The density ratio between two layers is eMgΔh/RT, so it depends only on the gap Δh. (a) A factor e needs Δh = RT/Mg = km. (b) A difference of η = % needs Δh = −(RT/Mg) ln(1 − η) = m. For 0 °C air and η = 1.0% these are 7.98 km and 80.2 m.
Problem statement
Assume the temperature and the molar mass of air are the same at every height. Find the height difference between two places where the air density at 0 °C differs (a) e times, (b) by η = 1.0%.
We assume the air is at 0 °C, so T = 273 K. This means the temperature is the same at both layers. It is a slider.
We assume air has molar mass M = 29 g/mol and g = 9.81 m/s² at both layers. This means g does not change over the gap.
We assume “differs by η” means the upper density is the fraction (1 − η) of the lower one. This means ρ2 = 0.99 ρ1 for 1%.
Symbols: h is height, ρ is density, M is molar mass, R = 8.314 J/(mol K), T is temperature, Δh is the gap between the two layers.
What the constant Mg/RT means
The factor Mg/RT is the fraction by which the density falls for each metre of climb. It has units of 1 per metre. Its inverse H = RT/Mg is the scale height: climbing one scale height divides the density by e. Here H = km. Case (a) asks exactly for one scale height. Case (b) asks for a small part of it.
The gas law alone, pV = nRT, links pressure, volume and temperature of one sample. It cannot say how the density changes with height. For that we also need the weight of the air above, which is why g and M enter.
The big idea in plain words
Each thin layer of air holds up the air above it. The pressure drop across the layer equals its weight. At constant temperature the density is proportional to the pressure, so a layer with half the density has half the pressure and weighs half as much per metre. This gives a constant fractional fall in density per metre, which is an exponential law. Equal climbs divide the density by the same factor, wherever you start.
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
Take a thin layer of thickness dh and area S. The pressure below is p and above is p + dp. The layer is at rest, so:
The gas law gives p = ρRT/M. At constant T, dp = (RT/M) dρ. Put this in:
Integrate between a lower layer (h1, ρ1) and an upper layer (h2, ρ2):
The scale height: RT/Mg = 8.314 × 273 / (0.029 × 9.81) = 7978 m, about 7.98 km.
(a) ρ1/ρ2 = e, so ln e = 1 and Δh = RT/Mg = km. This is about 8 km for the standard case.
(b) ρ2 = (1 − η) ρ1, so ρ1/ρ2 = 1/(1 − η) and Δh = −(RT/Mg) ln(1 − η). For η = 1%: 7978 m × 0.01005 = 80.2 m. For a small η, −ln(1 − η) is close to η, so Δh ≈ η RT/Mg = m. For the values on screen the exact answer is m.
A separate check does not use the exponential formula. It builds the pressure step by step upward (fourth-order Runge-Kutta, 6000 steps) from dp/dh = −ρg with ρ = Mp/(RT), and finds where the density has fallen to 1/e and to 1 − η of its starting value. It gets km and m, against the formula values km and m.
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 answer directly.
3. Graph method
Plot ln ρ against h. It is a straight line with slope −Mg/RT. A drop of 1 in ln ρ (factor e) takes one scale height; a drop of 0.01 takes 1/100 of it. The page shows the density curve and the ratio against gap.
4. Calculus method
This is the integration above: dρ/ρ = −(Mg/RT) dh.
5. Check method
Units: kg/mol × m/s² × m divided by J/mol is a pure number. Limits: η → 0 gives Δh → 0, and η → 1 gives Δh → infinity, since the density never reaches zero. Small η matches ηH.
Fastest: Δh = H ln(ρ1/ρ2). Most reliable: the step-by-step integration.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Temperature T | Yes, both | Scale height, both answers, shading and all graphs |
| Molar mass M | Yes, both | Scale height, both answers and all graphs |
| Free-fall acceleration g | Yes, both | Scale height, both answers and all graphs |
| Density difference η | Yes, case (b) | Answer (b), its target mark and the graphs |
| Height of the lower layer | No | Where the layers sit and the densities there. Only the gap matters. |
| Rise of the upper layer (a), (b) | No | The probe layer, the ratio now and the dots |
| Given-answer mode | Sets T and η | The page finds T and η that give the chosen gaps |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Temperature | T | K |
| Molar mass | M | g/mol |
| Free-fall acceleration | g | m/s² |
| Density difference | η | % |
| Scale height | RT/Mg | km |
| Gap for a factor e | Δh | km |
| Gap for a difference η | Δh | m |
Exam traps
- Using M in grams per mole. The formula needs kilograms per mole.
- Using Δh = η H for a large η. It is only a small-η rule; at 10% it is already 5% off.
- Thinking the gap depends on the height of the lower layer. Only the gap matters.
- Using degrees Celsius. The gas law needs kelvin: 273 K, not 0.
- Reading “differs by 1%” as a pressure drop of 1% per kilometre. It is a total difference between two layers.