*Mg*δ

*h*. The work available per kilogram of gas in the four cells is simply

*g*δ

*h*/4. Today we show that we can obtain this same answer with a calculation based upon enthalpy.

If we ignore kinetic energy, the specific enthalpy of a gas (that's the enthalpy per kilogram) is Cp

*T*+

*gz*, where

*z*is the altitude of the cell. When we heat up cell

**A**by δ

*T*we increase its specific enthalpy by ΔH1.

ΔH1 = δ

*T*Cp+

*g*δ

*h*/2

This is the energy required to bring about the change in the temperature of cell

**A**, and includes the work required to raise up the cells above

**A**as well as the work required to increase the internal energy of

**A**, and the work required to raise

**A**'s own weight.

For an ideal gas cell, we have

*pV*=

*MRT*, and from this we deduce that δ

*h*=

*m*Rδ

*T*/

*p*, where

*m*is the mass per square meter of the cell's bottom surface. Thus we have the following expression for ΔH1.

ΔH1 = δ

*T*(Cp+

*mg*R/2

*p*)

When cell

**A**rises, it cools by adiabatic expansion. It is now warmer by δ

*T*' than cell

**B**, and taller by δ

*h*'. Suppose we now allow

**A**to cool by δ

*T*', so that it becomes identical to cell

**B**. Its enthalpy decreases by ΔH2, and the four cells and the gas above are in an identical state to the one they were in before we warmed cell

**A**.

ΔH2 = δ

*T*'Cp+

*g*δ

*h*'/2

In buoyancy work we showed that δ

*T*' = δ

*T*(1−R

*mg*/

*p*Cp). We also showed that δ

*h*' ≈

*m*Rδ

*T*(1 +

*mg*Cv/

*p*Cp)/

*p*. Using these relations, and the fact that

*mg*≪

*p*, we obtain the following expression for ΔH2 after several lines of simplification.

ΔH2 = δ

*T*(Cp−

*mg*R/2

*p*)

The expressions for ΔH1 and ΔH2 are identical, except the sign of the gravitational term is opposite. When subtract ΔH2 from ΔH1, we obtain the enthalpy per kilogram of cell

**A**that has gone missing during the rotation.

ΔH1−ΔH2 =

*mg*Rδ

*T*/

*p*=

*g*δ

*h*

Enthalpy cannot go missing. Our expression for enthalpy ignores the kinetic energy of the cells when it should not. The kinetic energy of the cells after the rotation is equal to the enthalpy that is missing in our calculation. When we divide it among the mass of the four cells, the missing enthalpy is

*g*δ

*h*/4, which is our impetus for circulation.

We performed the same enthalpy calculation for incompressible cells and obtained the same answer. We have arrived at the same expression for impetus by two different means, for both incompressible and compressible cells. We are now confident that our calculation is correct.