Atmospheric physics - change of temperature of parcel

[deadringer]

"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.

 

[Jhero]

Hello there!

I can offer some insights into your calculations. First of all, your assumption of an adiabatic process both ways is correct. This means that there is no exchange of heat with the surroundings during the lifting and returning of the air parcel.

When the air parcel is lifted slowly to 5km, it will experience a decrease in pressure due to its expansion. This decrease in pressure will cause the air parcel to cool down, following the adiabatic lapse rate. However, when the air parcel is returning rapidly to the ground, it will experience a sudden increase in pressure, causing it to warm up.

The adiabatic lapse rate for dry air is approximately 9.8°C per kilometer of altitude. Using this rate, we can estimate the temperature of the air parcel at 5km to be approximately 246K, as you have calculated.

However, when the air parcel returns to the ground, it will experience a sudden increase in pressure. This means that the air parcel will undergo a process known as adiabatic compression, where it will warm up due to the increase in pressure. The temperature change can be calculated using the ideal gas law, as you have done. However, in this case, we need to consider the specific heat capacity at constant volume (Cv) instead of the specific heat capacity at constant pressure (Cp). For dry air, Cv is approximately 0.718 KJ per K per mole.

Using this value for Cv, we can calculate the change in temperature for the air parcel as it returns to the ground. Assuming the same molar mass of 34*10^-3 kg per mole for diatomic nitrogen, the change in temperature would be approximately -59K. This is still a significant temperature change, but it is not as extreme as the one you have calculated.

One possible reason for the discrepancy in our calculations could be the assumption of constant gravity. In reality, the force of gravity decreases as the air parcel moves away from the Earth's surface. This could affect the temperature change during the adiabatic compression process.

I hope this helps to clarify the calculations for the air parcel's temperature at 5km and after it returns to the ground. Remember, these are just estimations based on the assumptions we have made. In reality, there are many factors that could affect the temperature change, so these numbers should be taken with a grain of salt.