Atmospheric Temperature and Air Molecule Density Calculation

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Discussion Overview

The discussion revolves around calculating the average number of air molecules per cm³ at an altitude of 50 km, given the average atmospheric temperature of 0°C. Participants explore the application of relevant formulas and address unit conversion issues while attempting to solve a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating pressure at altitude, noting difficulties with unit conversions and errors encountered on their calculator.
  • Another participant points out that the equation used should be dimensionless and suggests using meters instead of kilometers for distance to align with the units of the gas constant R.
  • Concerns are raised regarding the molar mass of air being expressed in cm³/mol, with a suggestion to use kg/mol instead.
  • A later reply clarifies that the logarithmic equation presented is based on base-10 logarithm, not natural logarithm, and explains the conversion factor for altitude.
  • Further calculations are provided using a scale height of 8 km, leading to a derived pressure at that altitude, which is then compared to the original problem at 50 km.
  • Another participant confirms the calculations presented in the final posts, indicating that they seem correct.

Areas of Agreement / Disagreement

Participants express various viewpoints on the correct application of formulas and unit conversions, with no clear consensus reached on the best approach to solve the problem. Disagreements exist regarding the interpretation of the equations and the appropriate units to use.

Contextual Notes

Limitations include unresolved unit conversion issues, dependence on the interpretation of the gas constant R in different units, and the application of scale height in calculations. The discussion reflects a range of assumptions and interpretations that have not been fully reconciled.

Jilly
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Homework Statement


At an altitude of 50 km, the average atmospheric temperature is 0C. what is the average number of air molecules per cm^3 of air at this atmosphere?

Given formula:
P(sub h) =
P(sub 0) x e^(-Mgh/RT) =
-(Mgh x 10^5)/2.303RT
Assuming P = 1 atm at sea level and h in km

M=28.97g/mol=28.97cm^3/mol
g = 9.81 m/s^2 = .0981 km/s^2
h = 50 km
R = 8.314 J/K•mol =
82.06cm^3•atm/K•mol
T = 0 C = 273.15 K

I'm having trouble because using the scale height form of the equation (second form) I get lost in the units, and I'm receiving an error on my TI using the first.

The Attempt at a Solution



Using second equation:[/B]

Log P(sub h) =
(-Mgh x 10^5)/2.303RT =
-(28.97)(.0981)(50)(10^5)/
(2.303)(82.057)(273.15) =
-275.28
10^(-275.28) = 5.23... But the units after the conversion don't add up ... Suggestions?
 
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*5.23x10^-276
 
Jilly said:

Homework Statement


At an altitude of 50 km, the average atmospheric temperature is 0C. what is the average number of air molecules per cm^3 of air at this atmosphere?

Given formula:
P(sub h) =
P(sub 0) x e^(-Mgh/RT) =
-(Mgh x 10^5)/2.303RT
Assuming P = 1 atm at sea level and h in km

M=28.97g/mol=28.97cm^3/mol
g = 9.81 m/s^2 = .0981 km/s^2
h = 50 km
R = 8.314 J/K•mol =
82.06cm^3•atm/K•mol
T = 0 C = 273.15 K

I'm having trouble because using the scale height form of the equation (second form) I get lost in the units, and I'm receiving an error on my TI using the first.

The Attempt at a Solution



Using second equation:[/B]

Log P(sub h) =
(-Mgh x 10^5)/2.303RT =
-(28.97)(.0981)(50)(10^5)/
(2.303)(82.057)(273.15) =
-275.28
10^(-275.28) = 5.23... But the units after the conversion don't add up ... Suggestions?

Whew! What a dog's breakfast of units! Let's see if we can clean this mess up a bit.

R can be expressed in a variety of different units:

http://en.wikipedia.org/wiki/Gas_constant

-Mgh/RT should be dimensionless. Unfortunately, you picked km as your distance unit, which really doesn't fit in with your units for R. Meters would be the better choice.

The molar mass M for air should be expressed in proper units, which is kg/mol. It's not clear how you obtained M = 28.97 cm3/mol.

You use the formula:

P(sub h) =
P(sub 0) x e^(-Mgh/RT) =
-(Mgh x 10^5)/2.303RT

where did the last part -(Mgh x 10^5)/2.303RT) come from? That's not equivalent to raising the number e to the power of (-Mgh/RT).

See what you can do to fix this and we can talk again.
 
SteamKing said:
Whew! What a dog's breakfast of units! Let's see if we can clean this mess up a bit.

R can be expressed in a variety of different units:

http://en.wikipedia.org/wiki/Gas_constant

-Mgh/RT should be dimensionless. Unfortunately, you picked km as your distance unit, which really doesn't fit in with your units for R. Meters would be the better choice.

The molar mass M for air should be expressed in proper units, which is kg/mol. It's not clear how you obtained M = 28.97 cm3/mol.

You use the formula:

P(sub h) =
P(sub 0) x e^(-Mgh/RT) =
-(Mgh x 10^5)/2.303RT

where did the last part -(Mgh x 10^5)/2.303RT) come from? That's not equivalent to raising the number e to the power of (-Mgh/RT).

See what you can do to fix this and we can talk again.

Okay so, this second equation was derived from the first using an 8 km scale height and is true under the conditions of atm pressure units, according to my textbook, that's whereni acquired all of my information, I've attached a photo of the page, maybe you can see what I'm not understanding... Or rather misunderstanding.
image.jpg
 
The site autorotated the photo, here's an adjustment.
 

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The textbook image answers a lot of questions.

log (Ph) = (-Mgh x 105) / 2.303 RT, where log here is the base-10 log, not the natural log. ln (10) ≈ 2.303

The factor 105 converts altitude in kilometers into altitude in centimeters.

The molecular weight of air M = 28.97 grams / mol, as specified in the text.
g = 981 cm / s2, as specified in the text
R = 8.314 * 107 erg / K-mol, as specified in the text.

If we use h = 8 km as the scale height, then the pressure at this altitude is

log (Ph) = (-28.97 * 981 * 8 x 105) / (2.303 * 8.314 * 107 * 288)

log (Ph) = -0.4124

Ph = 10-0.4124

which means Ph = 0.3869, or about 39% of the pressure at sea level, which is as described in the text.

Now, you should be able to work the original HW problem for an altitude of 50 km.
 
SteamKing said:
The textbook image answers a lot of questions.

log (Ph) = (-Mgh x 105) / 2.303 RT, where log here is the base-10 log, not the natural log. ln (10) ≈ 2.303

The factor 105 converts altitude in kilometers into altitude in centimeters.

The molecular weight of air M = 28.97 grams / mol, as specified in the text.
g = 981 cm / s2, as specified in the text
R = 8.314 * 107 erg / K-mol, as specified in the text.

If we use h = 8 km as the scale height, then the pressure at this altitude is

log (Ph) = (-28.97 * 981 * 8 x 105) / (2.303 * 8.314 * 107 * 288)

log (Ph) = -0.4124

Ph = 10-0.4124

which means Ph = 0.3869, or about 39% of the pressure at sea level, which is as described in the text.

Now, you should be able to work the original HW problem for an altitude of 50 km.

Thank you!
 
SteamKing said:
The textbook image answers a lot of questions.

log (Ph) = (-Mgh x 105) / 2.303 RT, where log here is the base-10 log, not the natural log. ln (10) ≈ 2.303

The factor 105 converts altitude in kilometers into altitude in centimeters.

The molecular weight of air M = 28.97 grams / mol, as specified in the text.
g = 981 cm / s2, as specified in the text
R = 8.314 * 107 erg / K-mol, as specified in the text.

If we use h = 8 km as the scale height, then the pressure at this altitude is

log (Ph) = (-28.97 * 981 * 8 x 105) / (2.303 * 8.314 * 107 * 288)

log (Ph) = -0.4124

Ph = 10-0.4124

which means Ph = 0.3869, or about 39% of the pressure at sea level, which is as described in the text.

Now, you should be able to work the original HW problem for an altitude of 50 km.

Log P(sub h) =
(-28.97*981*50*10^5)/
(2.303*[8.314x10^7]*273.15)
LogP(sub h) = -2.72
p(sub h) = 0.00192 atm

n=PV/RT=(.00192atm*1cm^3)/ (273.15K*82.057cm^3atm•K*-1•mol^-1)

n = 8.566 x 10^-8 mol
= 5.16 x 10^16 molecules,

Yes?
 
Your calculation seems to be OK.
 

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