Temperature change with altitude

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SUMMARY

The discussion focuses on calculating atmospheric pressure at 7500 meters using the barometric formula and the temperature lapse rate of 0.0065 K/m. The user initially attempted to apply the ideal gas law but encountered discrepancies in their results. The correct approach involves integrating temperature as a function of altitude, utilizing the lapse rate to derive pressure accurately. The final calculated pressure should be approximately 38.3 kPa, contrasting with the user's result of 41.7 kPa.

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  • Understanding of the barometric formula
  • Knowledge of the ideal gas law
  • Familiarity with temperature lapse rates
  • Basic calculus for integration
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Alright guys, just looking for a point in the right direction for a tutorial question.

GIVEN VALUES:
Temperature change with increasing altitude: 0.0065 K/m
Temp at sea level: 15ºC
Pressure at sea level: 101.5 kPa

FIND:
Atmospheric pressure at 7500m.


So what I've done is found the temperature decrease between sea level and 7500m, so that temperature is 239.4K.

And then tried (P1/T1)=(P2/T2), was a bit sceptical over this working.. it didn't.

After that tried:

∫(1/P) dP = -∫(g/RT) dz


Integrated that, giving:

ln(P2/P1) = (-g/RT)(z2-z1)


Placed in the values given/calculated, leading to:

P = 101500*e^[(-9.81*7500)/(287*288.15)]

P = 41.7 kPa

... which is a wee bit off the stated answer of 38.3 kPa.




Any ideas?
 
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Welcome to PF!

In your second integral you should integrate T as a function of z (you have treated it as a constant). Usually one makes the substitution dz = (1/(dT/dz)) dT, where dT/dz is the constant lapse rate so that you get pressure as a function of temperature.

Edit: I see now that you have posted outside the homework section of this forum. Please post future homework questions in the homework section to get better responses :smile:.
 
Fantastic, thanks Filip.
 

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