For a very tall vessel the centre of gravity is at h = RT/Mg, one scale height above the bottom. For T = K and M = g/mol this is km. For the standard case (273 K, air) it is 7.98 km, about 8 km.
Problem statement
An ideal gas of molar mass M is in a very tall vertical cylindrical vessel, in a uniform gravitational field with free-fall acceleration g. The temperature is T at every height. Find the height at which the centre of gravity of the gas is located.
We assume T = 273 K, M = 29 g/mol and g = 9.81 m/s² for the drawing, since the problem gives no numbers. This means the page starts from air at 0 °C. All are sliders.
We assume “very tall” means the vessel is so tall that the gas thins out to almost nothing at the top. The page shows a vessel of 40 km at first, and the exact value for any chosen height.
Symbols: h is height above the bottom, ρ is density, H = RT/Mg is the scale height, R = 8.314 J/(mol K) is the gas constant.
What the constant Mg/RT means
The factor Mg/RT is the fraction by which the density falls for each metre of climb. Its inverse H = RT/Mg is the scale height, here km. Climbing one scale height divides the density by e. The answer of this problem turns out to be exactly H.
The gas law alone cannot place a centre of gravity. It describes one sample at one pressure. Here the density changes with height, so we also need the weight of the gas above each layer, which brings in g.
The big idea in plain words
The centre of gravity is the average height of the gas, where each slice counts in proportion to its mass. Low slices are dense and heavy. High slices are thin and light, but they are far up. The balance between the two gives the average. Because the density falls by the same factor for equal climbs, the sums are simple exponential integrals. The surprise is that the average height is exactly one scale height, even though the mass goes on to infinity.
The solution in each frame
Lab frame
The gas 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 in balance gives dp = −ρ g dh. The gas law gives p = ρRT/M, so at constant T we have dp = (RT/M) dρ. Put these together and divide by ρ:
The field is uniform, so the centre of gravity is the same as the centre of mass. A slice of area S and thickness dh has mass dm = ρ S dh. The average height is:
Both integrals run from 0 to infinity. The second is 1/a. The first is found by parts: ∫ h e−ah dh = 1/a². Dividing:
Numbers: RT/Mg = 8.314 × 273 / (0.029 × 9.81) = 7978 m, about 8 km. For the values on screen it is km.
For a vessel of finite height L, the integrals stop at L. The result is hcm = H [1 − x/(ex − 1)] with x = L/H. For a small x this is L/2, as for uniform gas. For a large x it tends to H. With the vessel on screen (L = km, x = ) it is km.
The centre is not the place with half the mass below it. That place is at H ln 2 = km. About % of the mass lies below the centre, because the thin gas at the top is far away and pulls the average up.
A separate check does not use the exponential formula. It builds the pressure step by step from the bottom (fourth-order Runge-Kutta), using dp = −ρg dh, and adds up the slices to find the mass-weighted average height. For this vessel it gets km, against the finite-vessel formula km. For a vessel 30 scale heights tall it gets km, against RT/Mg = km.
Five methods
1. Force method
Not the natural route. The balance of forces on one thin layer gives the density law, which is the first step above.
2. Energy method
The average gravitational potential energy per molecule is kT for the vertical motion in this isothermal gas, and it equals mg hcm. With m = M/NA this gives hcm = kT/(mg) = RT/(Mg), the same answer.
3. Graph method
The page shows the density curve and the shaded area. In units of H the curve is e−y/H, and the centre is at y = 1, where 63% of the mass is below.
4. Calculus method
This is the pair of integrals above.
5. Check method
Units: J/(mol K) × K divided by (kg/mol × m/s²) is metres. Limits: a colder gas, a heavier gas or stronger gravity makes H smaller, so the centre is lower. A very hot, light gas has a high centre. A short vessel gives L/2.
Fastest: the energy method. Most reliable: the integrals checked by the step-by-step sum.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Height of the vessel L | Not the answer H | The vessel, the centre for this vessel, the mass fractions and the graphs |
| Temperature T | Yes | H, both centres and the marks on all graphs |
| Molar mass M | Yes | H and both centres |
| Free-fall acceleration g | Yes | H and both centres |
| Probe slice height | No | The slice, the mass below and above it, and the dots |
| Given-answer mode | Sets T | The page finds the temperature that puts the centre at the chosen height |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Temperature | T | K |
| Molar mass | M | g/mol |
| Free-fall acceleration | g | m/s² |
| Vessel height | L | km |
| Scale height | RT/Mg | km |
| Centre, very tall vessel | hcm | km |
| Centre, this vessel | hcm | km |
Exam traps
- Putting the centre at half the height. That is true only for uniform density.
- Putting the centre where half the mass is below it (H ln 2). That is the median, not the average.
- Using M in grams per mole. The formula needs kilograms per mole.
- Thinking an infinite vessel has an infinite centre height. The mass above falls off so fast that the average stays at H.
- Using degrees Celsius. The gas law needs kelvin: 273 K, not 0.