1. At about 11:00 am on day in this summer, we measured the averaged air tempera

Question

1. At about 11:00 am on day in this summer, we measured the averaged air temperature of 300K near the ground. Assume the air temperature changes with height at a rate of 6.5K per km in the troposphere. One air parcel had its temperature of 302K near the ground and it then rose. It is assumed that the temperature change for the air parcel followed the dry adiabatic lapse rate. Using Archimedes’ law and the ideal gas law, discuss the buoyancy force and gravity force acting on the air parcel along the way when it rose. What height was the air parcel likely to reach? why?

Explanation / Answer

Near the ground, temperatur eof outside air is 300K and temperature of air parcel = 302K.

We know that, as temperature increases, the density of air decreases. Hence, the density of ait inside parcel is less than that of air outside.

Also, by ideal gas law, pressure is inversly proportional to volume. Density is inversly proportional to volume. Hence, we can say pressure is directly proportional to density.

Hence, the pressure of outside air is more and hence it exerts pressure on the air parcel. This results in a buoyancy force. The buoyancy force is opposite to the gravitational force. Because of buoyancy force the parcel moves up. The parcel moves up till the point when the pressure of air inside is equal to pressure of air outside. This can also be intepreted as the point at which the temperature of air outside is equal to temperature of parcel.

Air inside parcel follows dry adiabatic lapse rate. That means, temperature reduces by 1 unit every 100m of lift.

Temp reduction per km for air parcel = 10 K

Temp reduction per km for air outside = 6.5 K

Suppose, the maximum height till which parcel moves = x km from ground

Temperature at x km for air parcel = 302-10x K and for air outside = 300 - 6.5x K

302 - 10x = 300 - 6.5x

3.5x = 2

x = 4/7 km

Thus, height air parcel likely to reach = 4/7 km

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