p = p0 e−Mgh/RT, with h negative for a mine. At km above the ground the pressure is atm. At km down a mine it is atm. For the standard case (273 K, air, 5 km) these are 0.534 atm and 1.87 atm.
Problem statement
Assume that air is under standard conditions close to the Earth's surface. Take the temperature and the molar mass of air to be the same at every height. Find the air pressure at a height of 5.0 km above the surface, and in a mine at a depth of 5.0 km below the surface.
We assume standard conditions mean T = 273 K and p0 = 1 atm at the surface. This means the pressure at the ground is 1 atm. Both are sliders.
We assume air has molar mass M = 29 g/mol and g = 9.81 m/s² at every depth and height. This means g does not change from the mine bottom to 5 km up.
We assume the air in the mine is at the same temperature as at the surface, as the problem says. This means the answer for the mine is an idealised value.
Symbols: h is height above the ground (negative for a mine), p is pressure, ρ is density, M is molar mass, R = 8.314 J/(mol K) is the gas constant, T is temperature.
What the constant Mg/RT means
The factor Mg/RT in the exponent tells us how fast the pressure changes with height. It has units of 1 per metre. Its inverse, H = RT/Mg, is the scale height. Going up one scale height multiplies the pressure by 1/e, about 0.37. For the standard case H = km. Going the same distance down multiplies the pressure by e.
A heavier gas (larger M), stronger gravity (larger g) or colder air (smaller T) makes H shorter, so the pressure falls off faster with height. Hot, light air has a long scale height and thins out slowly.
The big idea in plain words
Air lower down holds up the air above it, so the pressure falls with height. A thin layer of air must be held up by the extra pressure from below, so the drop in pressure across the layer equals its weight. The weight depends on the density, and the density depends on the pressure itself (the gas law). So the lower the pressure, the lighter the air, and the slower the pressure falls. This gives an exponential law. In a mine, the same law runs the other way: there is more air above, so the pressure is higher.
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 at height h. The pressure below is p and above is p + dp. The layer is at rest, so the net upward force from the pressures equals its weight:
The gas law for a gas of molar mass M gives the density: ρ = Mp/(RT). Put this in and divide by p:
T, M and g are constant, so integrate from the ground (h = 0, pressure p0) to height h:
The standard numbers: Mg/RT = 0.029 × 9.81 / (8.314 × 273) = 1.253 × 10−4 per metre. At h = +5000 m the exponent is −0.627, so p = e−0.627 = 0.534 atm. In the mine h = −5000 m, the exponent is +0.627, and p = e+0.627 = 1.87 atm. For the values now on screen, the pressure is atm up and atm down.
A separate check does not use the exponential formula. It builds the pressure step by step from the ground (fourth-order Runge-Kutta, 500 steps) using dp/dh = −ρg with ρ = Mp/(RT). Above the ground it gets atm, against the formula value atm. In the mine it gets atm, against atm.
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(p/p0) against h. It is a straight line with slope −Mg/RT, and one reading of the line at +5 km and one at −5 km give both answers. The page shows this line.
4. Calculus method
This is the integration above: dp/p = −(Mg/RT) dh integrated from 0 to h.
5. Check method
Units: kg/mol × m/s² × m divided by J/mol gives a pure number, so the exponent has no units. Limits: at h = 0 the formula gives p0. For small h it gives p ≈ p0(1 − Mgh/RT), a straight-line fall. Going up and going down the same distance give factors e−x and e+x, whose product is 1: 0.534 × 1.87 = 1.
Fastest: the formula p = p0 e−Mgh/RT. Most reliable: the step-by-step integration.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Height above the ground | Yes, case 1 | The upper answer, its box and mark, and the curves |
| Depth of the mine | Yes, case 2 | The lower answer, its box and mark, and the curves |
| Temperature T | Yes, both | Both answers, the scale height, the shading and all graphs |
| Molar mass M | Yes, both | Both answers, the scale height and all graphs |
| Free-fall acceleration g | Yes, both | Both answers, the scale height and all graphs |
| Ground pressure p0 | Yes, both | Both answers (in proportion) and all graphs |
| Height of the probe layer | No | The layer, the pressure and density there, and the dots |
| Given-answer mode | Sets the height and depth | The page finds the height and depth that give the chosen pressures |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Temperature | T | K |
| Molar mass | M | g/mol |
| Free-fall acceleration | g | m/s² |
| Ground pressure | p0 | atm |
| Scale height | RT/Mg | km |
| Pressure above the ground | p | atm at km |
| Pressure in the mine | p | atm at km |
Exam traps
- Using M in grams per mole. The formula needs kilograms per mole, or the exponent is 1000 times too large.
- Using the height in kilometres in the exponent without changing to metres.
- Subtracting a fixed pressure per kilometre. The pressure falls by a fixed factor per kilometre, not by a fixed amount.
- Forgetting that the sign of h flips for a mine. Depth gives a plus sign in the exponent and a pressure above p0.
- Using a temperature in degrees Celsius. The gas law needs kelvin: 273 K, not 0.