With n = Mg/(aRT0): (a) p = p0(1 − ah)n, valid for h < 1/a; (b) p = p0(1 + ah)−n. For the values on screen (n = ) at h = km the pressure is p0 in (a) and p0 in (b).
Problem statement
Air is in a uniform gravitational field with free-fall acceleration g. Find the gas pressure as a function of height h, if p = p0 at h = 0 and the temperature varies with height as (a) T = T0(1 − ah), (b) T = T0(1 + ah), where a is a positive constant.
We assume the gas is air with M = 29 g/mol and g = 9.81 m/s². The problem gives no numbers, so these make the drawing concrete. Both are sliders.
We assume T0 = 300 K and a = 0.02 per km for the drawing. This means the temperature changes by a T0 = 6 K per km, close to the real fall in the lower atmosphere. Both are sliders.
We assume the gas stays an ideal gas and M does not change with height. This means the gas law p = ρRT/M holds at every height.
Symbols: h is height, p is pressure, p0 is the pressure at the ground, ρ is density, T is temperature, R = 8.314 J/(mol K).
What the constants a and n mean
The constant a tells how fast the temperature changes with height. The temperature changes by a T0 kelvin for each unit of height, here K per km. In case (a) the air cools as you climb, so the temperature would reach 0 K at h = 1/a = km. The formula cannot be used above that height, which is why the problem says h < 1/a.
The exponent n = Mg/(aRT0) compares two things: the pressure change from the weight of the air (Mg) and the thermal push (RT0) changing at rate a. A big n means the pressure follows the temperature closely, as p = p0(T/T0)n in case (a). A small a makes n huge, and the result then tends to the constant-temperature law.
The gas law alone cannot give p(h). It links p, ρ and T at one point only. We need the extra fact that the air layers are in balance, so the pressure change across a layer equals its weight.
The big idea in plain words
For constant temperature the pressure falls by the same factor for equal climbs. Here the temperature itself changes with height. Cold air is denser for the same pressure, so it weighs more and the pressure falls faster. Warm air is lighter, so the pressure falls more slowly. So in case (a) the pressure falls below the constant-temperature curve, and in case (b) it stays above it. Both cases can be solved by adding up thin layers, with the temperature of each layer put in.
The solution in each frame
Lab frame
The air is at rest. The full solution is in the next section.
Other frames
No other frame is needed. Nothing moves, so one frame is enough.
The solution step by step
A thin layer of thickness dh in balance gives dp = −ρ g dh. The gas law gives ρ = pM/RT. Put these together:
(a) Put T = T0(1 − ah) and integrate from the ground to height h:
This needs 1 − ah > 0, so h < 1/a.
(b) Put T = T0(1 + ah). The integral of dh/(1 + ah) is ln(1 + ah)/a, and the sign in front stays negative:
Numbers for the standard case: n = 0.029 × 9.81 / (8.314 × 300 × 2 × 10−5 per m) = 5.70. At 10 km, ah = 0.2, so (a) p = 0.85.70 = 0.280 p0 and (b) p = 1.2−5.70 = 0.354 p0. At constant T0 the pressure would be 0.320 p0. For the values now on screen: n = , so (a) p = p0, (b) p = p0 at h = km, and p0 for constant T0.
A separate check does not use the power laws. It steps up from the ground in 4000 thin layers (fourth-order Runge-Kutta) using dp/dh = −(Mg/R) p/T(h) with the temperature of each layer. At h = km it gets p0 in (a), against the formula value p0, and p0 in (b), against p0.
Five methods
1. Force method
This is the first step above: the balance of forces on a thin layer.
2. Energy method
Not needed. The balance of one layer and the gas law give the result directly.
3. Graph method
Plot ln p against ln(1 ± ah). It is a straight line with slope ±n. The page also shows p against T/T0, where the curves are power laws.
4. Calculus method
This is the integration above: dp/p = −(Mg/R) dh/T(h), with T(h) put in.
5. Check method
Units: n is a pure number, since Mg has units of J/m and aRT0 has units of J/(mol m) × mol. Limits: at h = 0 both give p0. When a → 0, (1 − ah)n tends to e−Mgh/RT0, the constant-temperature law, and so does (1 + ah)−n. Case (a) lies below this curve, case (b) above it.
Fastest: the integral of dp/p. Most reliable: the layer-by-layer numerical sum.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Ground temperature T0 | Yes | Both temperature profiles, n, both curves and all answers |
| Rate a | Yes | How fast T changes, n, the curves and the answers |
| Molar mass M | Yes | n, both curves and the answers |
| Free-fall acceleration g | Yes | n, both curves and the answers |
| Height of interest | Yes | The target marks, the answers at that height and the dots |
| Probe heights (a), (b) | No | The probe bands and the T and p read there |
| Given-answer mode | Sets a | The page finds the rate a that gives the chosen pressure |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Ground temperature | T0 | K |
| Rate of change | a | per km |
| Temperature gradient | a T0 | K per km |
| Molar mass | M | g/mol |
| Free-fall acceleration | g | m/s² |
| Exponent | n | |
| Limit of case (a) | 1/a | km |
Exam traps
- Using the constant-temperature law e−Mgh/RT when T changes. The result is a power law, not an exponential.
- Using a in units of 1 per km in n. n needs a in 1 per metre, or the exponent is 1000 times too small.
- Using case (a) above h = 1/a. The temperature would be 0 K or negative there.
- Mixing the signs of the two exponents. (a) has +n on (1 − ah); (b) has −n on (1 + ah).
- Forgetting that M is in kilograms per mole inside n.