recently my thermodynamics tutor and I came across a problem while discussing one of the exam questions.

Let me introduce you to the task first:

- An air parcel is located at the surface, where the pressure is

**p = 1000 mbar**. The ambient temperature at the surface is

**T_a = 300K**.

The parcel is now heated to a temperature of

**T_p = 302K**, but the pressure of the parcel is still equal to the ambient pressure of

**p = 1000 mbar**.

- The atmospheric

**lapse rate**is

**1K / 100 m**. The air parcel rises adiabatically because of the temperature difference between the surrounding air and the parcel. There is no humidity involved, so the

**change in temperature of the parcel**is

**1K / 100 m**, too.

Calculate the vertical speed and the upward acceleration of the parcel due to buoyancy at given heights (z = 2000 m, 5000 m, 10000 m), assuming that the initial vertical motion at the surface is w = 0 m/s.

So far, so good.

The problem lies in the calculation of the acceleration.

The students have tried to solve this in two different ways.

First of all, the buoyancy acceleration is given by this equation:

where the first term is the buoyancy acceleration, ρ_p and ρ_a are the densities of the parcel and the ambient air, and g is the earth's gravitational acceleration.

If you use the ideal gas law:

you can replace the densities with temperatures and you get:

This equations shows that if the temperature difference (T_p - T_a = 2 K) is constant - and we assume that it is as both the parcel and the atmosphere cool by 1 K every 100 m - the

**buoyancy increases**with

**decreasing ambient temperature**or increasing height.

Now, some students used the fact that during dry-adiabatic processes the potential temperature doesn't change. The potential temperature is defined as the temperature a parcel would have if it was lifted or lowered to 1000 mbar adiabatically. Because we start at a pressure level of 1000 mbar, the potential temperature of the atmosphere is equal to its absolute temperature, θ_a = 300 K. The potential temperature of the air parcel is θ_p = 302 K. Due to the given lapse rate and the adiabatic process, those values stay constant even at different height levels.

These students then swapped out the temperature variables from the buoyancy equation by replacing them with the help of this form of the adiabatic equation:

where θ is the potential temperature, T is the absolute temperature, p_0 is the surface pressure (1000 mbar), p the actual pressure and κ is a constant.

They now get the same buoyancy formula, but with θ instead of T.

Maybe you can see where this is going. This equation shows that because both the potential temperature difference (θ_p - θ_a = 2 K) and the ambient potential temperature (θ_a = 300 K) are constant, the

**buoyancy acceleration stays the same for every height**.

This is contrary to the result of the buoyancy equation that relies on the absolute temperature and I fail to see what we are missing or misunderstanding here. The following graphics illustrating the vertical temperature and potential temperature profiles should clarify what the problem with this task is (the height of 5000 m is just an example, the problem occurs at every level).

Vertical temperature profile

Vertical potential temperature profile

Maybe one of you has an idea where we are thinking wrong or why any of those equations cannot be used to solve this problem.

Any help would be greatly appreciated.