = (n0) e^-E/kBT is it possible just to say that since the atmosphere is assumed to have uniform temperature and composition that (assuming no heat loss or gain through interaction with the ground or space) all the molecules have the same kinetic energy and therefore the only change in E would occur through a change in potential energy, therefore the E in the Boltzmann distribution eq can be subsituted for U and U can be substituted for mgy?? Seems far too easy this way. Help anyone?

Correct title should have been Bolztmann distribution law
 
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  • #2
Anybody please?
 
  • #3
amcca064 said:
Assume that the Earth's atmosphere has a uniform temperature of 20ºC and uniform composition, with an effective molar mass of 28.9 g/mol. a) show that the number density of molecules depends on height according to:

nv (y) = (n0) e^ -(mgy)/kBT

b) commercial jetliners typically cruise at an altitude of 11.0 km. Find the ratio of the atmospheric density there to the density at sea level.


Ok, so I really am kind of stuck here, for the normal Boltzmann distribution equation, where nv(E) = (n0) e^-E/kBT is it possible just to say that since the atmosphere is assumed to have uniform temperature and composition that (assuming no heat loss or gain through interaction with the ground or space) all the molecules have the same kinetic energy and therefore the only change in E would occur through a change in potential energy, therefore the E in the Boltzmann distribution eq can be subsituted for U and U can be substituted for mgy?? Seems far too easy this way. Help anyone?

Correct title should have been Bolztmann distribution law

I think you're good, and here is a nice little paper on the subject

http://www.shef.ac.uk/physics/people/rjones/PDFs/PHY101/PHY101_RALJ_lecture6.pdf
 
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  • #4
Ok thanks, but my prof just emailed me and said to skip that part! haha all that was just for fun then. I am still having problems with the question though, I tried finding the ratio, but I am confused for what I should use for m in the equation, should I use the molar mass and find it on a per mole basis or should I find the mass of one molecule? I really don't understand it too much, but I'll read the link you sent, maybe it will help? Thanks!
 
  • #5
Alright thank you! That link was very very helpful in understanding what's going on! The equation given below from that link uses "psi" to represent the potential energy of one molecule, so I think from this that I should find the mass of one molecule and use that in the mgy of my equation. Awesome, thanks again!
 

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  • #6
amcca064 said:
Alright thank you! That link was very very helpful in understanding what's going on! The equation given below from that link uses "psi" to represent the potential energy of one molecule, so I think from this that I should find the mass of one molecule and use that in the mgy of my equation. Awesome, thanks again!

I can't see your diagram yet, but the m in the derivation is the molecular mass.
 
  • #7
oh really? ok yeah that makes sense i didn't see in the equation that its n(psi) and not just nv(y). Well if I'm going to use the molecular mass should i use kg/mol or g/mol? I converted to kilograms and used that but i ended up with the ratio equal to e^-7.27X10^22 which my calculator automatically rounds to zero since it is such a small number.
 
  • #8
Should I leave the answer expressed as e^-7.27X10^22 or should i redo with grams per mole?
 
  • #9
Well I redid it with g/mol and it still gave me 0 so I think i'll leave it with kg
 
  • #10
amcca064 said:
oh really? ok yeah that makes sense i didn't see in the equation that its n(psi) and not just nv(y). Well if I'm going to use the molecular mass should i use kg/mol or g/mol? I converted to kilograms and used that but i ended up with the ratio equal to e^-7.27X10^22 which my calculator automatically rounds to zero since it is such a small number.

The argument of an exponential, like the argument of a trig function, must be dimensionless. The units you use for m must be consistent with the units of everything else that is there. When you multiply everything in the exponential together you must have a pure number. This is also true of logarithms. That is why you almost always see logarithms of ratios when they appear in formulas. It would be incorrect to write

ln(I/Io) = ln(I) - ln(Io)

where I and Io are quantities with dimensions. What would be correct is if represents the units of I and Io then

ln(I/Io) = ln(I/) - ln(Io/)

so the the arguments of the ln() are dimensionless.
 

Related to Number Density in the Atmosphere

1. What is number density in the atmosphere?

Number density in the atmosphere refers to the number of gas molecules present in a given volume of air. It is a measure of how many gas particles are present in a specific area and is typically expressed in units of particles per cubic meter.

2. How is number density in the atmosphere measured?

Number density in the atmosphere is typically measured using instruments such as spectrometers or mass spectrometers. These instruments use various techniques to count the number of gas particles present in a specific volume of air.

3. What factors can affect number density in the atmosphere?

The main factors that can affect number density in the atmosphere include altitude, temperature, and pressure. As altitude increases, the number density decreases due to the decrease in air pressure. Similarly, as temperature and pressure decrease, number density also decreases.

4. Why is number density in the atmosphere important?

Number density in the atmosphere is important because it provides information about the composition and behavior of the atmosphere. It can help scientists understand the distribution of different gases in the atmosphere and how they interact with each other and the environment.

5. How does number density in the atmosphere impact weather and climate?

Number density in the atmosphere plays a significant role in weather and climate patterns. Different gases have different number densities, which can affect their ability to absorb and emit radiation, leading to changes in temperature and weather patterns. Changes in number density can also impact the overall composition of the atmosphere, which can have long-term effects on the Earth's climate.

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