Solved thermodynamics exercise

Isothermal atmosphere and chemical potential

Exercise 41 · Lesson 8 — Direction of Evolution, Equilibrium and Stability

chemical potential
atmosphere
barometric formula
gravity
ideal gas

Statement

For one mole of air at altitude $z$ , the generalized chemical potential is

μ (T, P, z) = μ^{\circ} (T) + R T ln (\frac{P}{P ^{\circ}}) + M g z,

where $M$ is the molar mass of air. The atmosphere is assumed to be isothermal.

Explain why $μ$ must be independent of $z$ at equilibrium.
Deduce the differential equation satisfied by $P (z)$ , then the barometric formula.
Estimate $P (h) / P (0)$ for $h = 8, 85 km$ , $T = 260 K$ , $M = 29, 0 g mo l^{- 1}$ and $g = 9, 81 m s^{- 2}$ .
Give one important limitation of this model.

Hint

At equilibrium, moving a small amount of matter from one altitude to another must not change the free energy to first order.

Detailed solution

Solution

1. If

μ

were higher at one altitude than at another, a transfer of matter towards the region of lower potential would decrease the free energy. Equilibrium therefore requires

d μ = 0

. 2. At constant temperature,

0 = d μ = R T \frac{d P}{P} + M g d z .

Thus $d P / P = - (M g / R T) d z$ . Integrating from $0$ to $z$ ,

\int_{P (0)}^{P (z)} \frac{d P}{P} = - \frac{M g}{R T} \int_{0}^{z} d z,

hence

P (z) = P (0) exp (- \frac{M g z}{R T}) .

3. In SI units, the molar mass is $M = 2, 90 \times 1 0^{- 2} kg mo l^{- 1}$ . The exponent is

\frac{M g h}{R T} = \frac{2 , 90 \times 1 0 ^{- 2} \times 9 , 81 \times 8 , 85 \times 1 0 ^{3}}{8 , 314 \times 260} ≃ 1, 17.

Therefore $P (h) / P (0) = e^{- 1, 17} ≃ 0, 31$ .

4. The actual temperature of the atmosphere varies with altitude. The isothermal model gives a useful order of magnitude, but not an accurate profile; it also neglects humidity and changes in composition.