How to Calculate Barometric Pressure at an Altitude

Barometric pressure is the pressure exerted by the air above a location. To calculate how it changes with altitude, model a thin vertical layer of air and relate the pressure difference across it to the weight of the air. Combining that hydrostatic balance with the ideal-gas law produces a first-order differential equation.

The simplest useful model assumes the air has a constant temperature. Under that assumption, pressure falls exponentially with height. The formula and a worked example follow; a later section explains how the result changes when temperature varies with altitude.

Derive the barometric formula

Let $h$ denote altitude measured upward from a reference level, and let $P(h)$ be the absolute pressure at that altitude. For a stationary atmosphere, hydrostatic balance gives

$$\frac{dP}{dh}=-\rho g,$$

where $\rho$ is air density and $g$ is gravitational acceleration. The minus sign appears because pressure decreases as you move upward.

For air treated as an ideal gas, density and pressure are related by

$$\rho=\frac{MP}{RT},$$

where $M$ is the molar mass of air, $R$ is the universal gas constant, and $T$ is absolute temperature in kelvins. Substituting this expression for density gives

$$\frac{dP}{dh}=-\frac{Mg}{RT}P.$$

If $g$, $M$, and $T$ are constant over the altitude range, separate variables and integrate from the reference altitude $h=0$, where pressure is $P_0$:

$$\begin{aligned} \frac{dP}{P}&=-\frac{Mg}{RT}\,dh,\\ \ln\left(\frac{P(h)}{P_0}\right)&=-\frac{Mgh}{RT}. \end{aligned}$$

Thus the isothermal barometric formula is

$$P(h)=P_0\exp\left(-\frac{Mgh}{RT}\right).$$

The quantity $H=RT/(Mg)$ is called the scale height, so the same result can be written $P(h)=P_0e^{-h/H}$. For dry air near 15°C, $H$ is about 8.43 km. The model is an approximation: real atmospheric temperature and gravity are not perfectly constant.

Worked example: pressure at 1,000 m

Estimate the pressure 1,000 m above a location where the pressure is 101.325 kPa. Use an isothermal temperature of 288.15 K (15°C), molar mass of dry air $M=0.0289644\ \text{kg/mol}$, gravitational acceleration $g=9.80665\ \text{m/s}^2$, and $R=8.31446\ \text{J/(mol·K)}$.

First calculate the scale height:

$$H=\frac{RT}{Mg} =\frac{(8.31446)(288.15)}{(0.0289644)(9.80665)} \approx 8435\ \text{m}.$$

Then substitute $h=1000\ \text{m}$:

$$P(1000)=101.325\exp\left(-\frac{1000}{8435}\right) \approx 90.00\ \text{kPa}.$$

So the model predicts about 90.0 kPa. This is lower than the reference pressure, as expected. If pressure is known and you want to estimate altitude instead, rearrange the formula:

$$h=H\ln\left(\frac{P_0}{P(h)}\right).$$

For example, using the same model, a pressure of 90.0 kPa corresponds to approximately 1,000 m above the reference level.

When temperature varies with altitude

The isothermal assumption is convenient but not always realistic. In the troposphere, a common simplified model assumes temperature falls linearly with altitude: $T(h)=T_0-Lh$, where $L$ is a positive lapse rate. Integrating the hydrostatic equation with this temperature profile gives

$$P(h)=P_0\left(\frac{T_0-Lh}{T_0}\right)^{Mg/(RL)}.$$

This expression applies only over a layer where the linear temperature model is appropriate and $T_0-Lh$ remains positive. For more accurate atmospheric calculations, use a standard-atmosphere model with separate layers and specified temperature profiles.

Common mistakes

• Using Celsius in the exponential. Convert temperature to kelvins; 15°C is 288.15 K.
• Mixing units. With $R$ in joules per mole-kelvin, use $M$ in kilograms per mole, $g$ in metres per second squared, and altitude in metres. The exponent must be dimensionless.
• Treating the result as exact weather pressure. The formula gives an idealized pressure profile. Actual pressure also depends on local weather, temperature structure, and the reference pressure at the time and place of measurement.
• Using gauge pressure. The barometric law uses absolute pressure, measured relative to a vacuum, not pressure relative to the surrounding atmosphere.
• Assuming pressure drops linearly. Under the isothermal ideal-gas model, pressure follows an exponential curve, not a straight line in altitude.