How Does Adiabatic Wind Affect Temperature Changes in the Rocky Mountains?

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SUMMARY

The discussion focuses on the effects of adiabatic wind on temperature changes in the Rocky Mountains, specifically the phenomenon of chinook winds. The calculations involve the pressure dependency on altitude using the formula p = p_0e^(-ay), with p_0 set at 1.00 atm and a = 1.16 x 10^-4 m^-1. The initial temperature of a descending air parcel at 4267 m is -5.00℃, and the calculations show that the temperature at 1567 m is incorrectly computed as -5.41℃. The error was identified as a result of not converting to the Kelvin scale, which corrected the temperature increase expected from the chinook effect.

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  • Knowledge of specific heat ratios, particularly γ (gamma)
  • Basic skills in logarithmic and exponential functions
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Homework Statement


Adiabatic wind. The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressur p depends on altitude y according to p=p_0e-ay where p_0=1.00atm and a=1.16*10-4m-1. Also assume that the ratio of the molar specific heats is: γ=4/3. A parcel of air with an initial temperature of -5.00℃ descends adiabatically from y_1=4267m to y=1567m. What is its temperature at the end of the descent?


Homework Equations


pVγ=a constant
TVγ-1=a constant


The Attempt at a Solution


I really don't know if I went right here.. but here it goes.

p_1=1*e^{-a*4267}
p_1=0.609587975
p_2=1*e^{-a*4267}
p_2=0.833791423

p_1V_1^γ=p_2V_2^γ
V_2^γ=(p_1/p_2) as v_1^γ=1
ln(V_2)=(ln(p_1/p_2))/γ
e^{ln(V_2)}=e^{ln(p_1/p_2)/γ}
plug in all the numbers and V_2=0.790455348

T_2V_2^{γ-1}=T_1V_1^{γ-1}
V_1=1 V_2=0.790455348 T_1=-5 T_2=?

T_2=\frac{T_1V_1^{γ-1}}{V_2^{γ-1}} as V_1^{γ-1}=1
T_2=\frac{T_1}{V_2^{γ-1}}
T_2=\frac{-5}{0.790455348^{4/3-1}}
T_2=-5.40767883

This answer doesn't make sense because the temperature is supposed to INCREASE.

So can anybody tell me where I went wrong? Maybe tell me I used the wrong formulas..


Thank you :)
 
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Try it using the Kelvin scale.
 
That solved it :) wow I'm an idiot
 

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