(a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure d$p$ over a differential height d$h$? (b) Considering the pressure $p$ to be proportional to the density, find the pressure $p$ at a height $h$ if the pressure on the surface of the earth is $p_0$. (c) If $p_0 = 1.03\times 10^{5}$ N m$^{-2}$, $\rho_0 = 1.29$ kg m$^{-3}$ and $g = 9.8$ m s$^{-2}$, at what height will the pressure drop to $(1/10)$ the value at the surface of the earth? (d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressu

NCERT Exemplar · Q21

\rho

(b) Considering the pressure

p

to be proportional to the density, find the pressure

p

at a height

h

if the pressure on the surface of the earth is

p_0

p_0 = 1.03\times 10^{5}

N m

^{-2}

\rho_0 = 1.29

kg m

^{-3}

and

g = 9.8

m s

^{-2}

, at what height will the pressure drop to

(1/10)

the value at the surface of the earth?

(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

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Concept understanding — Atmospheric Pressure Scale Height

What is Atmospheric Scale Height? The Intuition

Imagine you're standing at sea level. The air above you is a column of gas, and every square centimetre of your body is being pushed by that entire column. That's atmospheric pressure — roughly 1 kg per square centimetre at the ground.

Now climb a mountain. As you go up, there's less air above you, so pressure drops. But here's the key question: how fast does pressure fall with height?

If the atmosphere were a liquid (incompressible), pressure would drop linearly — every metre up, the same amount of pressure lost. But air is a gas: it compresses under its own weight. The lower layers are squished denser by the weight of the layers above. So near the ground, pressure falls quickly with height; higher up, it falls more slowly.

The scale height is the single number that captures this behaviour. It tells you: "If you go up by this height, the pressure drops by a factor of $e$ (about 2.718)."

Note

A factor of $e$ means the new pressure is roughly 37% of the original. So after one scale height, pressure is about one-third of what it was.

The Precise Definition

For an isothermal atmosphere (constant temperature), the pressure $P$ at height $z$ follows an exponential decay:

P(z) = P_0 \, e^{-z/H}

where $P_0$ is the pressure at $z=0$ (usually sea level), and $H$ is the scale height.

Formally: The scale height $H$ is the vertical distance over which the atmospheric pressure decreases by a factor of $e$ .

H = \frac{k_B T}{m g}

For Earth's lower atmosphere at 15°C (288 K), with $m \approx 4.8 \times 10^{-26}$ kg and $g = 9.8$ m/s²:

H = \frac{(1.38 \times 10^{-23})(288)}{(4.8 \times 10^{-26})(9.8)} \approx 8.5 \text{ km}

So every 8.5 km you go up, pressure drops to 37% of its value at that starting point.

Why Does This Formula Make Sense?

Think about the forces on a thin horizontal slice of air of thickness $dz$ at height $z$ :

Weight of the slice = $\rho g \, dz$ (density × gravity × thickness)
Pressure difference across the slice = $P(z) - P(z+dz) = -dP$

For equilibrium: $-dP = \rho g \, dz$

Now use the ideal gas law: $P = \rho \frac{k_B T}{m}$ (since $\rho = \frac{mN}{V}$ and $P = \frac{N}{V}k_B T$ )

Substitute $\rho = \frac{P m}{k_B T}$ into the equilibrium equation:

-dP = \frac{P m}{k_B T} \, g \, dz

Rearrange:

\frac{dP}{P} = -\frac{mg}{k_B T} \, dz

Integrate from $z=0$ to $z$ :

\ln\frac{P}{P_0} = -\frac{mg}{k_B T} \, z

Exponentiate:

P = P_0 \, e^{-z/H}, \quad H = \frac{k_B T}{mg}

Tip

A quick way to remember: $H$ is roughly $RT/(Mg)$ where $R$ is the gas constant (8.314 J/mol·K) and $M$ is the molar mass of air (~0.029 kg/mol). For Earth: $H \approx \frac{8.314 \times 288}{0.029 \times 9.8} \approx 8400$ m.

What Scale Height Tells You

Planet	Approx. Scale Height	What it means
Earth	8.5 km	Pressure halves every ~5.9 km
Mars	11 km	Thin CO₂ atmosphere, but falls off slowly
Venus	15.9 km	Very thick CO₂ atmosphere, pressure drops slowly
Jupiter	~27 km	Massive planet, but low mean molecular weight (H₂) gives large H

Key insight: A larger $H$ means the atmosphere "stretches" higher — pressure falls more gradually. A smaller $H$ means the atmosphere is more compact, hugging the surface.

Common Mistake to Avoid …

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Pressure falls exponentially with altitude when density is proportional to pressure. The scale height $h_0 = p_0/(\rho_0 g)$ characterizes the atmosphere; pressure drops to one-tenth at $h \approx 16\,000$ m. The model assumes constant temperature and uniform gravity.

The barometric formula and atmospheric scale height

When you climb a mountain, your ears pop because the air pressure drops. The reason is simple: the column of air above you weighs less. But air is compressible—unlike water, its density changes with pressure—so the relationship between height and pressure is not linear. If we assume the atmosphere stays at constant temperature (so pressure and density remain proportional), we arrive at an exponential decay law. The characteristic distance over which pressure falls by a factor of $e$ is called the scale height, and it tells us how "thick" the atmosphere is.

(a) Pressure change over a differential height

Hydrostatic equilibrium. Consider a thin horizontal slab of air at height $h$ with thickness $dh$ and cross-sectional area $A$ . The weight of this slab is $\rho g A \, dh$ , where $\rho$ is the local air density.
Pressure difference. The pressure at the bottom of the slab is $p$ ; at the top it is $p + dp$ . The net upward force from pressure is $p A - (p + dp)A = -dp \cdot A$ .
Balance. For the slab to be in equilibrium, the net upward pressure force must support the weight:

-dp \cdot A = -\rho g A \, dh.

Canceling $A$ and rearranging,

dp = -\rho g \, dh.

The negative sign reflects that pressure decreases as height increases.

(b) Pressure as a function of height

Proportionality assumption. We are told $p \propto \rho$ . At the surface, $p_0$ corresponds to $\rho_0$ , so

\frac{p}{p_0} = \frac{\rho}{\rho_0} \quad \Rightarrow \quad \rho = \rho_0 \frac{p}{p_0}.

Substitute into the hydrostatic equation. From step 3,

dp = -\rho g \, dh = -\rho_0 \frac{p}{p_0} g \, dh.

Rearrange to separate variables:

\frac{dp}{p} = -\frac{\rho_0 g}{p_0} dh.

Integrate. Let $h_0 = \frac{p_0}{\rho_0 g}$ be the scale height. Then

\frac{dp}{p} = -\frac{dh}{h_0}.

Integrating from the surface ( $h=0$ , $p=p_0$ ) to height $h$ (pressure $p$ ):

\ln p - \ln p_0 = -\frac{h}{h_0} \quad \Rightarrow \quad \ln\frac{p}{p_0} = -\frac{h}{h_0}.

Exponentiating both sides,

p(h) = p_0 \, e^{-h/h_0}, \quad \text{where } h_0 = \frac{p_0}{\rho_0 g}.

This is the barometric formula for an isothermal atmosphere.

(c) Height at which pressure drops to one-tenth

Set up the equation. We want $p = \frac{p_0}{10}$ :

\frac{p_0}{10} = p_0 \, e^{-h/h_0} \quad \Rightarrow \quad e^{-h/h_0} = \frac{1}{10}.

Taking natural logarithms,

-\frac{h}{h_0} = \ln\frac{1}{10} = -\ln 10 \quad \Rightarrow \quad h = h_0 \ln 10.

Compute the scale height. With $p_0 = 1.03 \times 10^5 \, \text{N m}^{-2}$ , $\rho_0 = 1.29 \, \text{kg m}^{-3}$ , and $g = 9.8 \, \text{m s}^{-2}$ :

h_0 = \frac{1.03 \times 10^5}{1.29 \times 9.8} = \frac{1.03 \times 10^5}{12.642} \approx 8148 \, \text{m}.

Find the height. Using $\ln 10 \approx 2.303$ :

h = 8148 \times 2.303 \approx 18\,765 \, \text{m} \approx 1.88 \times 10^4 \, \text{m}.

Tip

The scale height $h_0 \approx 8$ km is a useful benchmark: pressure falls by a factor of $e \approx 2.7$ every 8 km. To drop by a factor of 10, you need roughly $2.3 h_0 \approx 18$ km.

(d) Limitations of the model …

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Q.(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ\rhoρ, what is the change in pressure dppp over a differential height dhhh?

The barometric formula and atmospheric scale height

(a) Pressure change over a differential height

(b) Pressure as a function of height

(c) Height at which pressure drops to one-tenth

(d) Limitations of the model …

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Q.(a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$ , what is the change in pressure d $p$ over a differential height d $h$ ?

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