m = (p0 S/g)(1 − e−Mgh/RT). For the values on screen (h = km, S = m², T = K) the vessel holds kg. The mass can never exceed p0 S/g = kg.
Problem statement
An ideal gas of molar mass M is in a tall vertical cylindrical vessel of base area S and height h. The temperature of the gas is T. Its pressure on the bottom base is p0. Take T and the free-fall acceleration g to be the same at every height. Find the mass of the gas in the vessel.
We assume air-like numbers for the drawing: M = 29 g/mol, T = 273 K, g = 9.81 m/s², p0 = 1 atm, S = 1 m² and h = 10 km. This means the page starts from a realistic case. All are sliders.
We assume the gas is in balance, so it is at rest and its density depends only on height. This means the pressure at one height is the same across the whole slice.
Symbols: h is the vessel height, y is a height inside it, ρ is density, p is pressure, 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 pressure and the density fall for each metre of climb. Its inverse H = RT/Mg is the scale height, here km. The ratio x = h/H = Mgh/RT tells us how tall the vessel is compared with H. It decides the answer: for a small x the gas is almost uniform; for a large x the top part holds almost nothing.
The gas law alone, pV = nRT, would give the mass if the pressure were the same everywhere in the vessel. Here the pressure falls with height, so the gas law cannot be applied to the whole vessel at once. We need the weight of the gas above each layer as well, which is why g appears.
The big idea in plain words
Cut the vessel into thin slices. Each slice is so thin that its density is almost constant. The lower slices carry the weight of the gas above them, so they are squeezed and denser. Find the mass of one slice, then add all slices. The density falls by the same factor for equal climbs, so the sum is a simple exponential integral. A neat result follows: the total weight of the gas is the pressure force on the base minus the small force on the top.
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
From the barometric law (a thin layer in balance gives dp = −ρ g dy, and the gas law p = ρRT/M turns this into dp/p = −(Mg/RT) dy), the pressure at height y is:
The gas law gives the density: ρ = Mp/RT. At constant temperature the density falls with height in the same way as the pressure:
A slice of thickness dy and area S has mass dm = ρ S dy. Add all slices from y = 0 to y = h:
Put ρ0 = Mp0/RT. The factors M and RT cancel:
Numbers for the standard case: p0 S/g = 101325 × 1 / 9.81 = 10329 kg. The exponent is 0.029 × 9.81 × 10000 / (8.314 × 273) = 1.253, so 1 − e−1.253 = 0.7145 and m = 7380 kg. For the values now on screen: m = × = kg.
Two limits show the result makes sense. For a short vessel (x small), 1 − e−x is close to x, so m is close to ρ0 S h, the mass of uniform gas. For a very tall vessel, e−x goes to zero and m goes to p0 S/g, which is the weight balance: the whole gas weight is carried by the base.
A separate check does not use the exponential formula. It builds the pressure and the mass together from the bottom, slice by slice (fourth-order Runge-Kutta, 4000 steps), using dp = −ρg dy and dm = ρS dy with ρ = Mp/RT. It gets kg, against the formula value kg. The weight balance (p0 − ptop) S/g gives kg.
Five methods
1. Force method
The weight of the gas is carried by the difference of the pressure forces on the base and on the top: mg = (p0 − ptop) S. With ptop = p0 e−Mgh/RT this gives the same answer in one line.
2. Energy method
Not needed. The balance of one slice and the gas law give the answer directly.
3. Graph method
Plot ρ/ρ0 against height. The mass is S ρ0 times the area between the curve and the axis, up to h. The page shades that area in blue and the uniform-gas box in orange.
4. Calculus method
This is the integration above: m = ∫ ρ S dy from 0 to h.
5. Check method
Units: (Pa × m²) / (m/s²) = N / (m/s²) = kg. Limits: h → 0 gives m → 0; a short vessel gives ρ0 S h; a very tall vessel gives p0 S/g.
Fastest: the force method, mg = (p0 − ptop) S. Most reliable: the slice-by-slice integration.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Height of the vessel h | Yes | The vessel, the mass, the top pressure and the graphs |
| Base area S | Yes | The vessel width and the mass in proportion |
| Temperature T | Yes | Scale height, mass and all graphs |
| Molar mass M | Yes | Scale height, mass and all graphs |
| Free-fall acceleration g | Yes | Scale height, mass and the limit p0 S/g |
| Bottom pressure p0 | Yes | Mass and limit in proportion, and the pressures drawn |
| Probe slice height | No | The slice, the mass below and above it, and the dots |
| Given-answer mode | Sets h | The page finds the vessel height that holds the chosen mass |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Vessel height | h | km |
| Base area | S | m² |
| Temperature | T | K |
| Molar mass | M | g/mol |
| Free-fall acceleration | g | m/s² |
| Bottom pressure | p0 | atm |
| Scale height | RT/Mg | km |
| Mass of gas | m | kg |
Exam traps
- Using m = ρ0 S h. It is true only for a short vessel. For a tall one it overshoots badly (about 75% too high here).
- Using M in grams per mole. The formula needs kilograms per mole inside the exponent.
- Putting h in kilometres in the exponent without changing to metres.
- Thinking a taller vessel always holds more. The mass levels off at p0 S/g.
- Using p0 as the pressure at the top. It is the pressure on the bottom base.