dT/dh = −(n − 1) M g / (n R). For air (M = 29 g/mol) with your n = , the temperature gradient is K/km. For n = 1.4 the gradient is K/km.
Problem statement
Suppose the pressure p and the density ρ of air are related by p/ρn = constant, whatever the height (n is a constant). Find the temperature gradient dT/dh.
We assume air has molar mass M = 29 g/mol and g = 9.81 m/s². This means the answer can be given in kelvin per kilometre. The sliders let you change both.
We assume a ground temperature T0 = 288 K and ground pressure 101.3 kPa. This means the vessels show real temperatures and pressures. They do not change the gradient.
We assume the air is an ideal gas at rest. This means p = ρRT/M and every thin layer is in balance.
Symbols: h is height, T is temperature, p is pressure, ρ is density, M is molar mass, R = 8.314 J/(mol K) is the gas constant, C is the constant in p = Cρn.
What n and C mean
The law p = Cρn says how the pressure of the air changes when its density changes. The constant C sets the pressure at one density. The exponent n sets how steeply the pressure follows the density.
n = 1. Pressure is in proportion to density. With p = ρRT/M this means T is the same everywhere. This is the isothermal atmosphere, and the gradient is zero.
n = 1.4. This is the value for air when it expands or is lifted without exchanging heat (an adiabatic change). It gives a drop of about 9.8 K for each kilometre, close to what is seen in real rising air.
Larger n. The pressure changes faster for a given change of density. The drop per kilometre grows and tends to Mg/R = K/km. This is the uniform-density result, where the density does not change at all.
n below 1. The temperature rises with height, so the gradient is positive.
The big idea in plain words
Gas low down holds up the gas above it, so the pressure falls with height. The law p = Cρn then tells us how the density falls too. The gas law joins p, ρ and T, so once p and ρ are known as functions of height, so is T. The only job left is to find how fast T changes, and the chain rule does it in one line. On the page, the three vessels show n = 1, n = 1.4 and your own n side by side, with the same layer lifted in each.
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 with area S. The gas below pushes up with pS, the gas above pushes down with (p + dp)S, and the weight is ρgS dh. The layer is at rest, so
The gas law for a gas of molar mass M is p = ρRT/M, so T = (M/R)(p/ρ). Put p = Cρn:
Differentiate with respect to ρ. Also differentiate the law itself:
Height changes ρ and p together, so by the chain rule dT/dh = (dT/dρ)(dρ/dp)(dp/dh):
C and ρ cancel (ρn−2 × ρ = ρn−1):
For n = 1.4 and air: −(0.4/1.4) × 0.029 × 9.81 / 8.314 = −0.00978 K/m = −9.78 K/km. For your n = , dT/dh = K/km.
A separate check does not use the chain rule. It starts from the ground, and builds the atmosphere step by step (fourth-order Runge-Kutta, 800 steps) from dp/dh = −ρg with ρ = ρ0(p/p0)1/n. At the end it gets T from p, ρ and the gas law and takes the average gradient. This gives a gradient of K/km, the same as the formula value K/km.
Five methods
1. Force method
This is the first step above: the balance of forces on a thin layer gives dp/dh = −ρg.
2. Energy method
Not needed. Only the balance of one layer and the gas law matter.
3. Graph method
The graph of p against ρ is a curve p = Cρn. Moving up the atmosphere moves the state down this curve. The temperature graph is a straight line whose slope is the answer.
4. Calculus method
This is the chain rule used above. A second route is to solve for p(h) and ρ(h) first. This gives T = T0 − [(n − 1)Mg/(nR)] h, a straight line.
5. Check method
Units: kg/mol × m/s² divided by J/(mol K) gives K/m. Limits: at n = 1 the answer is 0, correct for an isothermal atmosphere. As n grows large the answer tends to −Mg/R, the uniform-density result. For n = 1.4 the answer matches the dry-air lapse rate of about 9.8 K/km.
Fastest: the chain rule. Most reliable: the numerical atmosphere.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Exponent n | Yes | The gradient, the third vessel, its lines and the answer graph |
| Molar mass M | Yes | The gradient and all three vessels |
| Free-fall acceleration g | Yes | The gradient and all three vessels |
| Ground temperature T0 | No | Where the temperature lines start, and the pressure and density |
| Height of the layers | No | The layer positions and the dots on the graphs |
| Given-answer mode | Sets n | The page finds the n that gives the chosen drop per kilometre |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Exponent | n | |
| Molar mass | M | g/mol |
| Free-fall acceleration | g | m/s² |
| Ground temperature | T0 | K |
| Temperature gradient | dT/dh | K/km |
| Limit for large n | −Mg/R | − K/km |
Exam traps
- Using M in grams per mole. The formula needs kilograms per mole, or the answer is 1000 times too big.
- Forgetting that T depends on p/ρ, not on p alone. A change of height changes both.
- Dropping the factor (n − 1)/n. At n = 1 the gradient must be zero.
- Confusing n with the ratio of specific heats. They are equal only for the adiabatic case, n = 1.4 for air.
- Giving the answer as a positive number. The temperature falls with height for n above 1.