May3-07, 12:32 PM
"A parcel of air is lifted slowly from the ground, where the temperature is 295K, to an elevation of 5km, and then returned rapidly to the ground. Estimate the air parcel temperature at 5km and after it returns to the groundm explanation any assumptions."
I assumed an adiabatic process both ways. Upwards the process is "slow" so we can assume reversibility, so the parcel should follow the adiabatic lapse rate. Therefore dT = -g/cp * 5km
assume g is constant, assume dry air and use cp = 1.01KJ per K per mole
This gives T = 246K at the top.
On the way down we can't assume reversibility. I tried setting dU = dW + dQ, where dQ is equal to zero. dW = mg*dz, where we assume that g is constant again, and this time dz = -5km. Therefore assuming the ideal gas law,
n*Cv*dT = mg*dz
m equals the molar mass* no of moles. I assume the gas is diatomic Nitrogen, so the molar mass is 34*10^-3 kg per mole. This gives a change in temperature of -133K which seems far too much (in fact this is close to the temperature at which nitrogen condenses). I'm not sure where I have gone wrong. Any hints appreciated.
|Register to reply|
|Kelvin Temperature - Celsius Temperature Change||General Physics||12|
|correcting for room temperature and atmospheric pressure||Introductory Physics Homework||8|
|Atmospheric Physics||Forum Feedback & Announcements||5|
|temperature change||Introductory Physics Homework||6|